USING GROEBNER BASES

TO FIN

D MINIMAL

POLYN

OMIALS

I S A B E L ‘ G U N N E R ’ F U L C H E R , R E N E E ‘ T H E S I T U A T I O N ’

R U S S E L L , A M A N D A ‘ P O O K I E ’ C L E M M

OUTLINE- Definitions - Rings- Fields- Field Extensions- Algebraic - Minimal Polynomials- First Isomorphism Theorem- Main Theorem(for simple field extensions)- Example of Main Theorem- What if my field extension is not simple?!

WHAT IS A RING ?A ring is a set R equipped with two binary operations called

addition and multiplication such that:- (R,+) is an abelian (commutative) group- (R, x) has the following properties: - Closed - Associative- There is a multiplicative identity “1” - Multiplication distributes over the addition

Examples: - Z , Z4 , R , Q

WHAT IS A FIELD ?A field is commutative ring whose nonzero elements form a

group under multiplicationSo a field is commutative ring whose nonzero elements each

have a multiplicative inverse

Examples: - Zp where p is prime, R, Q, C

Non-examples:- Zn where n is not prime

WHAT IS A FIELD EXTENSION ?Let K be a field. If a set k⊆ K is closed under the field operations

and inverses in K (i.e. k is a subfield of K) then we call K an extension field of k. We say K/k, “K over k,” is a field extension.

Example: - R and Q are both fields where Q ⊆ R. Therefore, R is an extension field of Q. We could state this as: R/Q is a field extension. The same relationship holds for R ⊆C.

𝑹𝑸

Extension field of Q and Subfield of C

Subfield of R

𝑪 Extension field of R

CONSTRUCTING AN EXTENSION FIELD Consider a base field k. Adjoin with it an element α ∈ k to

construct k(α), the smallest field that contains the base field k and also α.

Example: - If we start with the field Q, we can build the field extension Q(√2)/Q by adjoining √2 to Q where Q(√2) = {a + b√2 | a, b ∈ Q} is an extension field.

Note: we call any extension of the form k(α) a simple extension field because only one element is adjoined. 𝑸 (√ 2)

𝑸Extension Field (K)

Subfield (k)

WHAT IS AN ALGEBRAIC ELEMENT ?If K is a field extension of k, then an element α∈ K is called

algebraic over k if there exists some non-zero polynomial g(x) ∈ k[x] such that g(α) = 0.

Example:- Consider Q(√2), an extension field of Q. √2 is algebraic over Q since it is a root of ∈ Q[x].

Non-Example:- Consider Q(π). π is a root of x – π, but this polynomial is not in Q[x]. So, π is not algebraic over Q.

WHAT IS A MINIMAL POLYNOMIAL ? Let α ∈ K be algebraic over k. The minimal polynomial of α is

the monic polynomial p ∈k[x] of least degree such that p(α) = 0.

Proposition: The minimal polynomial is irreducible over k and any other non-zero polynomial f(x) such that f(α) = 0 must be a multiple of p.

Example: α = √2 element of Q(√2) where √2 is algebraic over Q. The minimal polynomial of √2 is

WHAT ARE QUOTIENT RINGS ?A quotient ring is constructed out of a given ring R by a

process of “moding out” by an ideal I ⊆ R, denoted R/I.Ex.1

In general, Ex.2

where is the ideal Note, when you have a principal ideal, it is generator is the

minimal polynomial

CONSEQUENCE OF THE FIRST ISOMORPHISM THEOREMGiven K/k, with algebraic over k and the minimal polynomial

of , consider the homomorphism

x NBy the First Isomorphism Theorem,

Domain Kernel Range

Given K = k(α), with α algebraic over k, how do we compute the minimal

polynomial of any β ∈ K ?Main Theorem: Let K/k be a field extension, and let α ∈ K be algebraic over k. Let be the minimal polynomial of α over k. Let 0 ∈ k(α), say

where . Let and be the corresponding polynomials in Consider the ideal of Then the minimal polynomial of over k is the monic polynomial that generates the ideal

Note: since , we can compute in

Proof.

Since / is a field and there is a polynomial such that ) that is

Let . Note that .

Now consider the map

is in the kernel of if and only if Therefore, the minimal polynomial of will be in the kernel of

We now find the generator of the kernel of because this will be minimal

polynomial !!!

…By another theorem, the kernel of is Recall that we are considering the ideal

So, if we can show then we know that the kernel of the map is namely the minimal of is the monic polynomial that generates the ideal , and we are done!

To show this, use containment:

Now show:

RESULTS! The minimal polynomial of over k is the monic

polynomial that generates the ideal

This gives us an algorithm for finding the minimal polynomial: Given and as in the theorem, compute the Groebner basis G for the ideal of k[x,y] with respect to the lex term ordering x > y. The polynomial in G which is in y alone is the minimal polynomial of

LET’S FIND THE MINIMAL POLYNOMIAL! Consider the field Q() , where is a root of the irreducible polynomial: We want to find the minimal polynomial for an element )Let’s construct the ideal as the theorem/algorithm =

Now compute the Groebner basis for (with respect to the lex term order):

minimal polynomial of !

THIS IS TRUE !Recall, and is a root of .

(

…

So, is a root of the (irreducible over k) polynomial

❑⇒This is the minimal polynomial for

over k.

WHAT IF THE FIELD EXTENSION IS NON-SIMPLE?As defined, a simple field extension can be written with only

one element adjoined to the base field in order to make up the extension field.

A non-simple field extension has multiple elements adjoined to the base field to make up the extension field. Examples:- Q()- R/Q

In a non-simple extension field, we want to extend our above theorem in order to find the minimal polynomial of any element of

Given K = , with ’s algebraic over k, how do we compute the minimal polynomial of any β ∈ K ?

Main Theorem Revised: Let K= be a field extension, and let ’s ∈ K be algebraic over k. Let be the minimal polynomial of over . Let 0 ∈ , say

where . Consider the ideal contained in Then the minimal polynomial of over k is the monic polynomial that generates the ideal

Note: For and , we let be any polynomial in such that .