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Section 31 -- Algebraic Extensions

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Section 31 -- Algebraic Extensions Section 31 – Algebraic extensions Instructor: Yifan Yang Spring 2007 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. Linear independence and bases Definition Let V be a vector space over F. A set {α1, . , αn} is said to be linearly independent over F if a1α1 + ··· + anαn = 0 implies that ai = 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 {αi : 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 αi with coefficients in F. 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, . , αn−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 29.18 says that {1, α, . , αn−1} is a basis for F(α). • If α is transcendental over F, then the vectors 1, α, α2,... are linearly independent. Thus, F(α) is not finite-dimensional. 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. 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, . , αn+1 are linearly dependent over F. Thus, there are n elements ai , not all zero, such that a0 + a1α + ··· + anα = 0, that is, α is algebraic over F. 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 31.4. Let {α1, . , αm} be a basis for E over F, and {β1, . , βn} be a basis for K over E. It suffices to prove that • every element of K is a linear combination of αi βj , i = 1,..., m, j = 1,..., n, with coefficients in F, and • αi βj are linearly independent over F. Proof of Theorem 31.4, continued Pm • Let γ ∈ K . Then γ = i=1 bi αi for some bi ∈ E since {αi } is a basis for K over E. Each bi is a linear combination Pn bi = j=1 cij βj over F since {βj } is a basis for E over F. P Then γ = i,j cij αi βj . Thus, every element of K is a linear combination of αi βj over F. • Now suppose that cij are elements in F such that P i,j cij αi βj = 0. Write it in the form m n X X cij βj αi = 0, i=1 j=1 Pn where j=1 cij βj are elements of E. Since αi are linearly Pn independent over E, j=1 cij βj = 0 for each i. Then cij = 0 for all j since βj are linearly independent over F. Degrees of field extensions Corollary (31.6) If F1 ≤ F2 ≤ ... ≤ Fn is a series of finite extension of fields, then [Fn : F1] = [Fn : Fn−1] ... [F2 : F1]. 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 3 3 not contain√ 2 since deg( 2, Q) = 3 does not divide deg( 2, Q) = 2. Example √Problem.√ Find the√ degree and the irreducible polynomial for 2 + 3 over Q[ 3]. Solution. √ √ √ 2 • Let α = 2√+ 3. It is clear that α satisfies (x − 3) = 2, i.e., x2 − 2 3x + 1 = 0. √ √ • Thus, [ (α): ( 3)] ≤ 2. Then [ (α): ( 3)] = 1 or 2. Q √Q Q √Q √ • If [ (α): ( 3)] = 1, then (α) = ( 3) and α ∈ ( 3). Q Q √ √ Q Q Q • This implies 2 = a + b 3 for some a, b ∈ . √ Q 2 • We have√ irr( 2, Q) = x − 2 and irr(a + b 3, Q) = x2 − 2ax + (a2 − 3b2). • We find a = 0 and 3b2 = 2, which is impossible when b ∈ . Q √ • Thus, [Q√(α): Q( 3)] =√2 and irr(α, Q( 3)) = x2 − 2 3x + 1. 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, . , αn−1 is a basis for F(α) over F. • Since β ∈ F(α), this implies that βαi is again a linear combination of αj . • In other words, we have 1 1 . . β . = M . αn−1 αn−1 for some n × n matrix over F. • Thus, β is a zero of the characteristic polynomial of M. • We then factor the characteristic polynomial to get irr(β, F). Example √ √ Problem. Find the irreducible polynomial of 3 22 + 3 2 − 1 over Q. √ Solution. Set α = 3 2 and β = α2 + α − 1. • We have α3 − 2 = 0. Thus, β1 = −1 + α + α2, βα = α3 + α2 − α = 2 − α + α2, βα2 = α4 + α3 − α2 = 2 + 2α − α2. • Thus, β is a zero of the characteristic polynomial of −1 1 1 2 −1 1 . 2 2 −1 • We find β is a zero of x3 + 3x2 − 3x − 11. Example Problem. Let α be a zero of x4 − 10x2 + 1. Find the irreducible polynomial for β = α3 − 9α. Solution. • We have β1 = 0 − 9α + 0α2 − α3 βα = α4 − 9α2 = (10α2 − 1) − 9α2 = −1 + 0α + α2 + 0α3 βα2 = α5 − 9α3 = 0 − α + 0α2 + α3 βα3 = −α2 + α4 = −1 + 0α + 9α2 + 0α3. • Thus, β is a zero of the characteristic polynomial of 0 −9 0 −1 −1 0 1 0 . 0 −1 0 1 −1 0 9 0 Example, continued • We find β is zero of x4 − 18x2 + 80. • This time, x4 − 18x2 + 80 = (x2 − 8)(x2 − 10) is not irreducible. • To determine which factor is the irreducible polynomial for β over Q, we compute β2. • We find β2 = (α3 − 9α)2 = α6 − 18α4 + 81α2. • Using the division algorithm, we find that this is equal to (α2 − 8)(α4 − 10α2 + 1) + 8 = 8. √ √ • Therefore, irr(β, Q) = x2 − 8, i.e., β is either 2 2 or −2 2. (The exact value of β depends on which zero of x4 − 10x2 + 1 we take as α.) In-class exercises √ Let α = 3 2. 1. Find irr(α2 − α, Q). 2. Find irr(α2 + 1, Q). 3 Let α be a zero of x + x + 1 ∈ Z2[x]. 2 1. Find irr(α + 1, Z2). 2 2. Find irr(α + α, Z2). Algebraic closure Lemma Let E be an extension field of a field F. If α, β 6= 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. 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. 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 31.15.) • 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 31.16.) Algebraic closure of a field Definition An algebraic closure F of a field F is an algebraic extension of F that is algebraically closed.
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