Using Groebner bases to find minimal polynomials

1. Using Groebner bases to find minimal polynomials Isabel ‘gunner’ fulcher, renee ‘The Situation’ russell, amanda ‘Pookie’ clemm

2. 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?!

3. 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

4. What is a Field ? • A field is commutative ring whose nonzero elements form a group under multiplication • So 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

5. 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. Extension field of R Extension field of Q and Subfield of C Subfield of R

6. 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 Qwhere 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. Extension Field (K) Subfield (k)

7. 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, πis not algebraic over Q.

8. What is a minimal polynomial ? • Let α∈K be algebraic over k. The minimal polynomial of αis 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

9. 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

10. Consequence of the First Isomorphism Theorem • Given K/k, with algebraic over k and the minimal polynomial of , consider the homomorphism • x • N • By the First Isomorphism Theorem, Kernel Range Domain

11. 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 αover 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

12. 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 !!!

13. … 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:

14. 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 Groebnerbasis 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

15. 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 Groebnerbasis for (with respect to the lex term order): minimal polynomial of !

16. 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.

17. 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

18. 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 .