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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 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 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 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. (√2) Extension 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 THEOREM Given K/k, with ∈ algebraic over k and the minimal polynomial of , consider the homomorphism : x N: ker = By 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 + 1 + ⋯ + = 0 + 1 + ⋯ + where , ∈ , 0 ≤ ≤ , 0 ≤ ≤ . Let = + 1 + ⋯ + and () = 0 + 1 + ⋯ + 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 ≠ 0, there is a polynomial ∈ such that ≡ 1 ( ) that is − 1 ∈ Let ℎ = . Note that ℎ = . Now consider the map ∶ / ℎ+ ≅ () is in the kernel of if and only if = 0. 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 , − ℎ are considering the ideal =< , − >. [] . Recall that we So, if we can show , − ℎ =< , − > 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: , − ⊇ , − ℎ − ℎ = − ≡ − − ℎ ∈ , − Now show: , − ⊆ , − ℎ − ≡ − = − ℎ − ∈ , − ℎ , − ℎ = , − 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: 5 − − 2 We want to find the minimal polynomial for an element = () () = 1− −2 3 ∈ () Let’s construct the ideal as the theorem/algorithm = , () − (x) = 5 − − 2, − (1 − − 2 3 ) = 5 − − 2, + 2 3 + minimal polynomial of ! THIS IS TRUE ! Recall, β = 1− α −2α3 α and α is a root of x 5 − x − 2. 5 + β5 + 1− α −2α3 5 ) α ( + 11 4 + 4 3 − 5 2 + 95y + 259 2 11 4 β 2 11 1− α −2α3 4 ( ) 2 α + 4β3 − 5β2 + 95β + 259 = + 4( 1− α −2α3 3 ) α − 5( 1− α −2α3 2 ) α 1− α −2α3 + 95( α ) + 259 = … =0 So, β is a root of the (irreducible over k) polynomial + + − + + This is the minimal polynomial for over k. WHAT IF THE FIELD EXTENSION IS NONSIMPLE? 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= 1 , … , be a field extension, and let ’s ∈ K be algebraic over k. Let ∈ 1 , … , −1 be the minimal polynomial of over 1 , … , −1 . Let 0 ≠ ∈ 1 , … , −1 , say (1 , … , ) = (1 , … , ) where , ∈ 1 , … . , . Consider the ideal =< 1 , … , , − > contained in [1 , … , , ]. Then the minimal polynomial of over k is the monic polynomial that generates the ideal . Note: For = 2, … , and ∈ 1 , … , −1 , we let be any polynomial in 1 , … , such that 1 , … , −1 , = .