PURELY INSEPARABLE, MODULAR EXTENSIONS …...PURELY INSEPARABLE, MODULAR EXTENSIONS OF UNBOUNDED...

TRANSACTIONS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 176, February 1973

PURELY INSEPARABLE, MODULAR EXTENSIONS

OF UNBOUNDED EXPONENT

BY

LINDA ALMGREN KIME

ABSTRACT. Let K be a purely inseparable extension of a field k of char-

acteristic p * 0. Sweedler has shown in [2, p. 4031 that if K over k is of

finite exponent, then 7v is modular over k if and only if K can be written as the

tensor product of simple extensions of k. This paper grew out of an attempt to

find an analogue to this theorem if K is of unbounded exponent over k. The

definition of a simple extension is extended to include extensions of the form

k [x, xl/i>, x^-IP , • • •][%1/P°Û]. If K is the tensor product of simple exten-

sions, then K is modular. The converse, however, is not true, as several

counterexamples in §4 illustrate. Even if we restrict [k : r ] < oo, the converse

is still shown to be false.

Given 7Í over k modular, we construct a field \\c_ kKp ®M (= Q) that

always imbeds in K where M is the tensor product of simple extensions in the

old sense. In general K * Q. For K to be the tensor product of simple exten-

sions, we need K = Q, andf)°° kKpl = k(C\°°_,Kpl). If for some finite ,V, kKpN

ftN + 1= kKp , then we have (by Theorem 11) that K = Q. This finiteness condition

guarantees that M is of finite exponent. Should ff?, kKp = k, then we would

have the condition of Sweedler's original theorem.

The counterexamples in §4 will hopefully be useful to others interested in

unbounded exponent extensions. Of more general interest are two side theorems

on modularity. These state that any purely inseparable field extension has a

unique minimal modular closure, and that the intersection of modular extensions

is again modular.

1. Modular extensions. All fields under consideration will be purely insepa-

rable extensions of a common ground field k. They are to be viewed as contained

in a common algebraically closed field (e.g. the algebraic closure of k). If two

fields are said to be linearly disjoint, we mean that they are linearly disjoint over

their intersection. A field extension K of k is said to be modular if Kp and k

ate linearly disjoint for all positive i. K is of exponent n over k if ra is the

minimal positive number such that Kp C k. If no such ra exists, then K is of

unbounded or infinite exponent over k. Unless otherwise indicated, the intersec-

tion and unions throughout are taken from i (or ra) = 1 to °°.

Received by the editors February 14, 1972.

AMS (MOS) subject classifications (1970). Primary 12F15-

Key words and phrases. Purely inseparable, modular, modular closure, infinite or

unbounded exponent, tensor product of simple extensions.

Copyright © 1973, American Mathematical Society

335

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

336 L. A. KIME [February

Lemma 1. Let {Mj be a set of fields such that, for any Ma.,, Ma £ \Mj,

there exists an Mn £ \Mn] such that both M„ and M„ lie in M„ . If each M„a.-, y a.' a, a2 clt ' a

and Q are linearly disjoint, then KJaMa and 0 are linearly disjoint.

Proof. Given any set |x,,•«•, x ¡ C Ua Ma, there exists an a. such that

\xv • ■ • , xn]C M £ \Ma]. If ixp • • - , x ¡ is linearly dependent over 0, then by

the linear disjointness of Ma and 0, \x . ■ . , x [ is linearly dependent over

Ma n Q. Hence fx,,..., x f is linear dependent over \JaMaC\ Q (D Ma n Q).

Q.E.D.

Lemma 2. Let \Ma] be as in Lemma 1. If each Ma is modular over k, then

Ua ^a zs m°dulaT over k.

Proof. We must show that (|Ja*la)i' (= Ua ^a ) anc^ ^ are linearly disjoint

for all « > 1. Given any (x., ••., x ] C (Ja Mpa there exists an aQ such that

\xv ... , xs]C /M£o". if {xv. .., xj is linearly independent over k n (\J a Mpa"),

it is linearly independent over k O Mpa . The modularity of M over k guaran-

tees that \xx, • • •, x ] is linearly independent over k. Q.E.D.

Proposition 3. K is modular over k «=> k p O K is modular over k for all

positive n.

Proof. Assume K modular over k. We want to show that (k /p O K)p and

k ate linearly disjoint for all positive i.

Case 1. n > i. (k1/p" n K)p' = kl/p"~' O Kp!. If jxp..., xj C /fe is

linearly independent over Kp" n k = (k p n Kp ) n k, by the modularity of K

over k, \x .,•••, x ] is linearly independent over Kp , hence over k ' p Ci Kp .

Case 2. « < z. (*">" n K)"'' = kpi~" n K^'. Since kpÍ~" O Kp¿ n zfe =

/fep! "n K''1, (^^^"n 7C)i' and k are trivially linearly dis joint.

Conversely, if k p n K is modular over k for any positive «, then

\J(kl/p" n K) = K is modular over ¿ by Lemma 2. Q.E.D.

Sweedler has shown in [2, p. 4081 that any purely inseparable extension 7<

over k oí finite exponent is contained in a unique minimal field extension L,

where L is modular over k. L is called the modular closure of K over k. We

now extend this definition to the infinite exponent case.

Theorem 4. Let K be a purely inseparable field extension of k.

(1) There exists a unique minimal field extension L D K where L is modular

over k.

(2) L is purely inseparable over k.

(3) If K is of exponent «, then L has exponent « over k.


1973] PURELY INSEPARABLE, MODULAR EXTENSIONS 337

Proof. Since kl/p' O K has exponent p* over k, by [2, p. 408] kup' n K

has a corresponding modular closure L. satisfying (1)—(3). The minimality of

each L . guarantees that L. CL, CL, C ••• . By Lemma 2, \JL . = L is modular

over k. L clearly contains K, since kl/p' Í1KCL. and K = \J(kUpt n K).

Any field extension of K that is modular over k must contain all the 7. . (by the

minimality of each L .). Thus L must be the unique minimal such field. Q.E.D.

Naturally L is called the modular closure of 7< over k.

Corollary 5. Let \Ma\ be a set of fields such that each Ma is modular over

a common ground field k. Then Ç\aMa is modular over k.

Proof. Let N be the modular closure of Mae/l Ma. By minimality, N C Ma

for all a e A. Hence NCf)aMa. Q.E.D.

2. A structure theorem for field extensions of unbounded exponent.

Lemma.6, If K over k is modular, then k(kl/p' n Kp) = k)/p' O k.Kp for

any positive i.

Proof. K = \J(kupi O K)-K'=UU1/<,IVl Kp)=*kKp =\Jk(kx/pi C\ Kp).

If \xn\ is a p-basis for kl/p" O Kp over A1/p"~ ' Pi Kp, \xj is ^-indepen-

dent over k1/pn~l n A/<p <=» S*£"_1! is p-independent over An kp"~l Kp".

\xp"~ j (C (k p n Kp") C Kp") is by assumption p-independent over

U1'^*" : O KO""" '^nK^.soby the linear disjointness of k and Kp",

\xPn | is p-independent over k, hence p-independent over kC\kf Kp . So \x \ is p-

independent over kl/p"~lnkKp. We also have that \xp\c(knKp)[\xl\,\x2\,. ■ ■ ,\xn__l\]

C ^Lix,!,---,ixn_]il where each ¡x.( isa p-basis for k1/p'r\Kp over A17*'"1 O Kp.

AK* - \Jk(kUp' n K*) - = A[|x,l, |x2¡,.. ]

where each |x.| £ A1^' o AK" is p-independent over kU'p'~ O kKp,

\xp\ c k[\x x\, \x2\,..., ix;._,n

=»each ix.i is a p-basis for A1''"' n AK" over kUpl~l C\ kK"

=>k1/pl n *K» = *[{*,!, íx2¡, •••, «*fn

= it [(A n **)[{*, |,..., \x.\] = k[kUpl n Kp\. Q.E.D.

I'roposition 7. 7/ K over A /s zz modular extension then kKp is modular

over A for any positive ra.

1 ' ti '+ IProof. Let iy^, ly2!, •■-,!)'•+[! be a modular basis for A p n K over

A, i.e., kl/pi + l O K = A[!yj|l ® A[!y2!l® • •• ® A[{y,+1ll where, for each y. £

|y.|, y. has exponent / over A. Then



kl/pi C\kKp = k[kUpi n Kp] (by Lemma 6)

= *tU1/í,í+1n K)p] = k[\yp],...,\yp+l]]

= k[\yp]]®...®k[\yp+A]

=> k p Cl k Kp is modular over & for any z > 0,

=» (by Proposition 3) that k Kp is modular over k.

Now replacing K by kKp, we have that k(kKp)p = /fezK^2 is modular over k,

hence in general that kKp" is modular over ¿. Q.E.D.

Lemma 8. If K is relatively perfect over k, i.e. K = kKp - kKp' for all posi-

tive i, and K is modular over k, then k1/p O K = kl/p n kKp" = k[kl/p n K»"]

/or a«y positive n.

.

Proof. Since K is modular over k, we have from Lemma 6 that k^^pl n kKp =

k[k1/pl n Kp] for any i > 0. Using this and the fact that K = kKp" fot any «>0,

we have that

kvp n K = /fe17i,n akâU17» n k»1

= k[(kp)up2 n a»k»21 = A . *'[(**)"^ n a»V31 = k[k1/p n K"2].

Applying this argument repeatedly we get that kl/p n K = kl/p n £K?* =

AUÔn K*"]. Q.E.D.

Proposition 9. If K over k is a modular extension then the relatively per-

fect subfield (\ kKp is modular over k. If \x\ is a p-basis for k p H

(0 kKpi) over k1/p" n (fï AKP¿), then \xpn] is a p-basis for kl/p n (flAK»*) oz,er k.

Proof. Apply Corollary 5 and Proposition 7 to obtain modularity. For sim-

plicity denote C\kKpl by L. The modularity of L over k =» ixp 1 is p-indepen-

dent over A.

A17? n L = kUp n 4L*" = AtA17" n L»"l (by Lemma 6)

..AU1'*"*1 n Ll»" = AftA17»" n DM»"

= A(AnL»")[lx»"|] = A[{x»"l]. Q.E.D.

We note that a relatively perfect field extension K over k (K = kK = II îC )

is not necessarily modular, as the following example shows.

Let k = Z (x», y», z» ), x, y, z indeterminates. Denote any set

\w,w1/p,wl/p2,..,\ byW1/p°°. Set K = *[*l""\ (** + y)1""]. 41/<>2 C K

over /fe has diagram

z, xz + y

(cf. [2, p. 4021) -* fc17»2 n K over k is not a modularextension. By Proposition

3, K is not modular over k.


19731 PURELY INSEPARABLE, MODULAR EXTENSIONS 339

Lemma 10. If A, B are field extensions of k, such that A is modular over

A and A p O A, B are linearly disjoint over k, then A, B are linearly disjoint

over k.

Proof. If ix i is a p-basis for Al7p* O A over A17p" n A, then jxj is

p-independent over (kl/pn O A)(b). Otherwise \xp"\ (C kl/p n A), which is p-

independent over A, would be p-dependent over (A O Ap )(BP ) C B, contradicting

the linear disjointness of A p D A and B. So the basis for A over A consisting

of monomials of finite products of the x . is a basis for A • B over B =* A ®, B

~ AB =» A, B are linearly disjoint over A. Q.E.D.

Theorem 11. Aray modular field extension K over k, where for some finite

N, kKp' = kKp , is isomorphic to (I AKP ® M where M is a modular subfield

of K of finite exponent.

Proof. Let ixjj be a p-basis for kX/p Ci K over kl/p n K. By modularity

\xpA is p-independent/A. Let A , be a completion of {x^j to a p-basis for

A p O K over A. A is not unique, but the choice of A. is independent of the

choice of [x |, since A. is precisely a p-basis for A pn K over k[k p C\ Kp\

(which equals kl,p n AKP by Lemma 6). Hence Al7p n K = (A17" n AKP) ®

A[Aj] => AKP® A[Aj1 c-, k (by Lemma 10).

Let \x2\ be a p-basis for A17p n K over kl/p O K. jx"j is p-independent

over A p n K. Choose A as any completion of \xp\ to a p-basis for A p (~) K

over Al7p O K, i.e., a p-basis for kUp2 O K over (A1/p n K)(A17p2 n Kp).

A[ixf }1 = AU17" nKp2l = Al7pn kKp2,so kupn K = (A17" n kKp2) ® k[Ap]

® A[A,1 =»AKp2® A[A21 ® A[Aj] c_ k.

In general, we let A be a p-basis for A p H K over

(¿i/i>*- 1 n K)(A17p* n Kp)" Al7pnK= (Al7pn AKP") ® A[Af " 'l ® • • • ® AU/!

-» AKP" ® A [A J ® • ■ • ® A [A J c» K. So for any finite ra, AKP" and

A[Aj U •• • U A 1 are linearly disjoint over A => fl AKP' and A[Aj u • • • U A\

ate linearly disjoint over A. By Lemma 1, (J A[Aj U ... ij A J = k[A ^ A^J.-A

and H AKP' are linearly disjoint over A, i.e., D AKP'® A[Aj u A2 u • • •]

imbeds in K.ftN ftN + 1

By assumption, there exists a finite TV such AKP = AKP , so A . = 0 tot

all i > N. Set Al = A[Aj U . . . u Avl. M is modular since A [A j u • • • U A N] =

A[A1]®A[A21® ... ®A[AN1 (cf. §3). We assert that K= f|AKp'® M. Let

\xv ■.. , xs\ be a p-basis for A1/p'+1 n K over Al7p! n K, where z > N. Then

A1/pn K= A[xp¿, ..., xp!l®A[A£/V-1l® ...® k[AA\, where *»',..., *J* are

p-independent over A. t[xf',•••, x*'l = A[A17p n Kp'] = Al7p O AKP* = A17p n

(M AKP ) (since i > N). Proposition 9 says that we may assume that {*,, ••., xs\

CkUpi + 1n (f|AKp¿). Q.E.D.



Remark 12. Either condition that [A p O K: k] < os or [A : kp] < oo implies

that, for some N, kKpN = kKpN + l. Since if [kUp n K : k] < o. (note [k:kp] < œ

implies that [k1/p O K : k] < oo), then ¿N = 0 for some /V => kKpN = kKpN + \

With only the restriction that K over A is modular we always have that

I 1 kKp ® A [A j u A2 u • • •! imbeds in K. Since there may be infinitely many non-

empty A .'s, A [A j ij A2 u • •.] is not, in general, of finite exponent. We can

describe k[A x u A2 u • . •] as being of locally finite exponent.

Definition. K/k, a purely inseparable extension, is said to be of locally

finite exponent if, for any field L, A C L C K, where [k p n L : k] < oo, then L

has finite exponent over A.

A[A j u A u • • •] is easily seen to be of locally finite exponent. Given any

intermediate field L, such that [k p C\ L : k] < oo, there must exist a finite «2

such that LCA[A1uA2u-,-U^ uAp + .u Ap +. . . .] => L is of exponent n

over A.

Though we have C\kKp' ® A [A . U A2 U •.• <] C-» K, counterexample (d) in §4

shows that we need not have (~\ kKp' ® A[A j u A2 U . . .] = K. We do have the

following:

Proposition 13. Given K over k modular, set Q= (\ kKp ® k[A .u Aaj...].

Then K is modular and relatively perfect over Q (i.e., K = (~)QKP ).

Proof. Let \xj be a p-basis for kl/p" O Q over k1/p" l n Q. \xj is p-

independent over A p n K (since Q is a modular subfield of 7<). If \yn] is

a completion of {x | toa p-basis for A p C\ K over A p OK, then iy^S is

a possible choice for iy„_1l for all positive « and hence \y j | u iy2| u íy?i u •••

is a modular basis for K over Q. Q.E.D.

Counterexample (e) in §4 shows that Theorem 11 cannot be refined by

"tensoring off" k[C\ Kp'] from f]kKp'.

3. A generalization of simple extensions and a tensor product theorem.

Sweedler has shown in [2, p. 403] that any purely inseparable extension K over

A of finite exponent is modular if and only if it is isomorphic to the tensor pro-

duct (over A) of simple extensions of A. In this section we will study the problem

as to whether this theorem has an analogue in the infinite exponent case. We must

first extend our definition of a simple extension to include those of the form

k[x,xp,x p ,•••]= A [x p ]. We may now ask whether a modular extension

of infinite exponent is modular if and only if it is isomorphic to the tensor product

of simple extensions of A.

The following theorem gives us the implication in one direction.



Theorem 14. If K is isomorphic to the tensor product of simple extensions

(of A), then K is modular over A.

Proof. By assumption K Ot (®a k[xla/p°°]) ® (®ßk[yß]). To prove that K

is modular we need to show that Kp and A are linearly disjoint for all positive

ra. Given any \zv ..., *f| C K, J*,,.... *,! C [«g>*=1 k[x\/pu]) ® (®J=,*[yy])](=ß)

for some finite 5, r and «. Q is modular since it is the tensor product of simple

(in the old sense) extensions. Hence if \zp , . . . , zp j C Kp is linearly depen-

dent over A, then, by the linear disjointness of Qp and A, \zp , • •., zp j is

linearly dependent over Qp" n A, and hence over Kp" D A D ÖP" n A). Thus Kp"

and fe are linearly disjoint. Q.E.D.

It is not true, however, that if K is modular, that it is the tensor product of

simple extensions, as counterexamples (a) and (e) of §4 demonstrate. Counter-

examples (b) and (c) show that even with the added restriction that [A : Ap] < t»,

K may still not be the tensor product of simple extensions.

If we restrict K over A such that AKP = AKP for some N

(or [A17p n K:A] < « or [A :kp] < <*>) by Theorem 11 we know that K » flAKp'

® M, where M is modular (and of finite exponent). Hence M is the tensor product

of simple extensions. Since f IAKP is modular, the following structure theorem

shows that if (l AKP = A[flKp ] then (1 AKP is the tensor product of simple

extensions. Thus K will be the tensor product of simple extensions.

Theorem 15. Given a field extension K over k, if A[f|Kp 1 is modular, then

A[flKp!] S* <g>aA[zi7p00l for some {zJ C C\KP'.

Proof. Let \zJ C f|Kp' be a p-basis for AI/pn A[fÏKp!l over A. Then

\zla/p"\ must be p-independent over A17p"o A[OKp!], If \zla/p"\ were not a

p-basis for Al7p" + 1 n A[f|Kpi] over Al7p"n A[fÏKp'l, then the modularity of

A[f|Kp!] would imply that \za\ was not a p-basis for k p C\ A[IIKP ] over A.

Hence AI7p"n A[flKp'l » ®ak[z)¿pn~ M - A[flKpîl * ®ak[z\/p°°]. Q.E.D.

Theorem 16. Given K over A a modular extension, K is the tensor product

of simple extensions of A if and only if (in the terminology of Theorem 11) K =

ÓaKp¿® A[A,u A2 U---1 and f| AKP¿ = A[D Kp'].

Proof. "«=". The remarks preceding Theorem 15 showed that if K S

A[flKp'l ® A[Aj U Au •. .], then K is the tensor product of simple extensions.

"-»". Assume that K S «8>ak[xla/p°°]) ® (^^Aty^l). Since each yß has

finite exponent over A, A[DKp!l C flAKp! s <8>a^^Í/p°°] c ^[riKp,l, i.e.

riAKpî'= AtflK"'].


r342 L. A. KIME [February

By Proposition 13 we have that

k= n(r\kKpi®k[A1u a2u...i) ((®A[x^n) ® (<s>k[yJ\y

= /Dak^'saUjU a2u...1) (®k[xvp°°]\

= fi A/C»'® A[Aj U A2 u ...1i

(since ®aA[x^»~] C A[f|¿ Kpl] C f\ Ail*1). Q.E.D.

We have shown (Remark 12) that if Afv^ = AT^' for some finite TV, or

[k1/pn K:k] <oo, or [A: A»] < oo, then we will have K^C\kKp' ® k[AxU A2U...].

The following theorem gives some conditions under which M A7<^ = k[f\Kp ].

Theorem 17. Let K over k be a modular extension.

(1) If K over Ç\Kpl is finitely generated as a field, then DkKp' = AtflK^l.

(2) If there exists a generating set for K over Kp that generates K over

C\Kpi, then C\kKpi = A[flKpi].

Proof. (1) Assume K = (f|. Kp ){x ,,--•, x ). There exists a finite n such

that xpn £ k, i=l,..., m. Hence Kpl = (f| - Kp') (xpl ,.. ., xp!) C k(f) ■ Kp') for

any / > « => AT^7 = A(f\ Kpl) for any ; > «. Therefore f| . vfe/C^* = A(f|¿ K»').

(2) Assume that \x] is a generating set for K over 7<p such that K =

(C\{ Kpi)(\x]). For any positive «, kKp" = Atifl,■ Kpi) i\xp"]]]. Hence if as

nBAK»", a EOjkKCii Kpi)[{xp"]]]) = kin, Kpi]. Q.E.D.

4. Some counterexamples, (a) The following counterexample illustrates that

a modular field extension of unbounded exponent need not be a tensor product of

simple extensions. In particular we will construct a modular relatively perfect

field extension K over A (i.e. K = C\kKp'), where K ¿ k[f\Kpl].

Let Z be the prime field of characteristic p, and x., x2, x^,. • • be

indeterminates.

Let

A = Z ÂX y X2, Xj, • • •)

A^n K=k[x\/p]

kl/p2 O K = A[x^2x^]

Ä^"n K=A[x^"x^"-1...x^l



It is easy to verify that K is of the form ÍI AKP . We claim that K 4 A[f|Kp 1,

in particular that MKP C A.

Assume not. Then there exists a y £ ÍIKP where y £ A - Ap (set difference).

Then y will be in all of the following fields:

A n KP = Z (x,, xp, xp,...)p i i y

A nKp2 =Zp(xlXP, xp, xf, xp2,...)

k nKp* =ZAx,x* ...xp'-\ *f , xpl, xp',...)ft 1 2 i 1 2 3 '

fl(An Kpl)CZp(xv x\, xp2,...),soy eZ^x,, xp,in — 1

, + 1finite ra. y £ A O KP"T1 ..t>n~KP" ,P"y £ ZAx, P

^■1A2

v" X ., X , , X_■*„ n + 1 1 2

) (= F) for some

n + 1 .. x"" + 1)' n + 1 '

,n + l(= G).. We claim that FnG = Zp(xpxp¿ ... xp", xpn, xf

y eFn GC Ap, which contradicts our hypothesis that y £ A - Ap.

,n + l). So

Consider the following diagram for ZAx., xp ,.,n-\

exte nsion of Z. (xp , xp , • • • ,xpô 1 2

n + 1 i + l

Cn + 1 ).

XP" ) as a fieldn + 1

(X1X2

(X¡xp

„«-1). x-e

*"- 1 „ A<p )p.xp

n 1

We want to describe F and G using the above diagram. Recall that the,n + l ,n + l j« + I

n + 1). If we deletedground field for this diagram is Z (xp.

xn + l ^rom tne diagram an(^ restricted the ground field to Z (xp , xp , ... ,xp ),

the above would be a diagram for F.



The elements of G ate those in the span of monomials in

((x{x* •••*Pnn~l)>**n)i and ((x,x»...xf-V)' where i = 0,. . ., p - 1,

j4o,...,p»-l-i.

Hence F H G can be described as polynomials in (x,x^ • • • xp ) with

coefficients in Z (x^ , x% . . . xp ), i.e.p 1 ¿ n '

FOG = Z (xpxp2 ...xp", xpn, xp"+1,..,,xpn + l). Q.E.D.p l 2 n 12' ' r? ^-

(b) The following argument shows that even with the restriction that

[A :kp] < oo, a modular extension may not always be a tensor product of simple

extensions.

Let A be any field of characteristic p / 0 such that [A : kp] > p. A by assump-

tion is not perfect, hence in particular |A| is infinite. Assume that |A| = N0, and

let L be the perfect closure of A.

A well-known theorem from Galois theory states that, given any field exten-

sion K over A, 7< is primitive over A <=> there are a finite number of intermediate

fields. In the case when K is purely inseparable over A and of exponent one,

there are a finite number of intermediate fields <=» [K : k] = p.

Lemma 18. Any purely inseparable extension K over A of the form (1 kKp ,

where [k1/p" + lD K : kl/p" D 7<] = p, is modular.

Proof. Any such 7< is of the form A[x., x2 ,• • •] where x . has exponent i

over A and x» e A[x.,..., x. ,]. A p n K = A[x,, x„ • •., x 1 = A[x ] isz 1' ' z— 1 1' 2' n n

modular by [2, p. 4031. By Proposition 3, Tí is modular over A. Q.E.D.

Lemma 19. There are at most |A|.= N0 purely inseparable modular extensions

K of the form k[C\Kp'], where [k1/p O K : k] = p (i.e., of the form k[z1/p°°]).

Proof. Trivial, since we may associate any field of the form A[z p ](z£k)

with z. Q.E.D.

Theorem 20. There exist 2 purely inseparable modular field extensions of

the form O W, where [k1/p n K : k] = p and [k:kp] > p.

Proof. The field extensions under consideration are of the form

Atxj, x x,...] where [A[x¿] : Al = pl and xp £ A[xp .. . , x._ A.

Given any Xj £(kl/p Ci L) - A, there exists a y £ (kl/p n L) - A such that

x. and y are p-independent over A (since [k:kp] > p). Consider Mj = A[xj p, y]

as a field extension of A[xj]. Since Mp C k[x A and [M : k[x A] = p , by an

earlier remark there are an infinite number of intermediate fields, all of which

(excepting A[x., y]) have exponent 2 over A. We associate with each such field

TV, an a such that TV = (Afx^Hal = A[a]. For any given a, [k[a] : k] = p2, and



if a 4 a!, k[x., a] = A[a) 4 k[d\ = A[xj, a]. Clearly the a's provide an infinite

number of candidates for an x

Having chosen an x we consider M, = A[xj, x2'p, y] as a field extension

of A[xj, %21. Mp2 C k[xj, x2], [M2 :A[xj, x2]] = p , so there exist an infinite num-

ber of intermediate fields of degree p over A[x xA and exponent 3 over A.

Each of these provides a candidate for x such that if x. 4 *?, then A[x.,x2, x,]

4 A[xj, x2, x'J.

We proceed in the above manner to choose an x ,, x_, etc. For each x. there

exist an infinite number of choices such that, if x. 4 x'., A[x,,- • •, x.] 4 k[x,,... , x'J.N0 K0 lit i \ i

Hence there exist at least NQ =2 choices for modular field extensions K/A

of the form f|AKpz, where [A1/p n K : A] = p and [A:AP] > p. Q.E.D.

(c) In the following example, we will explicitly construct a modular field

extension K over A, where K = f| AKP'^ A[fl Kp*] and [A : Ap] < <*,.

Let A = Z (x, y, Zj , z2 i • • •) where x, y, z,, z2, ■ . • are indetermi-

nates (note [A :AP1 = p ).

LetA1/p n K = A[xl7p]

k^p2n K = k[xup2 + ziyup]

A1/p3 nK = A[xI/p3 + z;7pyl7p2 + 22y17P]

A17p"n K = A[xl7pn + z;7p"-2yl7p"-1+... + Zn_1y1/p]

Assume that K = A[(lKp ]. Then there exists an a e k - kp such that a e

il Kp . a will lie in all of the following fields.

AnKp = Zp(x, yp, z\/p°°, z¥p°°,...)

A HKp2 = Z (x + zf yp, yp2, z}7*8*, z*7*0*, ...)

A n Kp¿ = Z (x + zp2yp + ... + zpi ,yp!_ \ yp\ z!/p~' ^1/p06 2l/p°

There exists a finite ra such that a £ Z (x, y, zj7p°°,. .. ,zl/p°°) (= F).

But a also £ A O Kp"+2 = Z (x + zp2yp + ... + zp"+2ypn+1 ypn+2 zl7p°° z17p°° ...)ft I ' n+1 ' ' ' ' i '2 '

(=G).

Consider the following diagram for Z (x, y, z!7p°°, z27p°°, . . .) as a fieldpx~' " "1 ' ~2

7I/p°° 7ft"+3 ,

Pv" " ' "1 '••• 'Zn ' Zn + I ' 2n+2 '"''extension of Z>p" + \ yp"+2, z |7*°",.. .', zI7p°°, zpn^,zx/p



"rz+1

rP"n + 1

„n+2

x-, zP yp

(x ~zp~yp

.." + ' *r,

'■l yp

yP )P

We would like to use the above diagram to describe F and G. If we deleted

the left-hand column (i.e. all pth roots of zp +. ) and restricted the ground field

to Z (xp , yp • zi > • " • > z ) the above would be a diagram for F.

G consists of the span of monomials in {((z^ + 1 ) ) ¡,

l(x + zÇ2y* + .. . + zPnn+lyP") + (zf+j2y^+1)]7' and ((x + zfy* + •• • + z»"+V»*

where z, /' = 0, • • • , p - 1, A = 0, • • •, p" - 1. Hence F O G can be described as

polynomials in (x + zp yp + • ■ • + zp yp ) with coefficients in

Zp(xpn + \ ypn+2, *}/»", • • • , z*70, i.e., F O C = Zp((x + zp2yp + ... + *»"">")»

Un ,«+2 ,/(,o. _ _ 2l/í,o. c ¿P ^ a e ¿P. Contradiction.* ' 1 n

(à) The next counterexample shows (using the terminology of Theorem 11)

that in general for a modular extension K over A, (] kKp ® M /= K.2 3

Set A = Z (x, ap, ap , ap , • • •) where x, a «-,... are indeterminates. Set

k1/p n K = A[x17^, «j, a», fl^2,...]

A1/p2n 7< = A[x1/i'2iz|/^ 8„ «„ 4, a»2,...]1 1" 2 3 4 '

A17"3 n^Atx17^}7"2^7", fl[, fl2- ay 4 <

7< is modular since

k'/pn n K= k[x^p\\/pn-1 ■■■ ̂ _pv av a2,..., an, apn + v ap2+2,...]

= Atx'^V/P""1 • • • a^fA ® Atfll ® A[a2l ® • • • ® A[anl ® A[a^ + 1l ® • • •

We will show that il kKp = A. Using the terminology of Theorem 11, we set



A.= \a.\. Hence we will have (f|AKp') ® A [A x u A2 u •••] = k[ay a ...] ¿ K.

Since it takes only one element to complete \av ap, ap , . ..\ to a p-basis

for A p Pi K over A, and since fl AKP is modular (by Proposition 9), if

MAKP j£ k, then |)AKP = A[xj, x2,. ..] where x{ has exponent i over A, and

x? £A[xj,..., x._jJ. Consider K as lying in the perfect closure L of A. %j £

A[x /p, a,, ap, • ■ • , apn~ ] for some finite ra, so1 2' ' n

xn,2£(kl/pn + 1n L)[X'/pn'2,a)/pn + \aYpn,...,ayp2}

= zypnn, aypn+\ ayp,...,ayp2, «n+1, a£+2, «&,•••> ua

We also have that

x +9£A17pr'+2n Kn + 2

= Z (x17p* + 2fl}7p" + 1fl,17p".,.aI7pV7¿ft 12 n n+1

UV «2''"' *„+2' flf,+3' fln+4'--° (=G)-

Consider the following diagram of Z(xl/pn+2, a!7p"+1, fli7p",. • •, fl1/p2, fl17?,° ° p 1 2 ' ' n n + 1

fl +T, ö +-,,•••) over Z (x, a,, a...... ,fl ... flp,-,, flp.,,•••).n+2 n + 3' p 1 2' ' n + 1 n + 2 n + 3'

,'V

uiV2ai/»-+I ,iV ,! '(,"+'

(,}'»n+1

,^"

fl'i>" ,'Vy.-1 (a '» )" ,>/»n + 1' n+2' n + 3'

The above would be for a diagram for F if we deleted the far right-hand box

(containing fl¿7p, fln+2, aPn+y...).

G consists of the span of monomials in ((x ' p a 'p" • • ■ a 'p ) . a^'p)1,

a„+2' K+3)¿'-" and (U1/pn+2«J/p" + 1---«yp2)V where i = 0,".. , p - 1 and

/=0,...,p" + I-l. So F n G = Zp((x^pn+2a\pn + l ... al/p2)p, av a2,. .:,

an + v aPn+2, fl^23,...). [F nG:A] = p" + 1 => [k[xn+2]:k] <p" + 1, with contradicts

our initial assumption that [A[x + ] :A] = p" 2

(e) This final counterexample shows that in general A[||KP ] does not have

a tensor complement over A in fl AKP (K over A a modular extension).

We let A = Z (x, y, z v z2,.. •) where x, y, zy z2,... are indeterminates.

Let


348 L. A. KIME [Febru! ■y

kup n K = k[\zyp], x1/p]

kup2 n K = k[{zyp2], x1/p2 + z\/pyl/p]

kl/pi n K = A[íz;17*3¡, x17"3 + 2J7i>2y17"2 + zl'Py1'*}

Clearly A[íz!7p°°S] C k[C\Kp']. We assert that A [{z^^f] has no tensor

complement in DA/<p . This will imply that A[jz.7p ]] = AtflTv*' ], and in partic-

ular be another counterexample showing that in general |) kKp / k[(]Kp ].

Assume that there exists an M C 7< such that M ®, Atiz:7'' ¡] = K. Sincetz J

it takes only one element to complete \z /p ] to a p-basis for A17^" O K over

k p O K, M must be of the form k[xx, x2, • •-1 where x. has exponent i

over A, and xp. £ A [x ... x. ,].z 1' ' z— i

xx £Z(xl/p, y, z\/p,. .. , zlJp) for some finite « =» xn+, 6

z.(. .1/ö"+2 1/ö"+1 ,l/p"+2 zl7""+2) (=F). ^ut

x ., 6 Al7^" + 2 O K = Z (x17»"+2 + zYP" + ly"p!: + l + • •.77 "• 7 O 1

+ zJfr1". y. *l/pn+2 _l/z>1 > *1

rz+2

x . eZ()77+2 Ö

l/p"+2 ,_ -l/i>n + 1vl/T>n + 1+ z

+ z17?v17*>, >, z}/Pn+2,...,zl/pn + 2, zl/p) (=C).« + 1-7 J 1 r? :-z+l

/~ -J i_ í m J- r "7 Z l/pn+2 l/i)"+1 l/i)n+2 l/i)"+2 l/fi\Consider the following diagram of L\x v . y ' zl '""s2 . z +?)

l/t."+2overZ(x,y,z^-,...,z^-,2n + 1).

l/pn+2

. P

(x'_« +2 2I-'>" + 1vl p" + !

.1 />"..! 'P

„1 0"

.1 P yl V)i> n + I,1 P ,! P

F consists of the Z (x, y, Zj p , • • • , z„ ? ) span of monomials in

which (in terms of the above modular basis) no power of z 'M occurs. G consists

of polynomials in l(xl/pn+2 + z1/pn + :yl/P" + 1 + .. . + zl/p2y Up2) + *$y1/»li,

(ZV*)' (l = 0,...,p-l) and (x17>"+2 + zYpn + 1y1/pn"+1 + ...+ zypn2y^'P2)P-i

(j JO,..., p" + 1 - 1). Hence


19731 PURELY INSEPARABLE, MODULAR EXTENSIONS 349

F O G-Z ((x;/p-"+2_z!7p" + 1y17p" + 1 + ... +z!/p2v1/p2)p,p

v _l/p"+2 ,l/p"+2)

Then

*p"^ £AU;/P,..,, z17p]=,x. £A[z}/P,.,., z1/pj^x, £A,n+2 1 'nil n 1

r

which contradicts our assumption that [A(xj) :A] = p.

BIBLIOGRAPHY

1. N. Jacobson, Lectures in abstract algebra. Vol. 3: Theory of fields and Galois

theory, Van Nostrand, Princeton, N. J., 1964. MR 30 #3087.

2. M. Sweedler, Structure of inseparable extensions, Ann. of Math. (2.) 87 (1968),

401-410. MR 36 #6391.

DEPARTMENT OF MATHEMATICS, BOSTON STATE COLLEGE, BOSTON, MASSACHUSETTS02115


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