Free Encyclopedia of Mathematics (0.0.1) volume 2

Chapter 242 16-00 General reference works (handbooks, dictionaries, bibliographies, etc.)242.1 direct product of modules

Let {Xi : i I } be a collection of modules in some category of modules. Then the direct product iI Xi of that collection is the module whose underlying set is the cartesian product of the Xi with componentwise addition and scalar multiplication. For example, in a category of left modules: (xi ) + (yi ) = (xi + yi), r (xi ) = (rxi ). For each j I we have a projection pj : iI Xi Xj dened by (xi ) xj , and an injection j : Xj iI Xi where an element xj of Xj maps to the element of iI Xi whose j th term is xj and every other term is zero. The direct product iI Xi satises a certain universal property. Namely, if Y is a module and there exist homomorphisms fi : Xi Y for all i I , then there exists a unique homomorphism : Y iI Xi satisfying i = fi for all i I . Xii iI fi

Y

Xi

The direct product is often referred to as the complete direct sum, or the strong direct sum, or simply the product. 1088

Compare this to the direct sum of modules. Version: 3 Owner: antizeus Author(s): antizeus

242.2

direct sum

Let {Xi : i I } be a collection of modules in some category of modules. Then the direct sum iI Xi of that collection is the submodule of the direct product of the Xi consisting of all elements (xi ) such that all but a nite number of the xi are zero. For each j I we have a projection pj : iI Xi Xj dened by (xi ) xj , and an injection j : Xj iI Xi where an element xj of Xj maps to the element of iI Xi whose j th term is xj and every other term is zero. The direct sum iI Xi satises a certain universal property. Namely, if Y is a module and there exist homomorphisms fi : Y Xi for all i I , then there exists a unique homomorphism : iI Xi Y satisfying pi = fi for all i I . Xipi iI fi

Y

Xi

The direct sum is often referred to as the weak direct sum or simply the sum. Compare this to the direct product of modules. Version: 3 Owner: antizeus Author(s): antizeus

242.3

exact sequence

If we have two homomorphisms f : A B and g : B C in some category of modules, then we say that f and g are exact at B if the image of f is equal to the kernel of g . A sequence of homomorphisms An+1 An An1 fn+1 fn

is said to be exact if each pair of adjacent homomorphisms (fn+1 , fn ) is exact in other words if imfn+1 = kerfn for all n. Compare this to the notion of a chain complex. Version: 2 Owner: antizeus Author(s): antizeus 1089

242.4

quotient ring

Denition. Let R be a ring and let I be a two-sided ideal of R. To dene the quotient ring R/I , let us rst dene an equivalence relation in R. We say that the elements a, b R are equivalent, written as a b, if and only if a b I . If a is an element of R, we denote the corresponding equivalence class by [a]. Thus [a] = [b] if and only if a b I . The quotient ring of R modulo I is the set R/I = {[a] | a R}, with a ring structure dened as follows. If [a], [b] are equivalence classes in R/I , then [a] + [b] := [a + b], [a] [b] := [a b]. Here a and b are some elements in R that represent [a] and [b]. By construction, every element in R/I has such a representative in R. Moreover, since I is closed under addition and multiplication, one can verify that the ring structure in R/I is well dened. properties. 1. If R is commutative, then R/I is commutative. Examples. 1. For any ring R, we have that R/R = {0} and R\{0} = R. 2. Let R = Z, and let I be the set of even numbers. Then R/I contains only two classes; one for even numbers, and one for odd numbers. Version: 3 Owner: matte Author(s): matte, djao

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Chapter 243 16D10 General module theory243.1 annihilator

Let R be a ring. Suppose that M is a left R-module. If X is a subset of M , then we dene the left annihilator of X in R: l.ann(X ) = {r R | rx = 0 for all x X }. If Z is a subset of R, then we dene the right annihilator of Z in M : r.annM (Z ) = {m M | zm = 0 for all z Z }. Suppose that N is a right R-module. If Y is a subset of N , then we dene the right annihilator of Y in R: r.ann(Y ) = {r R | yr = 0 for all y Y }. If Z is a subset of R, then we dene the left annihilator of Z in N : l.annN (Z ) = {n N | nz = 0 for all z Z }. Version: 3 Owner: antizeus Author(s): antizeus

243.2

annihilator is an ideal

The right annihilator of a right R-module MR in R is an ideal. 1091

y the distributive law for modules, it is easy to see that r. ann(MR ) is closed under addition and right multiplication. Now take x r. ann(MR ) and r R.B

Take any m MR . Then mr MR , but then (mr )x = 0 since x r. ann(MR ). So m(rx) = 0 and rx r. ann(MR ). An equivalent result holds for left annihilators. Version: 2 Owner: saforres Author(s): saforres

243.3

artinian

A module M is artinian if it satises the following equivalent conditions: the descending chain condition holds for submodules of M ; every nonempty family of submodules of M has a minimal element. A ring R is left artinian if it is artinian as a left module over itself (i.e. if R R is an artinian module), and right artinian if it is artinian as a right module over itself (i.e. if RR is an artinian module), and simply artinian if both conditions hold. Version: 3 Owner: antizeus Author(s): antizeus

243.4

composition series

Let R be a ring and let M be a (right or left) R-module. A series of submodules M = M0 M1 M2 Mn = 0 in which each quotient Mi /Mi+1 is simple is called a composition series for M . A module need not have a composition series. For example, the ring of integers, Z, condsidered as a module over itself, does not have a composition series. A necessary and sucient condition for a module to have a composition series is that it is both noetherian and artinian. If a module does have a composition series, then all composition series are the same length. This length (the number n above) is called the composition length of the module. If R is a semisimple Artinian ring, then RR and R R always have composition series. Version: 1 Owner: mclase Author(s): mclase 1092

243.5

conjugate module

If M is a right module over a ring R, and is an endomorphism of R, we dene the conjugate module M to be the right R-module whose underlying set is {m | m M }, with abelian group structure identical to that of M (i.e. (m n) = m n ), and scalar multiplication given by m r = (m (r )) for all m in M and r in R. In other words, if : R EndZ (M ) is the ring homomorphism that describes the right module action of R upon M , then describes the right module action of R upon M . If N is a left R-module, we dene N similarly, with r n = ((r ) n). Version: 4 Owner: antizeus Author(s): antizeus

243.6

modular law

Let R M be a left R-module with submodules A, B, C , and suppose C B . Then C + (B A) = B (C + A)

Version: 1 Owner: saforres Author(s): saforres

243.7

module

Let R be a ring, and let M be an abelian group. We say that M is a left R-module if there exists a ring homomorphism : R EndZ (M ) from R to the ring of abelian group endomorphisms on M (in which multiplication of endomorphisms is composition, using left function notation). We typically denote this function using a multiplication notation: [(r )](m) = r m = rm This ring homomorphism denes what is called a left module action of R upon M . If R is a unital ring (i.e. a ring with identity), then we typically demand that the ring homomorphism map the unit 1 R to the identity endomorphism on M , so that 1 m = m for all m M . In this case we may say that the module is unital. Typically the abelian group structure on M is expressed in additive terms, i.e. with operator +, identity element 0M (or just 0), and inverses written in the form m for m M . 1093

Right module actions are dened similarly, only with the elements of R being written on the right sides of elements of M . In this case we either need to use an anti-homomorphism R EndZ (M ), or switch to right notation for writing functions. Version: 7 Owner: antizeus Author(s): antizeus

243.8

proof of modular law

First we show C + (B A) B (C + A): Note that C B, B A B , and therefore C + (B A) B . Further, C C + A, B A C + A, thus C + (B A) C + A. Next we show B (C + A) C + (B A): Let b B (C + A). Then b = c + a for some c C and a A. Hence a = b c, and so a B since b B and c C B . Hence a B A, so b = c + a C + (B A). Version: 5 Owner: saforres Author(s): saforres

243.9

zero module

Let R be a ring. The abelian group which contains only an identity element (zero) gains a trivial R-module structure, which we call the zero module. Every R-module M has an zero element and thus a submodule consisting of that element. This is called the zero submodule of M . Version: 2 Owner: antizeus Author(s): antizeus

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Chapter 244 16D20 Bimodules244.1 bimodule

Suppose that R and S are rings. An (R, S )-bimodule is an abelian group M which has a left R-module action as well as a right S -module action, which satisfy the relation r (ms) = (rm)s for every choice of elements r of R, s of S , and m of M . A (R, S )-sub-bi-module of M is a subgroup which is also a left R-submodule and a right S -submodule. Version: 3 Owner: antizeus Author(s): antizeus

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Chapter 245 16D25 Ideals245.1 associated prime

Let R be a ring, and let M be an R-module. A prime ideal P of R is an annihilator prime for M if P is equal to the annihilator of some nonzero submodule X of M . Note that if this is the case, then the module annA (P ) contains X , has P as its annihilator, and is a faithful (R/P )-module. If, in addition, P is equal to the annihilator of a submodule of M that is a fully faithful (R/P )-module, then we call P an associated prime of M . Version: 2 Owner: antizeus Author(s): antizeus

245.2

nilpotent ideal

A left (right) ideal I of a ring R is a nilpotent ideal if I n = 0 for some positive integer n. Here I n denotes a product of ideals I I I . Version: 2 Owner: antizeus Author(s): antizeus

245.3

primitive ideal

Let R be a ring, and let I be an ideal of R. We say that I is a left (right) primitive ideal if there exists a simple left (right) R-module X such that I is the annihilator of X in R. We say that R is a left (right) primitive ring if the zero ideal is a left (right) primitive ideal 1096

of R. Note that I is a left (right) primitive ideal if and only if R/I is a left (right) primitive ring. Version: 2 Owner: antizeus Author(s): antizeus

245.4

product of ideals

Let R be a ring, and let A and B be left (right) ideals of R. Then the product of the ideals A and B , which we denote AB , is the left (right) ideal generated by the products {ab | a A, b B }. Version: 2 Owner: antizeus Author(s): antizeus

245.5

proper ideal

Suppose R is a ring and I is an ideal of R. We say that I is a proper ideal if I is not equal to R. Version: 2 Owner: antizeus Author(s): antizeus

245.6

semiprime ideal

Let R be a ring. An ideal I of R is a semiprime ideal if it satises the following equivalent conditions: (a) I can be expressed as an intersection of prime ideals of R; (b) if x R, and xRx I , then x I ; (c) if J is a two-sided ideal of R and J 2 I , then J I as well; (d) if J is a left ideal of R and J 2 I , then J I as well; (e) if J is a right ideal of R and J 2 I , then J I as well. Here J 2 is the product of ideals J J . The ring R itself satises all of these conditions (including being expressed as an intersection of an empty family of prime ideals) and is thus semiprime. A ring R is said to be a semiprime ring if its zero ideal is a semiprime ideal. 1097

Note that an ideal I of R is semiprime if and only if the quotient ring R/I is a semiprime ring. Version: 7 Owner: antizeus Author(s): antizeus

245.7

zero ideal

In any ring, the set consisting only of the zero element (i.e. the additive identity) is an ideal of the left, right, and two-sided varieties. It is the smallest ideal in any ring. Version: 2 Owner: antizeus Author(s): antizeus

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Chapter 246 16D40 Free, projective, and at modules and ideals246.1 nitely generated projective module

Let R be a unital ring. A nitely generated projective right R-module is of the form eRn , n N, where e is an idempotent in EndR (Rn ). Let A be a unital C -algebra and p be a projection in EndA (An ), n N. Then, E = pAn is a nitely generated projective right A-module. Further, E is a pre-Hilbert A-module with (A-valued) inner productn

u, v =i=1

u i vi ,

u, v E.

Version: 3 Owner: mhale Author(s): mhale

246.2

at module

A right module M over a ring R is at if the tensor product functor M R () is an exact functor. Similarly, a left module N over R is at if the tensor product functor () R N is an exact functor. Version: 2 Owner: antizeus Author(s): antizeus

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246.3

free module

Let R be a commutative ring. A free module over R is a direct sum of copies of R. In particular, as every abelian group is a Z-module, a free abelian group is a direct sum of copies of Z. This is equivalent to saying that the module has a free basis, i.e. a set of elements with the property that every element of the module can be uniquely expressed as an linear combination over R of elements of the free basis. In the case that a free module over R is a sum of nitely many copies of R, then the number of copies is called the rank of the free module. An alternative denition of a free module is via its universal property: Given a set X , the free R-module F (X ) on the set X is equipped with a function i : X F (X ) satisfying the property that for any other R-module A and any function f : X A, there exists a unique R-module map h : F (X ) A such that (h i) = f . Version: 4 Owner: mathcam Author(s): mathcam, antizeus

246.4

free module

Let R be a ring. A free module over R is a direct sum of copies of R. Similarly, as an abelian group is simply a module over Z, a free abelian group is a direct sum of copies of Z. This is equivalent to saying that the module has a free basis, i.e. a set of elements with the property that every element of the module can be uniquely expressed as an linear combination over R of elements of the free basis. Version: 1 Owner: antizeus Author(s): antizeus

246.5

projective cover

Let X and P be modules. We say that P is a projective cover of X if P is a projective module and there exists an epimorphism p : P X such that ker p is a superuous submodule of P . Equivalently, P is an projective cover of X if P is projective, and there is an epimorphism p : P X , and if g : P X is an epimorphism from a projective module P to X , then

1100

there exists an epimorphism h : P P such that ph = g . Ph g p

P

X 0

0

Version: 2 Owner: antizeus Author(s): antizeus

246.6

projective module

A module P is projective if it satises the following equivalent conditions: (a) Every short exact sequence of the form 0 A B P 0 is split; (b) The functor Hom(P, ) is exact; (c) If f : X Y is an epimorphism and there exists a homomorphism g : P Y , then there exists a homomorphism h : P X such that f h = g . Ph g f

X

Y

0

(d) The module P is a direct summand of a free module. Version: 3 Owner: antizeus Author(s): antizeus

1101

Chapter 247 16D50 Injective modules, self-injective rings247.1 injective hull

Let X and Q be modules. We say that Q is an injective hull or injective envelope of X if Q is both an injective module and an essential extension of X . Equivalently, Q is an injective hull of X if Q is injective, and X is a submodule of Q, and if g : X Q is a monomorphism from X to an injective module Q , then there exists a monomorphism h : Q Q such that h(x) = g (x) for all x X . 0 0 Xg i

Qh

Q Version: 2 Owner: antizeus Author(s): antizeus

247.2

injective module

A module Q is injective if it satises the following equivalent conditions: (a) Every short exact sequence of the form 0 Q B C 0 is split; (b) The functor Hom(, Q) is exact; 1102

(c) If f : X Y is a monomorphism and there exists a homomorphism g : X Q, then there exists a homomorphism h : Y Q such that hf = g . 0 Xg f

Yh

Q Version: 3 Owner: antizeus Author(s): antizeus

1103

Chapter 248 16D60 Simple and semisimple modules, primitive rings and ideals248.1 central simple algebra

Let K be a eld. A central simple algebra A (over K ) is an algebra A over K , which is nite dimensional as a vector space over K , such that A has an identity element, as a ring A is central: the center of A equals K (for all z A, we have z a = a z for all a A if and only if z K ) A is simple: for any two sided ideal I of A, either I = {0} or I = A By a theorem of Brauer, for every central simple algebra A over K , there exists a unique (up to isomorphism) division ring D containing K and a unique natural number n such that A is isomorphic to the ring of n n matrices with coecients in D . Version: 2 Owner: djao Author(s): djao

248.2

completely reducible

A module M is called completely reducible (or semisimple) if it is a direct sum of irreducible (or simple) modules. Version: 1 Owner: bwebste Author(s): bwebste

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248.3

simple ring

A nonzero ring R is said to be a simple ring if it has no (two-sided) ideal other then the zero ideal and R itself. This is equivalent to saying that the zero ideal is a maximal ideal. If R is a commutative ring with unit, then this is equivalent to being a eld. Version: 4 Owner: antizeus Author(s): antizeus

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Chapter 249 16D80 Other classes of modules and ideals249.1 essential submodule

Let X be a submodule of a module Y . We say that X is an essential submodule of Y , and that Y is an essential extension of X , if whenever A is a nonzero submodule of Y , then A X is also nonzero. A monomorphism f : X Y is an essential monomorphism if the image imf is an essential submodule of Y . Version: 2 Owner: antizeus Author(s): antizeus

249.2

faithful module

Let R be a ring, and let M be an R-module. We say that M is a faithful R-module if its annihilator annR (M ) is the zero ideal. We say that M is a fully faithful R-module if every nonzero R-submodule of M is faithful. Version: 3 Owner: antizeus Author(s): antizeus

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249.3

minimal prime ideal

A prime ideal P of a ring R is called a minimal prime ideal if it does not properly contain any other prime ideal of R. If R is a prime ring, then the zero ideal is a prime ideal, and is thus the unique minimal prime ideal of R. Version: 2 Owner: antizeus Author(s): antizeus

249.4

module of nite rank

Let M be a module, and let E (M ) be the injective hull of M . Then we say that M has nite rank if E (M ) is a nite direct sum of indecomposible submodules. This turns out to be equivalent to the property that M has no innite direct sums of nonzero submodules. Version: 3 Owner: antizeus Author(s): antizeus

249.5

simple module

Let R be a ring, and let M be an R-module. We say that M is a simple or irreducible module if it contains no submodules other than itself and the zero module. Version: 2 Owner: antizeus Author(s): antizeus

249.6

superuous submodule

Let X be a submodule of a module Y . We say that X is a superuous submodule of Y if whenever A is a submodule of Y such that A + X = Y , then A = Y . Version: 2 Owner: antizeus Author(s): antizeus

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249.7

uniform module

A module M is said to be uniform if any two nonzero submodules of M must have a nonzero intersection. This is equivalent to saying that any nonzero submodule is an essential submodule. Version: 3 Owner: antizeus Author(s): antizeus

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Chapter 250 16E05 Syzygies, resolutions, complexes250.1 n-chain

An n-chain on a topological space X is a nite formal sum of n-simplices on X . The group of such chains is denoted Cn (X ). For a CW-complex Y, Cn (Y ) = Hn (Y n , Y n1 ), where Hn denotes the nth homology group. The boundary of an n-chain is the (n 1)-chain given by the formal sum of the boundaries of its constitutent simplices. An n-chain is closed if its boundary is 0 and exact if it is the boundary of some (n + 1)-chain. Version: 3 Owner: mathcam Author(s): mathcam

250.2

chain complex

A sequence of modules and homomorphismsn An1 An+1 An

dn+1

d

is said to be a chain complex or complex if each pair of adjacent homomorphisms (dn+1 , dn ) satises the relation dn dn+1 = 0. This is equivalent to saying that im dn+1 ker dn . We often denote such a complex as (A, d) or simply A. Compare this to the notion of an exact sequence. Version: 4 Owner: antizeus Author(s): antizeus

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250.3

at resolution

Let M be a module. A at resolution of M is an exact sequence of the form Fn Fn1 F1 F0 M 0 where each Fn is a at module. Version: 2 Owner: antizeus Author(s): antizeus

250.4

free resolution

Let M be a module. A free resolution of M is an exact sequence of the form Fn Fn1 F1 F0 M 0 where each Fn is a free module. Version: 2 Owner: antizeus Author(s): antizeus

250.5

injective resolution

Let M be a module. An injective resolution of M is an exact sequence of the form 0 M Q0 Q1 Qn1 Qn where each Qn is an injective module. Version: 2 Owner: antizeus Author(s): antizeus

250.6

projective resolution

Let M be a module. A projective resolution of M is an exact sequence of the form Pn Pn1 P1 P0 M 0 where each Pn is a projective module. Version: 2 Owner: antizeus Author(s): antizeus

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250.7

short exact sequence

A short exact sequence is an exact sequence of the form 0 A B C 0. Note that in this case, the homomorphism A B must be a monomorphism, and the homomorphism B C must be an epimorphism. Version: 2 Owner: antizeus Author(s): antizeus

250.8

split short exact sequencef g

In an abelian category, a short exact sequence 0 A B C 0 is split if it satises the following equivalent conditions: (a) there exists a homomorphism h : C B such that gh = 1C ; (b) there exists a homomorphism j : B A such that jf = 1A ; (c) B is isomorphic to the direct sum A C . In this case, we say that h and j are backmaps or splitting backmaps. Version: 4 Owner: antizeus Author(s): antizeus

250.9

von Neumann regular

An element a of a ring R is said to be von Neumann regular if there exists b R such that aba = a. A ring R is said to be a von Neumann regular ring (or simply a regular ring, if the meaning is clear from context) if every element of R is von Neumann regular. Note that regular ring in the sense of von Neumann should not be confused with regular ring in the sense of commutative algebra. Version: 1 Owner: igor Author(s): igor

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Chapter 251 16K20 Finite-dimensional251.1 quaternion algebra

A quaternion algebra over a eld K is a central simple algebra over K which is four dimensional as a vector space over K . Examples: For any eld K , the ring M22 (K ) of 2 2 matrices with entries in K is a quaternion algebra over K . If K is algebraically closed, then all quaternion algebras over K are isomorphic to M22 (K ). For K = R, the well known algebra H of Hamiltonian quaternions is a quaternion algebra over R. The two algebras H and M22 (R) are the only quaternion algebras over R, up to isomorphism. When K is a number eld, there are innitely many nonisomorphic quaternion algebras over K . In fact, there is one such quaternion algebra for every even sized nite collection of nite primes or real primes of K . The proof of this deep fact leads to many of the major results of class eld theory. Version: 1 Owner: djao Author(s): djao

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Chapter 252 16K50 Brauer groups252.1 Brauer group

Let K be a eld. The Brauer group Br(K ) of K is the set of all equivalence classes of central simple algebras over K , where two central simple algebras A and B are equivalent if there exists a division ring D over K and natural numbers n, m such that A (resp. B ) is isomorphic to the ring of n n (resp. m m) matrices with coecients in D . The group operation in Br(K ) is given by tensor product: for any two central simple algebras A, B over K , their product in Br(K ) is the central simple algebra A K B . The identity element in Br(K ) is the class of K itself, and the inverse of a central simple algebra A is the opposite algebra Aopp dened by reversing the order of the multiplication operation of A. Version: 5 Owner: djao Author(s): djao

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Chapter 253 16K99 Miscellaneous253.1 division ring

A division ring is a ring D with identity such that 1=0 For all nonzero a D , there exists b D with a b = b a = 1 A eld is equivalent to a commutative division ring. Version: 3 Owner: djao Author(s): djao

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Chapter 254 16N20 Jacobson radical, quasimultiplication254.1 Jacobson radical

The Jacobson radical J (R) of a ring R is the intersection of the annihilators of irreducible left R-modules. The following are alternate characterizations of the Jacobson radical J (R): 1. The intersection of all left primitive ideals. 2. The intersection of all maximal left ideals. 3. The set of all t R such that for all r R, 1 rt is left invertible (i.e. there exists u such that u(1 rt) = 1). 4. The largest ideal I such that for all v I , 1 v is a unit in R. 5. (1) - (3) with left replaced by right and rt replaced by tr . Note that if R is commutative and nitely generated, then J (R) = {x R | xn = 0for some n N} = Nil(R) Version: 13 Owner: saforres Author(s): saforres

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254.2

a ring modulo its Jacobson radical is semiprimitive

Let R be a ring. Then J (R/J (R)) = (0).L et [u] J (R/J (R)). Then by one of the alternate characterizations of the Jacobson radical, 1 [r ][u] is left invertible for all r R, so there exists v R such that [v ](1 [r ][u]) = 1.

Then v (1 ru) = 1 a for some a J (R). So wv (1 ru) = 1 since w (1 a) = 1 for some w R. Since this holds for all r R, u J (R), then [u] = 0. Version: 3 Owner: saforres Author(s): saforres

254.3

examples of semiprimitive rings

Examples of semiprimitive rings:The integers Z: ince Z is commutative, any left ideal is two-sided. So the maximal left ideals of Z are the maximal ideals of Z, which are the ideals pZ for p prime. Note that pZ q Z = (0) if gcd(p, q ) > 1. Hence J (Z) = p pZ = (0).S

A matrix ring Mn (D ) over a division ring D :T

he ring Mn (D ) is simple, so the only proper ideal is (0). Thus J (Mn (D )) = (0).

A polynomial ring R[x] over a domain R:T

1. By one of the alternate characterizations of the Jacobson radical, 1 ax is a unit. But deg(1 ax) = max{deg(1), deg(ax)} 1. So 1 ax is not a unit, and by this contradiction we see that J (R[x]) = (0). Version: 5 Owner: saforres Author(s): saforres

ake a J (R[x]) with a = 0. Then ax J (R[x]), since J (R[x]) is an ideal, and deg(ax)

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254.4

proof of Characterizations of the Jacobson radical

First, note that by denition a left primitive ideal is the annihilator of an irreducible left Rmodule, so clearly characterization 1) is equivalent to the denition of the Jacobson radical. Next, we will prove cyclical containment. Observe that 5) follows after the equivalence of 1) - 4) is established, since 4) is independent of the choice of left or right ideals. 1) 2) We know that every left primitive ideal is the largest ideal contained in a maximal left ideal. So the intersection of all left primitive ideals will be contained in the intersection of all maximal left ideals. 2) 3) Let S = {M : M a maximal left ideal of R} and take r R. Let t M S M . Then rt M S M . Assume 1 rt is not left invertible; therefore there exists a maximal left ideal M0 of R such that R(1 rt) M0 . Note then that 1 rt M0 . Also, by denition of t, we have rt M0 . Therefore 1 M0 ; this contradiction implies 1 rt is left invertible. 3) 4) We claim that 3) satises the condition of 4). Let K = {t R : 1 rt is left invertible for all r R}. We shall rst show that K is an ideal. Clearly if t K , then rt K . If t1 , t2 K , then 1 r (t1 + t2 ) = (1 rt1 ) rt2 Now there exists u1 such that u1(1 rt1 ) = 1, hence u1 ((1 rt1 ) rt2 ) = 1 u1rt2 Similarly, there exists u2 such that u2 (1 u1 rt2 ) = 1, therefore u2 u1 (1 r (t1 + t2 )) = 1 Hence t1 + t2 K . Now if t K, r R, to show that tr K it suces to show that 1 tr is left invertible. Suppose u(1 rt) = 1, hence u urt = 1, then tur turtr = tr . So (1 + tur )(1 tr ) = 1 + tur tr turtr = 1. Therefore K is an ideal. Now let v K . Then there exists u such that u(1 v ) = 1, hence 1 u = uv K , so u = 1 (1 u) is left invertible. So there exists w such that wu = 1, hence wu(1 v ) = w , then 1 v = w . Thus 1117

(1 v )u = 1 and therefore 1 v is a unit. Let J be the largest ideal such that, for all v J , 1 v is a unit. We claim that K J. Suppose this were not true; in this case K + J strictly contains J . Consider rx + sy K + J with x K, y J and r, s R. Now 1 (rx + sy ) = (1 rx) sy , and since rx K , then 1 rx = u for some unit u R. So 1 (rx + sy ) = u sy = u(1 u1 sy ), and clearly u1sy J since y J . Hence 1 u1 sy is also a unit, and thus 1 (rx + sy ) is a unit. Thus 1 v is a unit for all v K + J . But this contradicts the assumption that J is the largest such ideal. So we must have K J . 4) 1) We must show that if I is an ideal such that for all u I , 1 u is a unit, then I ann(R M ) for every irreducible left R-module R M . Suppose this is not the case, so there exists R M such that I ann(R M ). Now we know that ann(R M ) is the largest ideal inside some maximal left ideal J of R. Thus we must also have I J , or else this would contradict the maximality of ann(R M ) inside J . But since I J , then by maximality I + J = R, hence there exist u I and v J such that u + v = 1. Then v = 1 u, so v is a unit and J = R. But since J is a proper left ideal, this is a contradiction. Version: 25 Owner: saforres Author(s): saforres

254.5

properties of the Jacobson radical

Theorem: Let R, T be rings and : R T be a surjective homomorphism. Then (J (R)) J (T ). e shall use the characterization of the Jacobson radical as the set of all a R such that for all r R, 1 ra is left invertible.W

Let a J (R), t T . We claim that 1 t(a) is left invertible: Since is surjective, t = (r ) for some r R. Since a J (R), we know 1 ra is left invertible, so there exists u R such that u(1 ra) = 1. Then we have (u) ((1) (r )(a)) = (u)(1 ra) = (1) = 1 So (a) J (T ) as required. Theorem: Let R, T be rings. Then J (R T ) J (R) J (T ). 1118

et 1 : R T R be a (surjective) projection. By the previous theorem, 1 (J (R T )) J (R).L

Similarly let 2 : R T T be a (surjective) projection. We see that 2 (J (R T )) J (T ). Now take (a, b) J (R T ). Note that a = 1 (a, b) J (R) and b = 2 (a, b) J (T ). Hence (a, b) J (R) J (T ) as required. Version: 8 Owner: saforres Author(s): saforres

254.6

quasi-regularity

An element x of a ring is called right quasi-regular [resp. left quasi-regular] if there is an element y in the ring such that x + y + xy = 0 [resp. x + y + yx = 0]. For calculations with quasi-regularity, it is useful to introduce the operation dened: x y = x + y + xy. Thus x is right quasi-regular if there is an element y such that x y = 0. The operation is easily demonstrated to be associative, and x 0 = 0 x = 0 for all x. An element x is called quasi-regular if it is both left and right quasi-regular. In this case, there are elements x and y such that x+ y + xy = 0 = x+ z + zx (equivalently, x y = z x = 0). A calculation shows that y = 0 y = (z x) y = z (x y ) = z. So y = z is a unique element, depending on x, called the quasi-inverse of x. An ideal (one- or two-sided) of a ring is called quasi-regular if each of its elements is quasiregular. Similarly, a ring is called quasi-regular if each of its elements is quasi-regular (such rings cannot have an identity element). Lemma 1. Let A be an ideal (one- or two-sided) in a ring R. If each element of A is right quasi-regular, then A is a quasi-regular ideal. This lemma means that there is no extra generality gained in dening terms such as right quasi-regular left ideal, etc. Quasi-regularity is important because it provides elementary characterizations of the Jacobson radical for rings without an identity element: The Jacobson radical of a ring is the sum of all quasi-regular left (or right) ideals. 1119

The Jacobson radical of a ring is the largest quasi-regular ideal of the ring. For rings with an identity element, note that x is [right, left] quasi-regular if and only if 1 + x is [right, left] invertible in the ring. Version: 1 Owner: mclase Author(s): mclase

254.7

semiprimitive ring

A ring R is said to be semiprimitive (sometimes semisimple) if its Jacobson radical is the zero ideal. Any simple ring is automatically semiprimitive. A nite direct product of matrix rings over division rings can be shown to be semiprimitive and both left and right artinian. The Artin-Wedderburn Theorem states that any semiprimitive ring which is left or right Artinian is isomorphic to a nite direct product of matrix rings over division rings. Version: 11 Owner: saforres Author(s): saforres

1120

Chapter 255 16N40 Nil and nilpotent radicals, sets, ideals, rings255.1 Koethe conjecture

The Koethe Conjecture is the statement that for any pair of nil right ideals A and B in any ring R, the sum A + B is also nil. If either of A or B is a two-sided ideal, it is easy to see that A + B is nil. Suppose A is a two-sided ideal, and let x A + B . The quotient (A + B )/A is nil since it is a homomorphic image of B . So there is an n > 0 with xn A. Then there is an m > 0 such that xnm = 0, because A is nil. In particular, this means that the Koethe conjecture is true for commutative rings. It has been shown to be true for many classes of rings, but the general statement is still unproven, and no counter example has been found. Version: 1 Owner: mclase Author(s): mclase

255.2

nil and nilpotent ideals

An element x of a ring is nilpotent if xn = 0 for some positive integer n. A ring R is nil if every element in R is nilpotent. Similarly, a one- or two-sided ideal is called nil if each of its elements is nilpotent. A ring R [resp. a one- or two sided ideal A] is nilpotent if Rn = 0 [resp. An = 0] for some positive integer n. 1121

A ring or an ideal is locally nilpotent if every nitely generated subring is nilpotent. The following implications hold for rings (or ideals): nilpotent locally nilpotent nil

Version: 3 Owner: mclase Author(s): mclase

1122

Chapter 256 16N60 Prime and semiprime rings256.1 prime ring

A ring R is said to be a prime ring if the zero ideal is a prime ideal. If R is commutative, this is equivalent to being an integral domain. Version: 2 Owner: antizeus Author(s): antizeus

1123

Chapter 257 16N80 General radicals and rings257.1 prime radical

The prime radical of a ring R is the intersection of all the prime ideals of R. Note that the prime radical is the smallest semiprime ideal of R, and that R is a semiprime ring if and only if its prime radical is the zero ideal. Version: 2 Owner: antizeus Author(s): antizeus

257.2

radical theory

Let x be a property which denes a class of rings, which we will call the x -rings. Then x is a radical property if it satises: 1. The class of x -rings is closed under homomorphic images. 2. Every ring R has a largest ideal in the class of x -rings; this ideal is written x (R). 3. x (R/x (R)) = 0. Note: it is extremely important when interpreting the above denition that your denition of a ring does not require an identity element. The ideal x (R) is called the x -radical of R. A ring is called x -radical if x (R) = R, and is called x -semisimple if x (R) = 0. If x is a radical property, then the class of x -rings is also called the class of x -radical rings. 1124

The class of x -radical rings is closed under ideal extensions. That is, if A is an ideal of R, and A and R/A are x -radical, then so is R. Radical theory is the study of radical properties and their interrelations. There are several well-known radicals which are of independent interest in ring theory (See examples to follow). The class of all radicals is however very large. Indeed, it is possible to show that any partition of the class of simple rings into two classes, R and S gives rise to a radical x with the property that all rings in R are x -radical and all rings in S are x -semisimple. A radical x is hereditary if every ideal of an x -radical ring is also x -radical. A radical x is supernilpotent if the class of x -rings contains all nilpotent rings. Version: 2 Owner: mclase Author(s): mclase

1125

Chapter 258 16P40 Noetherian rings and modules258.1 Noetherian ring

A ring R is right noetherian (or left noetherian ) if R is noetherian as a right module (or left module ), i.e., if the three equivalent conditions hold: 1. right ideals (or left ideals) are nitely generated 2. the ascending chain condition holds on right ideals (or left ideals) 3. every nonempty family of right ideals (or left ideals) has a maximal element. We say that R is noetherian if it is both left noetherian and right noetherian. Examples of Noetherian rings include any eld (as the only ideals are 0 and the whole ring) and the ring Z of integers (each ideal is generated by a single integer, the greatest common divisor of the elements of the ideal). The Hilbert basis theorem says that a ring R is noetherian i the polynomial ring R[x] is. Version: 10 Owner: KimJ Author(s): KimJ

258.2

noetherian

A module M is noetherian if it satises the following equivalent conditions:

1126

the ascending chain condition holds for submodules of M ; every nonempty family of submodules of M has a maximal element; every submodule of M is nitely generated. A ring R is left noetherian if it is noetherian as a left module over itself (i.e. if R R is a noetherian module), and right noetherian if it is noetherian as a right module over itself (i.e. if RR is an noetherian module), and simply noetherian if both conditions hold. Version: 2 Owner: antizeus Author(s): antizeus

1127

Chapter 259 16P60 Chain conditions on annihilators and summands: Goldie-type conditions , Krull dimension259.1 Goldie ring

Let R be a ring. If the set of annihilators {r. ann(x) | x R} satisies the ascending chain condition, then R is said to satisfy the ascending chain condition on right annihilators. A ring R is called a right Goldie ring if it satises the ascending chain condition on right annihilators and RR is a module of nite rank. Left Goldie ring is dened similarly. If the context makes it clear on which side the ring operates, then such a ring is simply called a Goldie ring. A right noetherian ring is right Goldie. Version: 3 Owner: mclase Author(s): mclase

259.2

uniform dimension

Let M be a module over a ring R, and suppose that M contains no innite direct sums of non-zero submodules. (This is the same as saying that M is a module of nite rank.)

1128

Then there exits an integer n such that M contains an essential submodule N where N = U1 U2 Un is a direct sum of n uniform submodules. This number n does not depend on the choice of N or the decomposition into uniform submodules. We call n the uniform dimension of M . Sometimes this is written u-dim M = n. If R is a eld K , and M is a nite-dimensional vector space over K , then u-dim M = dimK M . u-dim M = 0 if and only if M = 0. Version: 3 Owner: mclase Author(s): mclase

1129

Chapter 260 16S10 Rings determined by universal properties (free algebras, coproducts, adjunction of inverses, etc.)260.1 Ore domain

Let R be a domain. We say that R is a right Ore domain if any two nonzero elements of R have a nonzero common right multiple, i.e. for every pair of nonzero x and y , there exists a pair of elements r and s of R such that xr = ys = 0. This condition turns out to be equivalent to the following conditions on R when viewed as a right R-module: (a) RR is a uniform module. (b) RR is a module of nite rank. The denition of a left Ore domain is similar. If R is a commutative domain, then it is a right (and left) Ore domain. Version: 6 Owner: antizeus Author(s): antizeus

1130

Chapter 261 16S34 Group rings , Laurent polynomial rings261.1 support

Let R[G] be the group ring of a group G over a ring R. Let x = g xg g be an element of R[G]. The support of x, often written supp(x), is the set of elements of G which occur with non-zero coecient in the expansion of x. Thus: supp(x) = {g G | xg = 0}. Version: 2 Owner: mclase Author(s): mclase

1131

Chapter 262 16S36 Ordinary and skew polynomial rings and semigroup rings262.1 Gaussian polynomials

For an indeterminate u and integers n m 0 we dene the following: (a) (m)u = um1 + um2 + + 1 for m > 0, (b) (m!)u = (m)u (m 1)u (1)u for m > 0, and (0!)u = 1, (c)n m u (n!)u . (m!)u ((nm)!)u n m u n m u

=

If m > n then we dene

= 0.

The expressions

are called u-binomial coecients or Gaussian polynomials.

Note: if we replace u with 1, then we obtain the familiar integers, factorials, and binomial coecients. Specically, (a) (m)1 = m, (b) (m!)1 = m!, (c)n m 1

=

n m

.

Version: 3 Owner: antizeus Author(s): antizeus

1132

262.2

q skew derivation

Let (, ) be a skew derivation on a ring R. Let q be a central (, )-constant. Suppose further that = q . Then we say that (, ) is a q -skew derivation. Version: 5 Owner: antizeus Author(s): antizeus

262.3

q skew polynomial ring

If (, ) is a q -skew derivation on R, then we say that the skew polynomial ring R[; , ] is a q -skew polynomial ring. Version: 3 Owner: antizeus Author(s): antizeus

262.4

sigma derivation

If is a ring endomorphism on a ring R, then a (left) -derivation is an additive map on R such that (x y ) = (x) (y ) + (x) y for all x, y in R. Version: 7 Owner: antizeus Author(s): antizeus

262.5

sigma, delta constant

If (, ) is a skew derivation on a ring R, then a (, )-constant is an element q of R such that (q ) = q and (q ) = 0. Note: If q is a (, )-constant, then it follows that (q x) = q (x) and (q x) = q (x) for all x in R. Version: 3 Owner: antizeus Author(s): antizeus

262.6

skew derivation

A (left) skew derivation on a ring R is a pair (, ), where is a ring endomorphism of R, and is a left -derivation on R. Version: 4 Owner: antizeus Author(s): antizeus 1133

262.7

skew polynomial ring

If (, ) is a left skew derivation on R, then we can construct the (left) skew polynomial ring R[; , ], which is made up of polynomials in an indeterminate and left-hand coecients from R, with multiplication satisfying the relation r = (r ) + (r ) for all r in R. Version: 2 Owner: antizeus Author(s): antizeus

1134

Chapter 263 16S99 Miscellaneous263.1 algebra

Let A be a ring with identity. An algebra over A is a ring B with identity together with a ring homomorphism f : A Z (B ), where Z (B ) denotes the center of B . Equivalently, an algebra over A is an Amodule B which is a ring and satises the property a (x y ) = (a x) y = x (a y ) for all a A and all x, y B . Here denotes Amodule multiplication and denotes ring multiplication in B . One passes between the two denitions as follows: given any ring homomorphism f : A Z (B ), the scalar multiplication rule a b := f (a) b makes B into an Amodule in the sense of the second denition. Version: 5 Owner: djao Author(s): djao

263.2

algebra (module)

Given a commutative ring R, an algebra over R is a module M over R, endowed with a law of composition f :M M M which is R-bilinear. Most of the important algebras in mathematics belong to one or the other of two classes: the unital associative algebras, and the Lie algebras. 1135

263.2.1

Unital associative algebras

In these cases, the product (as it is called) of two elements v and w of the module, is denoted simply by vw or v w or the like. Any unital associative algebra is an algebra in the sense of djao (a sense which is also used by Lang in his book Algebra (Springer-Verlag)). Examples of unital associative algebras: tensor algebras and quotients of them Cayley algebras, such as the ring of quaternions polynomial rings the ring of endomorphisms of a vector space, in which the bilinear product of two mappings is simply the composite mapping.

263.2.2

Lie algebras

In these cases the bilinear product is denoted by [v, w ], and satises [v, v ] = 0 for all v M [v, [w, x]] + [w, [x, v ]] + [x, [v, w ]] = 0 for all v, w, x M The second of these formulas is called the Jacobi identity. One proves easily [v, w ] + [w, v ] = 0 for all v, w M for any Lie algebra M. Lie algebras arise naturally from Lie groups, q.v. Version: 1 Owner: karthik Author(s): Larry Hammick

1136

Chapter 264 16U10 Integral domains264.1 Pr ufer domain

An integral domain R is a Pr ufer domain if every nitely generated ideal I of R is invertible. Let RI denote the localization of R at I . Then the following statements are equivalent: i) R is a Pr ufer domain. ii) For every prime ideal P in R, RP is a valuation domain. iii) For every maximal ideal M in R, RM is a valuation domain. A Pr ufer domain is a Dedekind domain if and only if it is noetherian. If R is a Pr ufer domain with quotient eld K , then any domain S such that R S K is Pr ufer.

REFERENCES1. Thomas W. Hungerford. Algebra. Springer-Verlag, 1974. New York, NY.

Version: 2 Owner: mathcam Author(s): mathcam

264.2

valuation domain

An integral domain R is a valuation domain if for all a, b R, either a|b or b|a. 1137

Version: 3 Owner: mathcam Author(s): mathcam

1138

Chapter 265 16U20 Ore rings, multiplicative sets, Ore localization265.1 Goldies Theorem

Let R be a ring with an identity. Then R has a right classical ring of quotients Q which is semisimple Artinian if and only if R is a semiprime right Goldie ring. If this is the case, then the composition length of Q is equal to the uniform dimension of R. An immediate corollary of this is that a semiprime right noetherian ring always has a right classical ring of quotients. This result was discovered by Alfred Goldie in the late 1950s. Version: 3 Owner: mclase Author(s): mclase

265.2

Ore condition

A ring R satises the left Ore condition (resp. right Ore condition) if and only if for all elements x and y with x regular, there exist elements u and v with v regular such that ux = vy (resp.xu = yv ).

A ring which satises the (left, right) Ore condition is called a (left, right) Ore ring. Version: 3 Owner: mclase Author(s): mclase

1139

265.3

Ores theorem

A ring has a (left, right) classical ring of quotients if and only if it satises the (left, right) Ore condition. Version: 3 Owner: mclase Author(s): mclase

265.4

classical ring of quotients

Let R be a ring. An element of R is called regular if it is not a right zero divisor or a left zero divisor in R. A ring Q R is a left classical ring of quotients for R (resp. right classical ring of quotients for R) if it satisies: every regular element of R is invertible in Q every element of Q can be written in the form x1 y (resp. yx1 ) with x, y R and x regular. If a ring R has a left or right classical ring of quotients, then it is unique up to isomorphism. If R is a commutative integral domain, then the left and right classical rings of quotients always exist they are the eld of fractions of R. For non-commutative rings, necessary and sucient conditions are given by Ores theorem. Note that the goal here is to construct a ring which is not too dierent from R, but in which more elements are invertible. The rst condition says which elements we want to be invertible. The second condition says that Q should contain just enough extra elements to make the regular elements invertible. Such rings are called classical rings of quotients, because there are other rings of quotients. These all attempt to enlarge R somehow to make more elements invertible (or sometimes to make ideals invertible). Finally, note that a ring of quotients is not the same as a quotient ring. Version: 2 Owner: mclase Author(s): mclase

1140

265.5

saturated

Let S be multiplicative subset of A. We say that S is a saturated if ab S a, b S. When A is an integral domain, then S is saturated if and only if its complement A\S is union of prime ideals. Version: 1 Owner: drini Author(s): drini

1141

Chapter 266 16U70 Center, normalizer (invariant elements)266.1 center (rings)

If A is a ring, the center of A, sometimes denoted Z(A), is the set of all elements in A that commute with all other elements of A. That is, Z(A) = {a A | ax = xax A} Note that 0 Z(A) so the center is non-empty. If we assume that A is a ring with a multiplicative unity 1, then 1 is in the center as well. The center of A is also a subring of A. Version: 3 Owner: dublisk Author(s): dublisk

1142

Chapter 267 16U99 Miscellaneous267.1 anti-idempotent

An element x of a ring is called an anti-idempotent element, or simply an anti-idempotent if x2 = x. The term is most often used in linear algebra. Every anti-idempotent matrix over a eld is diagonalizable. Two anti-idempotent matrices are similar if and only if they have the same rank. Version: 1 Owner: mathcam Author(s): mathcam

1143

Chapter 268 16W20 Automorphisms and endomorphisms268.1 ring of endomorphisms

Let R be a ring and let M be a right R-module. An endomorphism of M is a R-module homomorphism from M to itself. We shall write endomorphisms on the left, so that f : M M maps x f (x). If f, g : M M are two endomorphisms, we can add them: f + g : x f (x) + g (x) and multiply them With these operations, the set of endomorphisms of M becomes a ring, which we call the ring of endomorphisms of M , written EndR (M ). Instead of writing endomorphisms as functions, it is often convenient to write them multiplicatively: we simply write the application of the endomorphism f as x f x. Then the fact that each f is an R-module homomorphism can be expressed as: f (xr ) = (f x)r for all x M and r R and f EndR (M ). With this notation, it is clear that M becomes an EndR (M )-R-bimodule. Now, let N be a left R-module. We can construct the ring EndR (N ) in the same way. There is a complication, however, if we still think of endomorphism as functions written on the left. In order to make M into a bimodule, we need to dene an action of EndR (N ) on the right of N : say x f = f (x) 1144 f g : x f (g (x))

But then we have a problem with the multiplication: x f g = f g (x) = f (g (x)) but (x f ) g = f (x) g = g (f (x))! In order to make this work, we need to reverse the order of composition when we dene multiplication in the ring EndR (N ) when it acts on the right. There are essentially two dierent ways to go from here. One is to dene the muliplication in EndR (N ) the other way, which is most natural if we write the endomorphisms as functions on the right. This is the approach taken in many older books. The other is to leave the muliplication in EndR (N ) the way it is, but to use the opposite ring to dene the bimodule. This is the approach that is generally taken in more recent works. Using this approach, we conclude that N is a R-EndR (N )op -bimodule. We will adopt this convention for the lemma below. Considering R as a right and a left module over itself, we can construct the two endomorphism rings EndR (RR ) and EndR (R R). Lemma 2. Let R be a ring with an identity element. Then R EndR (R R)op .D

EndR (RR ) and R

ene r EndR (R R) by x xr .

A calculation shows that rs = s r (functions written on the left) from which it is easily seen that the map : r r is a ring homomorphism from R to EndR (R R)op . We must show that this is an isomorphism. If r = 0, then r = 1r = r (1) = 0. So is injective. Let f be an arbitrary element of EndR (R R), and let r = f (1). Then for any x R, f (x) = f (x1) = xf (1) = xr = r (x), so f = r = (r ). The proof of the other isomorphism is similar. Version: 4 Owner: mclase Author(s): mclase

1145

Chapter 269 16W30 Coalgebras, bialgebras, Hopf algebras ; rings, modules, etc. on which these act269.1 Hopf algebra

A Hopf algebra is a bialgebra A over a eld K with a K-linear map S : A A, called the Denition 1. antipode, such that m (S id) = = m (id S ) , (269.1.1)

where m : A A A is the multiplication map m(a b) = ab and : K A is the unit map (k ) = k 1 I. In terms of a commutative diagram: A

AAS id

AAidS

C m m

AA

AA

A

1146

Example 1 (Algebra of functions on a nite group). Let A = C (G) be the algebra of complexvalued functions on a nite group G and identify C (G G) with A A. Then, A is a Hopf algebra with comultiplication ((f ))(x, y ) = f (xy ), counit (f ) = f (e), and antipode (S (f ))(x) = f (x1 ). Example 2 (Group algebra of a nite group). Let A = CG be the complex group algebra of a nite group G. Then, A is a Hopf algebra with comultiplication (g ) = g g , counit (g ) = 1, and antipode S (g ) = g 1 . The above two examples are dual to one another. Dene a bilinear form C (G) CG C by f, x = f (x). Then, f g, x 1, x (f ), x y (f ) S (f ), x = f g, (x) , = (x), = f, xy , = f, e , = f, S (x) .

Example 3 (Polynomial functions on a Lie group). Let A = Poly(G) be the algebra of complex-valued polynomial functions on a complex Lie group G and identify Poly(G G) with A A. Then, A is a Hopf algebra with comultiplication ((f ))(x, y ) = f (xy ), counit (f ) = f (e), and antipode (S (f ))(x) = f (x1 ). Example 4 (Universal enveloping algebra of a Lie algebra). Let A = U(g) be the universal enveloping algebra of a complex Lie algebra g. Then, A is a Hopf algebra with comultiplication (X ) = X 1 + 1 X , counit (X ) = 0, and antipode S (X ) = X . The above two examples are dual to one another (if g is the Lie algebra of G). Dene a d bilinear form Poly(G) U(g) C by f, X = d f (exp(tX )). t t=0 Version: 6 Owner: mhale Author(s): mhale

269.2

almost cocommutative bialgebraR(a) = op (a)R

A bialgebra A is called almost cocommutative if there is an unit R A A such that where op is the opposite comultiplication (the usual comultiplication, composed with the ip map of the tensor product A A). The element R is often called the R-matrix of A. The signicance of the almost cocommutative condition is that V,W = R : V W W V gives a natural isomorphism of bialgebra representations, where V and W are Amodules, making the category of A-modules into a quasi-tensor or braided monoidal category. Note that W,V V,W is not necessarily the identity (this is the braiding of the category). Version: 2 Owner: bwebste Author(s): bwebste 1147

269.3A

bialgebra

Denition 2. bialgebra is a vector space that is both a unital algebra and a coalgebra, such that the comultiplication and counit are unital algebra homomorphisms. Version: 2 Owner: mhale Author(s): mhale

269.4A

coalgebra

Denition 3. coalgebra is a vector space A over a eld K with a K-linear map : A A A, called the Denition 4. comultiplication, and a (non-zero) K-linear map : A K, called the ( id) = (id ) (coassociativity), ( id) = id = (id ) . In terms of commutative diagrams: A

Denition 5. counit, such that (269.4.1) (269.4.2)

AAid

AA AAA A id id

AAid

AAid

A

Let : A A A A be the ip map (a b) = b a. A coalgebra is said to be Denition 6. cocommutative if = . Version: 4 Owner: mhale Author(s): mhale 1148

269.5

coinvariant

Let V be a comodule with a right coaction t : V V A of a coalgebra A. An element v V is Denition 7. right coinvariant if t(v ) = v 1 IA . Version: 1 Owner: mhale Author(s): mhale (269.5.1)

269.6

comodule

Let (A, , ) be a coalgebra. A Denition 8. right A-comodule is a vector space V with a linear map t : V V A, called the Denition 9. right coaction, satisfying (t id) t = (id ) t, (id ) t = id. (269.6.1)

An A-comodule is also referred to as a corepresentation of A. Let V and W be two right A-comodules. Then V W is also a right A-comodule. If A is a bialgebra then V W is a right A-comodule as well (make use of the multiplication map A A A). Version: 2 Owner: mhale Author(s): mhale

269.7

comodule algebra

Let H be a bialgebra. A right H -comodule algebra is a unital algebra A which is a right H -comodule satisfying t(ab) = t(a)t(b) = for all h H and a, b A. There is a dual notion of a H -module coalgebra. Example 5. Let H be a bialgebra. Then H is itself a H -comodule algebra for the right regular coaction t(h) = (h). Version: 5 Owner: mhale Author(s): mhale 1149 a(1) b(1) a(2) b(2) , t(1 IA ) = 1 IA 1 IH , (269.7.1)

269.8

comodule coalgebra

Let H be a bialgebra. A right H -comodule coalgebra is a coalgebra A which is a right H -comodule satisfying ( id)t(a) = a(1)(1) a(2)(1) a(1)(2) a(2)(2) , ( id)t(a) = (a)1 IH , (269.8.1)

for all h H and a A. There is a dual notion of a H -module algebra. Example 6. Let H be a Hopf algebra. Then H is itself a H -comodule coalgebra for the adjoint coaction t(h) = h(2) S (h(1) )h(3) . Version: 4 Owner: mhale Author(s): mhale

269.9

module algebra

Let H be a bialgebra. A left H -module algebra is a unital algebra A which is a left H -module satisfying h (ab) = for all h H and a, b A. There is a dual notion of a H -comodule coalgebra. Example 7. Let H be a Hopf algebra. Then H is itself a H -module algebra for the adjoint action g h = g(1) hS (g(2) ). Version: 4 Owner: mhale Author(s): mhale (h(1) a)(h(2) b), h 1 IA = (h)1 IA , (269.9.1)

269.10

module coalgebra

Let H be a bialgebra. A left H -module coalgebra is a coalgebra A which is a left H module satisfying (h a) = (h(1) a(1) ) (h(2) a(2) ), (h a) = (h)(a), (269.10.1)

for all h H and a A. There is a dual notion of a H -comodule algebra. Example 8. Let H be a bialgebra. Then H is itself a H -module coalgebra for the left regular action g h = gh. Version: 5 Owner: mhale Author(s): mhale 1150

Chapter 270 16W50 Graded rings and modules270.1 graded algebra

An algebra A is graded if it is a graded module and satises Ap Aq Ap+q Examples of graded algebras include the polynomial ring k [X ] being an N-graded k -algebra, and the exterior algebra. Version: 1 Owner: dublisk Author(s): dublisk

270.2

graded module

If R = R0 R1 is a graded ring, then a graded module over R is a module M of the form M = i= Mi and satises Ri Mj Mi+j for all i, j . Version: 4 Owner: KimJ Author(s): KimJ

270.3

supercommutative

Let R be a Z2 -graded ring. Then R is supercommutative if for any homogeneous elements a and b R: ab = (1)deg a deg b ba. 1151

This is, even homogeneous elements are in the center of the ring, and odd homogeneous elements anti-commute. Common examples of supercommutative rings are the exterior algebra of a module over a commutative ring (in particular, a vector space) and the cohomology ring of a topological space (both with the standard grading by degree reduced mod 2). Version: 1 Owner: bwebste Author(s): bwebste

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Chapter 271 16W55 Super (or skew) structure271.1 super tensor product

If A and B are Z-graded algebras, we dene the super tensor product A su B to be the ordinary tensor product as graded modules, but with multiplication - called the super product - dened by (a b)(a b ) = (1)(deg b)(deg a ) aa bb where a, a , b, b are homogeneous. The super tensor product of A and B is itself a graded algebra, as we grade the super tensor product of A and B as follows: (A su B )n = Ap B q

p,q : p+q =n

Version: 4 Owner: dublisk Author(s): dublisk

271.2

superalgebra

A graded algebra A is said to be a super algebra if it has a Z/2Z grading. Version: 2 Owner: dublisk Author(s): dublisk

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271.3

supernumber

Let N be the Grassmann algebra generated by i , i = 1 . . . N , such that i j = j i and (i )2 = 0. Denote by , the case of an innite number of generators i . A Denition 10. supernumber is an element of N or . Any supernumber z can be expressed uniquely in the form 1 1 z = z0 + zi i + zij i j + . . . + zi1 ...in i1 . . . in + . . . , 2 n! where the coecients zi1 ...in C are antisymmetric in their indices. The Denition 11. body of z is dened as zB = z0 , and its Denition 12. soul is dened as zS = z zB . If zB = 0 then z has an inverse given by z1

1 = zB

k =0

zS zB

k

.

A supernumber can be decomposed into the even and odd parts 1 1 zeven = z0 + zij i j + . . . + zi1 ...i2n i1 . . . i2n + . . . , 2 (2n)! 1 1 zodd = zi i + zijk i j k + . . . + zi ...i i1 . . . i2n+1 + . . . . 6 (2n + 1)! 1 2n+1 Purely even supernumbers are called Denition 13. c-numbers, and odd supernumbers are called Denition 14. a-numbers. The superalgebra N thus has a decomposition N = Cc Ca , where Cc is the space of c-numbers, and Ca is the space of a-numbers. Supernumbers are the generalisation of complex numbers to a commutative superalgebra of commuting and anticommuting numbers. They are primarily used in the description of fermionic elds in quantum eld theory. Version: 5 Owner: mhale Author(s): mhale

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Chapter 272 16W99 Miscellaneous272.1 Hamiltonian quaternions

Denition of Q We dene a unital associative algebra Q over R, of dimension 4, by the basis {1, i, j, k} and the multiplication table 1 i i 1 j k k j j k 1 i k j i 1

(where the element in row x and column y is xy , not yx). Thus an arbitrary element of Q is of the form a1 + bi + cj + dk, a, b, c, d R (sometimes denoted by a, b, c, d or by a + b, c, d ) and the product of two elements a, b, c, d and , , , is w, x, y, z where w x y z = = = = a b c d a + b + c d a b + c + k a + b c + k

The elements of Q are known as Hamiltonian quaternions. Clearly the subspaces of Q generated by {1} and by {1, i} are subalgebras isomorphic to R and C respectively. R is customarily identied with the corresponding subalgebra of Q. (We

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shall see in a moment that there are other and less obvious embeddings of C in Q.) The real numbers commute with all the elements of Q, and we have a, b, c, d = a, b, c, d for R and a, b, c, d Q. norm, conjugate, and inverse of a quaternion Like the complex numbers (C), the quaternions have a natural involution called the quaternion conjugate. If q = a1 + bi + cj + dk, then the quaternion conjugate of q , denoted q, is simply q = a1 bi cj dk. One can readily verify that if q = a1 + bi + cj + dk, then qq = (a2 + b2 + c2 + d2 )1. (See Euler four-square identity.) This product is used to form a norm | | on the algebra (or the ring) Q: We dene q = s where qq = s1. If v, w Q and R, then 1. v 0 with equality only if v = 0, 0, 0, 0 = 0

2. v = || v 3. v + w v + w 4. v w = v w which means that Q qualies as a normed algebra when we give it the norm | |. Because the norm of any nonzero quaternion q is real and nonzero, we have qq qq = = 1, 0, 0, 0 2 q q 2 which shows that any nonzero quaternion has an inverse: q 1 = Other embeddings of C into Q One can use any non-zero q to dene an embedding of C into Q. If n(z ) is a natural embedding of z C into Q, then the embedding: z q n(z )q 1 is also an embedding into Q. Because Q is an associative algebra, it is obvious that: (q n(a)q 1 )(q n(b)q 1 ) = q (n(a)n(b))q 1 1156 q q2

.

and with the distributive laws, it is easy to check that (q n(a)q 1 ) + (q n(b)q 1 ) = q (n(a) + n(b))q 1 Rotations in 3-space Let us write U = {q Q : ||q || = 1} With multiplication, U is a group. Let us briey sketch the relation between U and the group SO (3) of rotations (about the origin) in 3-space. An arbitrary element q of U can be expressed cos 2 + sin (ai + bj + ck), for some real 2 2 2 2 numbers , a, b, c such that a + b + c = 1. The permutation v qv of U thus gives rise to a permutation of the real sphere. It turns out that that permutation is a rotation. Its axis is the line through (0, 0, 0) and (a, b, c), and the angle through which it rotates the sphere is . If rotations F and G correspond to quaternions q and r respectively, then clearly the permutation v qrv corresponds to the composite rotation F G. Thus this mapping of U onto SO (3) is a group homomorphism. Its kernel is the subset {1, 1} of U , and thus it comprises a double cover of SO (3). The kernel has a geometric interpretation as well: two unit vectors in opposite directions determine the same axis of rotation.

Version: 3 Owner: mathcam Author(s): Larry Hammick, patrickwonders

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Chapter 273 16Y30 Near-rings273.1 near-ring

A near-ring is a set N together with two binary operations, denoted + : N N N and : N N N , such that 1. (a + b) + c = a + (b + c) and (a b) c = a (b c) for all a, b, c N (associativity of both operations) 2. There exists an element 0 N such that a + 0 = 0 + a = a for all a N (additive identity) 3. For all a N , there exists b N such that a + b = b + a = 0 (additive inverse) 4. (a + b) c = (a c) + (b c) for all a, b, c N (right distributive law) Note that the axioms of a near-ring dier from those of a ring in that they do not require addition to be commutative, and only require distributivity on one side. Every element a in a near-ring has a unique additive inverse, denoted a. We say N has an identity element if there exists an element 1 N such that a 1 = 1 a = a for all a N . We say N is distributive if a (b + c) = (a b) + (a c) holds for all a, b, c N . We say N is commutative if a b = b a for all a, b N . A natural example of a near-ring is the following. Let (G, +) be a group (not necessarily abelian), and let M be the set of all functions from G to G. For two functions f and g in M dene f + g M by (f + g )(x) = f (x) + g (x) for all x G. Then (M, +, ) is a near-ring with identity, where denotes composition of functions. Version: 13 Owner: yark Author(s): yark, juergen 1158

Chapter 274 17A01 General theory274.1 commutator bracket[a, b] = ab ba

Let A be an associative algebra over a eld K . For a, b A, the element of A dened by is called the commutator of a and b. The corresponding bilinear operation is called the commutator bracket. [, ] : A A A

The commutator bracket is bilinear, skew-symmetric, and also satises the Jacobi identity. To wit, for a, b, c A we have [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0. The proof of this assertion is straightforward. Each of the brackets in the left-hand side expands to 4 terms, and then everything cancels. In categorical terms, what we have here is a functor from the category of associative algebras to the category of Lie algebras over a xed eld. The action of this functor is to turn an associative algebra A into a Lie algebra that has the same underlying vector space as A, but whose multiplication operation is given by the commutator bracket. It must be noted that this functor is right-adjoint to the universal enveloping algebra functor. Examples Let V be a vector space. Composition endows the vector space of endomorphisms End V with the structure of an associative algebra. However, we could also regard End V as a Lie algebra relative to the commutator bracket: [X, Y ] = XY Y X, 1159 X, Y End V.

The algebra of dierential operators has some interesting properties when viewed as a Lie algebra. The fact is that even though, even though the composition of dierential operators is a non-commutative operation, it is commutative when restricted to the highest order terms of the involved operators. Thus, if X, Y are dierential operators of order p and q , respectively, the compositions XY and Y X have order p + q . Their highest order term coincides, and hence the commutator [X, Y ] has order p + q 1. In light of the preceding comments, it is evident that the vector space of rst-order dierential operators is closed with respect to the commutator bracket. Specializing even further we remark that, a vector eld is just a homogeneous rst-order dierential operator, and that the commutator bracket for vector elds, when viewed as rst-order operators, coincides with the usual, geometrically motivated vector eld bracket. Version: 4 Owner: rmilson Author(s): rmilson

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Chapter 275 17B05 Structure theory275.1 Killing form

Let g be a nite dimensional Lie algebra over a eld k , and adX : g g be the adjoint action, adX Y = [X, Y ]. Then the Killing form on g is a bilinear map Bg : g g k given by Bg(X, Y ) = tr(adX adY ). The Killing form is invariant and symmetric (since trace is symmetric). Version: 4 Owner: bwebste Author(s): bwebste

275.2

Levis theorem

Let g be a complex Lie algebra, r its radical. Then the extension 0 r g g/r 0 is split, i.e., there exists a subalgebra h of g mapping isomorphically to g/r under the natural projection. Version: 2 Owner: bwebste Author(s): bwebste

275.3

nilradical

Let g be a Lie algebra. Then the nilradical n of g is dened to be the intersection of the kernels of all the irreducible representations of g. Equivalently, n = [g, g] rad g, the 1161

interesection of the derived ideal and radical of g. Version: 1 Owner: bwebste Author(s): bwebste

275.4

radical

Let g be a Lie algebra. Since the sum of any two solvable ideals of g is in turn solvable, there is a unique maximal solvable ideal of any Lie algebra. This ideal is called the radical of g. Note that g/rad g has no solvable ideals, and is thus semi-simple. Thus, every Lie algebra is an extension of a semi-simple algebra by a solvable one. Version: 2 Owner: bwebste Author(s): bwebste

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Chapter 276 17B10 Representations, algebraic theory (weights)276.1 Ados theorem

Every nite dimensional Lie algebra has a faithful nite dimensional representation. In other words, every nite dimensional Lie algebra is a matrix algebra. This result is not true for Lie groups. Version: 2 Owner: bwebste Author(s): bwebste

276.2

Lie algebra representation

A representation of a Lie algebra g is a Lie algebra homomorphism : g End V, where End V is the commutator Lie algebra of some vector space V . In other words, is a linear mapping that satises ([a, b]) = (a)(b) (b)(a), We call the representation faithful if is injective. A invariant subsspace or sub-module W V is a subspace of V satisfying (a)(W ) W for all a g. A representation is called irreducible or simple if its only invariant subspaces are {0} and the whole representation. 1163 a, b g

Alternatively, one calls V a g-module, and calls (a), a g the action of a on V .

The dimension of V is called the dimension of the representation. If V is innite-dimensional, then one speaks of an innite-dimensional representation. Given a representation or pair of representation, there are a couple of operations which will produce other representations: First there is direct sum. If : g End(V ) and : g End(W ) are representations, then V W has the obvious Lie algebra action, by the embedding End(V ) End(W ) End(V W ). Version: 9 Owner: bwebste Author(s): bwebste, rmilson

276.3

adjoint representation

Let g be a Lie algebra. For every a g we dene the adjoint endomorphism, a.k.a. the adjoint action, ad(a) : g g to be the linear transformation with action ad(a) : b [a, b], The linear mapping ad : g End(g) with action a ad(a), ag b g.

is called the adjoint representation of g. The fact that ad denes a representation is a straight-forward consequence of the Jacobi identity axiom. Indeed, let a, b g be given. We wish to show that ad([a, b]) = [ad(a), ad(b)], where the bracket on the left is the g multiplication structure, and the bracket on the right is the commutator bracket. For all c g the left hand side maps c to [[a, b], c], while the right hand side maps c to [a, [b, c]] + [b, [a, c]]. Taking skew-symmetry of the bracket as a given, the equality of these two expressions is logically equivalent to the Jacobi identity: [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0. Version: 2 Owner: rmilson Author(s): rmilson 1164

276.4

examples of non-matrix Lie groups

While most well-known Lie groups are matrix groups, there do in fact exist Lie groups which are not matrix groups. That is, they have no faithful nite dimensional representations. For example, let H be the real Heisenberg group 1 a b H = 0 1 c | a, b, c R , 0 0 1 1 0 n = 0 1 0 | n Z . 0 0 1

and the discrete subgroup

The subgroup is central, and thus normal. The Lie group H/ has no faithful nite dimensional representations over R or C. Another example is the universal cover of SL2 R. SL2 R is homotopy equivalent to a circle, and thus (SL2 R) = Z, and thus has an innite-sheeted cover. Any real or complex representation of this group factors through the projection map to SL2 R. Version: 3 Owner: bwebste Author(s): bwebste

276.5

isotropy representation

Let g be a Lie algebra, and h g a subalgebra. The isotropy representation of h relative to g is the naturally dened action of h on the quotient vector space g/h. Here is a synopsis of the technical details. As is customary, we will use b + h, b g to denote the coset elements of g/h. Let a h be given. Since h is invariant with respect to adg(a), the adjoint action factors through the quotient to give a well dened endomorphism of g/h. The action is given by b + h [a, b] + h, This is the action alluded to in the rst paragraph. Version: 3 Owner: rmilson Author(s): rmilson 1165 b g.

Chapter 277 17B15 Representations, analytic theory277.1 invariant form (Lie algebras)

Let V be a representation of a Lie algebra g over a eld k . Then a bilinear form B : V V k is invariant if B (Xv, w ) + B (v, Xw ) = 0. for all X g, v, w V . This criterion seems a little odd, but in the context of Lie algebras, : V V given by v B (, v ) is equivariant if and it makes sense. For example, the map B only if B is an invariant form. Version: 2 Owner: bwebste Author(s): bwebste

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Chapter 278 17B20 Simple, semisimple, reductive (super)algebras (roots)278.1 Borel subalgebra

Let g be a semi-simple Lie group, h a Cartan subalgebra, R the associated root system and R+ R a set of positive roots. We have a root decomposition into the Cartan subalgebra and the root spaces g g=h gR

.

Now let b be the direct sum of the Cartan subalgebra and the positive root spaces. This is called a Borel subalgebra. b=h R+

g .

Version: 2 Owner: bwebste Author(s): bwebste

278.2

Borel subgroup

Let G be a complex semi-simple Lie group. Then any maximal solvable subgroup B G is called a Borel subgroup. All Borel subgroups of a given group are conjugate. Any Borel group is connected and equal to its own normalizer, and contains a unique Cartan subgroup. The intersection of B with a maximal compact subgroup K of G is the maximal torus of K . If G = SLn C, then the standard Borel subgroup is the set of upper triangular matrices. 1167

Version: 2 Owner: bwebste Author(s): bwebste

278.3

Cartan matrix

Let R E be a reduced root system, with E a euclidean vector space, with inner product (, ), and let = {1 , , n } be a base of this root system. Then the Cartan matrix of the root system is the matrix 2(i , j ) Ci,j = . (i , i ) The Cartan matrix uniquely determines the root system, and is unique up to simultaneous permutation of the rows and columns. It is also the basis change matrix from the basis of fundamental weights to the basis of simple roots in E . Version: 1 Owner: bwebste Author(s): bwebste

278.4

Cartan subalgebra

Let g be a Lie algebra. Then a Cartan subalgebra is a maximal subalgebra of g which is selfnormalizing, that is, if [g, h] h for all h h, then g h as well. Any Cartan subalgebra h is nilpotent, and if g is semi-simple, it is abelian. All Cartan subalgebras of a Lie algebra are conjugate by the adjoint action of any Lie group with algebra g. Version: 3 Owner: bwebste Author(s): bwebste

278.5

Cartans criterion

A Lie algebra g is semi-simple if and only if its Killing form Bg is nondegenerate. Version: 2 Owner: bwebste Author(s): bwebste

278.6

Casimir operator

Let g be a semisimple Lie algebra, and let (, ) denote the Killing form. If {gi} is a basis of g, then there is a dual basis {g i} with respect to the Killing form, i.e., (gi , g j ) = ij . Consider the element = gi g i of the universal enveloping algebra of g. This element, called the Casimir operator is central in the enveloping algebra, and thus commutes with the g action on any representation. 1168

Version: 2 Owner: bwebste Author(s): bwebste

278.7

Dynkin diagram

Dynkin diagrams are a combinatorial way of representing the imformation in a root system. Their primary advantage is that they are easier to write down, remember, and analyze than explicit representations of a root system. They are an important tool in the classication of simple Lie algebras. Given a reduced root system R E , with E an inner-product space, choose a base or simple roots (or equivalently, a set of positive roots R+ ). The Dynkin diagram associated to R is a graph whose vertices are . If i and j are distinct elements of the root system, we 4(i ,j )2 add mij = (i ,i)( lines between them. This number is obivously positive, and an integer j ,j ) since it is the product of 2 quantities that the axioms of a root system require to be integers. By the Cauchy-Schwartz inequality, and the fact that simple roots are never anti-parallel (they are all strictly contained in some half space), mij {0, 1, 2, 3}. Thus Dynkin diagrams are nite graphs, with single, double or triple edges. Fact, the criteria are much stronger than this: if the multiple edges are counted as single edges, all Dynkin diagrams are trees, and have at most one multiple edge. In fact, all Dynkin diagrams fall into 4 innite families, and 5 exceptional cases, in exact parallel to the classication of simple Lie algebras. (Does anyone have good Dynkin diagram pictures? Id love to put some up, but am decidedly lacking.) Version: 1 Owner: bwebste Author(s): bwebste

278.8

Verma module

Let g be a semi-simple Lie algebra, h a Cartan subalgebra, and b a Borel subalgebra. Let F for a weight h be the 1-d dimensional b module on which h acts by multiplication by , and the positive root spaces act trivially. Now, the Verma module M of the weight is the g module M = F U(b) U(g). This is an innite dimensional representation, and it has a very important property: If V is any representation with highest weight , there is a surjective homomorphism M V . That is, all representations with highest weight are quotients of M . Also, M has a unique maximal submodule, so there is a unique irreducible representation with highest weight . Version: 1 Owner: bwebste Author(s): bwebste 1169

278.9

Weyl chamber

If R E is a root system, with E a euclidean vector space, and R+ is a set of positive roots, then the positive Weyl chamber is the set C = {e E |(e, ) 0 R+ }.

The interior of C is a fundamental domain for the action of the Weyl group on E . The image w (C) of C under the any element of the Weyl group is called a Weyl chamber. The Weyl group W acts simply transitively on the set of Weyl chambers. A weight which lies inside the positive Weyl chamber is called dominant Version: 2 Owner: bwebste Author(s): bwebste

278.10

Weyl group

The Weyl group WR of a root system R E , where E is a euclidean vector space, is the subgroup of GL(E ) generated by reection in the hyperplanes perpendicular to the roots. The map of reection in a root is given by r (v ) = v 2 . The Weyl group is generated by reections in the simple roots for any choice of a set of positive roots. There is a well-dened length function : WR Z, where (w ) is the minimal number of reections in simple roots that w can be written as. This is also the number of positive roots that w takes to negative roots. Version: 1 Owner: bwebste Author(s): bwebste (v, ) (, )

278.11

Weyls theorem

Let g be a nite dimensional semi-simple Lie algebra. Then any nite dimensional representation of g is completely reducible. Version: 1 Owner: bwebste Author(s): bwebste

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278.12

classication of nite-dimensional representations of semi-simple Lie algebras

If g is a semi-simple Lie algebra, then we say that an irreducible representation V has highest weight , if there is a vector v V , the weight space of , such that Xv = 0 for X in any positive root space, and v is called a highest vector, or vector of highest weight. There is a unique (up to isomorphism) irreducible nite dimensional representation of g with highest weight for any dominant weight W , where W is the weight lattice of g, and every irreducible representation of g is of this type. Version: 1 Owner: bwebste Author(s): bwebste

278.13

cohomology of semi-simple Lie algebras

There are some important facts that make the cohomology of semi-simple Lie algebras easier to deal with than general Lie algebra cohomology. In particular, there are a number of vanishing theorems. First of all, let g be a nite-dimensional, semi-simple Lie algebra over C. Theorem. Let M be an irreducible representation of g. Then H n (g, M ) = 0 for all n. Whiteheads lemmata. Let M be any representation of g, then H 1 (g, M ) = H 2 (g, M ) = 0. Whiteheads lemmata lead to two very important results. From the vanishing of H 1 , we can derive Weyls theorem, the fact that representations of semi-simple Lie algebras are completely reducible, since extensions of M by N are classied by H 1 (g, HomMN ). And from the vanishing of H 2 , we obtain Levis theorem, which states that every Lie algebra is a split extension of a semi-simple algebra by a solvable algebra since H 2 (g, M ) classies extensions of g by M with a specied action of g on M . Version: 2 Owner: bwebste Author(s): bwebste

278.14

nilpotent cone

Let g be a nite dimensional semisimple Lie algebra. Then the nilpotent cone N of g is set of elements which act nilpotently on all representations of g. This is a irreducible subvariety of g (considered as a k -vector space), which is invariant under the adjoint action of G on g (here G is the adjoint group associated to g).

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Version: 3 Owner: bwebste Author(s): bwebste

278.15

parabolic subgroup

Let G be a complex semi-simple Lie group. Then any subgroup P of G containg a Borel subgroup B is called parabolic. Parabolics are classied in the following manner. Let g be the Lie algebra of G, h the unique Cartan subalgebra contained in b, the algebra of B , R the set of roots corresponding to this choice of Cartan, and R+ the set of positive roots whose root spaces are contained in b and let p be the Lie algebra of P . Then there exists a unique subset P of , the base of simple roots associated to this choice of positive roots, such that {b, g }P generates p. In other words, parabolics containing a single Borel subgroup are classied by subsets of the Dynkin diagram, with the empty set corresponding to the Borel, and the whole graph corresponding to the group G. Version: 1 Owner: bwebste Author(s): bwebste

278.16

pictures of Dynkin diagrams

Here is a complete list of connected Dynkin diagrams. In general if the name of a diagram has n as a subscript then there are n dots in the diagram. There are four innite series that correspond to classical complex (that is over C) simple Lie algebras. No pan intended. An , for n 1 represents the simple complex Lie algebra sln+1 : A1 A2 A3 An Bn , for n Cn , for n

1 represents the simple complex Lie algebra so2n+1 : 1 represents the simple complex Lie algebra sp2n :

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B1 B2 B3 Bn C1 C2 C3 Cn Dn , for n

3 represents the simple complex Lie algebra so2n :

D3

D4

D5

Dn

And then there are the exceptional cases that come in nite families. The corresponding Lie algebras are usually called by the name of the diagram.

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There is the E series that has three members: E6 which represents a 78dimensional Lie algebra, E7 which represents a 133dimensional Lie algebra, and E8 which represents a 248dimensional Lie algebra.

E6

E7

E8

There is the F4 diagram which represents a 52dimensional complex simple Lie algebra:

F4

And nally there is G2 that represents a 14dimensional Lie algebra. G2

A1 = B1 = C1 sl2 = so3 = sp2 . B2 = C2 so5 = sp4 . A3 = D3 sl4 = so6 . 1174

Notice the low dimensional coincidences:

which reects the exceptional isomorphisms

Also reecting the isomorphism And, reecting

Remark 1. Often in the literature the listing of Dynkin diagrams is arranged so that there are no intersections between dierent families. However by allowing intersections one gets a graphical representation of the low degree isomorphisms. In the same vein there is a graphical representation of the isomorphism so4 = sl2 sl2 .

Namely, if not for the requirement that the families consist of connected diagrams, one could start the D family with

D2

which consists of two disjoint copies of A2 .

Version: 9 Owner: Dr Absentius Author(s): Dr Absentius

278.17

positive root

If R E is a root system, with E a euclidean vector space, then a subset R+ R is called a set of positive roots if there is a vector v E such that (, v ) > 0 if R+ , and (, v ) < 0 if R\R+ . roots which are not positive are called negative. Since is negative exactly when is positive, exactly half the roots must be positive. Version: 2 Owner: bwebste Author(s): bwebste

278.18

rank

Let lg be a nite dimensional Lie algebra. One can show that all Cartan subalgebras h lg have the same dimension. The rank of lg is dened to be this dimension. Version: 5 Owner: rmilson Author(s): rmilson

278.19

root lattice

If R E is a root system, and E a euclidean vector space, then the root lattice R of R is the subset of E generated by R as an abelian group. In fact, this group is free on the simple roots, and is thus a full sublattice of E . 1175

Version: 1 Owner: bwebste Author(s): bwebste

278.20

root system

Root systems are sets of vectors in a Euclidean space which are used classify simple Lie algebras, and to understand their representation theory, and also in the theory of reection groups. Axiomatically, an (abstract) root system R is a set of vectors in a euclidean vector space E with inner product (, ), such that: 1. R spans the vector space E . 2. if R, then reection in the hyperplane orthogonal to preserves R.(, ) 3. if , R, then 2 ( is an integer. ,)

Axiom 3 is sometimes dropped when dealing with reection groups, but it is necessary for the root systems which arise in connection with Lie algebras. Additionally, a root system is called reduced if for all R, if k R, then k = 1. We call a root system indecomposable if there is no proper subset R R such that every vector in R is orthogonal to R. Root systems arise in the classication of semi-simple Lie algebras in the following manner: If g is a semi-simple complex Lie algebra, then one can choose a maximal self-normalizing subalgebra of g (alternatively, this is the commutant of an element with commutant of minimal dimension), called a Cartan subalgebra, traditionally denote h. These act on g by the adjoint action by diagonalizable linear maps. Since these maps all commute, they are all simultaneously diagonalizable. The simultaneous eigenspaces of this action are called root spaces, and the decomposition of g into h and the root spaces is calle