Section 31 Algebraic extensions

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1 Section 31 Algebraic extensions Instructor: Yifan Yang Spring 2007

2 Vector spaces over a field Definition Let F be a field. A vector space over F is an additive group V, together with a scalar multiplication by elements of F, satisfying aα V, (ab)α = a(bα), (a + b)α = aα + bα, a(α + β) = aα + aβ, 1α = α, for all a, b F and all α, β V. Lemma Let E be an extension field of a field F. Then E is a vector space over F.

3 Linear independence and bases Definition Let V be a vector space over F. A set {α 1,..., α n } is said to be linearly independent over F if a 1 α a n α n = 0 implies that a i = 0 for all i. If a set of vectors is not linearly independent over F, then it is linearly dependent over F. Definition Let V be a vector space over F. A set of vectors {α i : i I} is a basis for V over F if it is linearly independent and every vector in V is a finite linear combination of α i with coefficients in F.

4 Basis for F (α) Theorem (30.23) Let E = F(α) be a simple extension of a field F. If α is algebraic over F with deg(α, F) = n, then E = F(α) is a finite-dimensional vector space over n, and a basis is given by {1, α,..., α n 1 }. If α is not algebraic over F, then E = F(α) is an infinite-dimensional vector space over F. Proof. If α is algebraic over F, then Theorem says that {1, α,..., α n 1 } is a basis for F(α). If α is transcendental over F, then the vectors 1, α, α 2,... are linearly independent. Thus, F(α) is not finite-dimensional.

5 Algebraic extensions and finite extensions Definition An extension field E of a field F is an algebraic extension of F if every element in E is algebraic over F. Definition If an extension field E of a field F is of finite dimension n as a vector space over F, then E is a finite extension of degree n over F. We let [E : F] denote the degree of E over F. Remark The finiteness in the definition of a finite extension refers to the degree. It does not mean that the field E is a finite field. For instance, Q[i] is a finite extension of degree 2 of Q, but it has infinitely many elements.

6 Finite extension algebraic extension Theorem (31.3) A finite extension E of a field F is an algebraic extension. Proof. Let α E. Assume that [E : F] = n. Then the n + 1 vectors 1, α,..., α n+1 are linearly dependent over F. Thus, there are elements a i, not all zero, such that a 0 + a 1 α + + a n α n = 0, that is, α is algebraic over F.

7 Theorem (31.4) If E is a finite extension of a field F, and K is a finite extension of E, then K is a finite extension of F and [K : F] = [K : E][E : F]. Proof of Theorem Let {α 1,..., α m } be a basis for E over F, and {β 1,..., β n } be a basis for K over E. It suffices to prove that every element of K is a linear combination of α i β j, i = 1,..., m, j = 1,..., n, with coefficients in F, and α i β j are linearly independent over F.

8 Proof of Theorem 31.4, continued Let γ K. Then γ = m i=1 b iα i for some b i E since {α i } is a basis for K over E. Each b i is a linear combination b i = n j=1 c ijβ j over F since {β j } is a basis for E over F. Then γ = i,j c ijα i β j. Thus, every element of K is a linear combination of α i β j over F. Now suppose that c ij are elements in F such that i,j c ijα i β j = 0. Write it in the form m n c ij β j α i = 0, i=1 j=1 where n j=1 c ijβ j are elements of E. Since α i are linearly independent over E, n j=1 c ijβ j = 0 for each i. Then c ij = 0 for all j since β j are linearly independent over F.

9 Degrees of field extensions Corollary (31.6) If F 1 F 2... F n is a series of finite extension of fields, then [F n : F 1 ] = [F n : F n 1 ]... [F 2 : F 1 ]. Corollary (31.7) Assume that E is an extension field of F and α E is algebraic over F. If β F(α), then deg(β, F) divides deg(α, F). Example Using Corollary 31.7, it is easy to see that the field Q[ 2] does not contain 3 2 since deg( 3 2, Q) = 3 does not divide deg( 2, Q) = 2.

10 Example Problem. Find the degree and the irreducible polynomial for over Q[ 3]. Solution. Let α = It is clear that α satisfies (x 3) 2 = 2, i.e., x 2 2 3x + 1 = 0. Thus, [Q(α) : Q( 3)] 2. Then [Q(α) : Q( 3)] = 1 or 2. If [Q(α) : Q( 3)] = 1, then Q(α) = Q( 3) and α Q( 3). This implies 2 = a + b 3 for some a, b Q. We have irr( 2, Q) = x 2 2 and irr(a + b 3, Q) = x 2 2ax + (a 2 3b 2 ). We find a = 0 and 3b 2 = 2, which is impossible when b Q. Thus, [Q(α) : Q( 3)] = 2 and irr(α, Q( 3)) = x 2 2 3x + 1.

11 Determining irr(β, F ) for β F (α) Problem. Let E = F (α) be a finite extension of a field F. Given β F(α), find irr(β, F) (and thus also deg(β, F)). Idea. Assume that deg(α, F) = n. Recall that 1, α,..., α n 1 is a basis for F(α) over F. Since β F(α), this implies that βα i is again a linear combination of α j. In other words, we have 1 β. α n 1 = M for some n n matrix over F. 1. α n 1 Thus, β is a zero of the characteristic polynomial of M. We then factor the characteristic polynomial to get irr(β, F).

12 Example Problem. Find the irreducible polynomial of over Q. Solution. Set α = 3 2 and β = α 2 + α 1. We have α 3 2 = 0. Thus, β1 = 1 + α + α 2, βα = α 3 + α 2 α = 2 α + α 2, βα 2 = α 4 + α 3 α 2 = 2 + 2α α 2. Thus, β is a zero of the characteristic polynomial of We find β is a zero of x 3 + 3x 2 3x 11.

13 Example Problem. Let α be a zero of x 4 10x Find the irreducible polynomial for β = α 3 9α. Solution. We have β1 = 0 9α + 0α 2 α 3 βα = α 4 9α 2 = (10α 2 1) 9α 2 = 1 + 0α + α 2 + 0α 3 βα 2 = α 5 9α 3 = 0 α + 0α 2 + α 3 βα 3 = α 2 + α 4 = 1 + 0α + 9α 2 + 0α 3. Thus, β is a zero of the characteristic polynomial of

14 Example, continued We find β is zero of x 4 18x This time, x 4 18x = (x 2 8)(x 2 10) is not irreducible. To determine which factor is the irreducible polynomial for β over Q, we compute β 2. We find β 2 = (α 3 9α) 2 = α 6 18α α 2. Using the division algorithm, we find that this is equal to (α 2 8)(α 4 10α 2 + 1) + 8 = 8. Therefore, irr(β, Q) = x 2 8, i.e., β is either 2 2 or 2 2. (The exact value of β depends on which zero of x 4 10x we take as α.)

15 In-class exercises Let α = Find irr(α 2 α, Q). 2. Find irr(α 2 + 1, Q). Let α be a zero of x 3 + x + 1 Z 2 [x]. 1. Find irr(α 2 + 1, Z 2 ). 2. Find irr(α 2 + α, Z 2 ).

16 Algebraic closure Lemma Let E be an extension field of a field F. If α, β 0 E are algebraic over F, then so are α + β and α/β. Proof. In view of Theorem 31.3, it suffices to prove that F(α, β) = F(α)(β) is a finite extension of F. Regarding f (x) = irr(β, F) as a polynomial in F(α)[x], we see that β is algebraic over F(α). Then [F(α, β) : F] = [F(α)(β) : F(α)][F(α) : F ] <. Therefore, F (α, β) is a finite extension of F.

17 Algebraic closure Corollary (31.12) Let E be an extension field of F. Then the set F E = {α E : α is algebraic over F} is a subfield of E, the algebraic closure of F in E. Corollary (31.13) The set Q of all algebraic numbers forms a field.

18 Algebraically closed Definition A field F is algebraically closed if every nonconstant polynomial in F[x] has a zero in F. Remarks An algebraically closed field F can be characterized by the property that every polynomial f (x) in F[x] factors into a product of linear factors over F. (Theorem ) This means that if F is algebraically closed, then we will not get anything new by joining zeros of polynomials in F[x] to F. (Corollary )

19 Algebraic closure of a field Definition An algebraic closure F of a field F is an algebraic extension of F that is algebraically closed. Remark An algebraic closure of a field F can be thought of as the largest field one may obtain by algebraic means. Example The field C is an algebraic closure of R. (Fundamental theorem of algebra, Theorem ) The set Q of algebraic numbers is an algebraic closure of Q. (Proved in the next page.)

20 Algebraic closure of Q Theorem The set Q of algebraic numbers is an algebraic closure of Q. Proof. We have proved earlier that if α and β are algebraic over Q, then so are α + β and α/β. In other words, Q is an algebraic extension of Q. It remains to prove that Q is algebraically closed. That is, if α is a zero of f (x) Q[x], then α is in Q. Assume that α is a zero of a n x n + + a 0, where a i are algebraic numbers. We need to show that [Q(α) : Q] <.

21 Proof of Theorem, continued We have [Q(α) : Q] [Q(α, a 0,..., a n ) : Q] = [Q(α, a 0,..., a n ) : Q(a 0,..., a n )][Q(a 0,..., a n ) : Q]. Here [Q(α, a 0,..., a n ) : Q(a 0,..., a n )] n. Also using the same argument as in the proof of an earlier lemma (with induction), we find [Q(a 0,..., a n ) : Q] <. Therefore, [Q(α) : Q] < and α is algebraic over Q.

22 Algebraic closure of a field Theorem (31.17) Every field has an algebraic closure. Remark 1. The proof uses Zorn s lemma, which is equivalent to the axiom of choice. 2. A field may have several algebraic closures, but they are all isomorphic.

23 Homeworks Problems 8, 10, 12, 24, 26, 28, 29, 30 of Section 31.

Section 33 Finite fields

Section 33 Finite fields Instructor: Yifan Yang Spring 2007 Review Corollary (23.6) Let G be a finite subgroup of the multiplicative group of nonzero elements in a field F, then G is cyclic. Theorem (27.19)