Algebra Homework Set 8 Hung Tran. 9.4.2.c Reducing Mod 2

Algebra Homework Set 8 Hung Tran. 9.4.2.c Reducing mod 2 yields: f2 (x) = x4 + 1 = (x + 1)4 . Then appyling Eisenstein’s criterion for the shifted polynomial g(x) = f (x + 1) and observing that g(0) = f (−1) = 1 − 4 + 6 − 2 + 1 = 2 6≡ 0(mod22 ). Thus, g(x) is irreducible and so is f (x) in Z[x]. QED 13.1.1 Since the polynomial is monic and primitive and Q is the fraction field of Z, it is irreducible in Q[x] iff it is irreducible in Z[x]. Then applying Eisenstein’s criterion and observing that 3 6≡ 0(mod32 ); thus it is irreducible in Z[x] and Q[x]. Using the Division Algorithm we have: x3 + 9x + 6 = (x2 + 8)(x + 1) − 2; that is, 1 2 3 3 ∼ 2 (x +8)(x+1) = 1 in Q[x]/(x +9x+6). By theorem 6, Q(θ) = Q[x]/(x +9x+6), so (θ + 1)−1 = 21 (θ2 + 8). QED 13.2.19 a. Obviously for a, b in F and x, y in K, α(ax+by) = α(ax)+α(by) = aα(x)+bα(y) since K is commutative. So α acting by left multiplication on K is an F-linear transformation on K. b. Consider K as a vector space over F.Since the extension is of degree n, there is a basis containing n vectors ei , i = 1, ..., n, for K as vector space over F. Consider the map: φ : K → Mn (F ) with φ(α) is the matrix associated with the linear transformation of α acting by left multiplication with respect to the fixed basis. Elementary algebra (as F is commutative) shows that this is a homomorphism of rings (addition is defined pointwise and composition of functions is corresponding to multiplication of matrices). Since for each nonzero α there exists α−1 as K is a field, φ(α) is invertible as matrix.(1) Since αβ = βα, φ(α)φ(β)(x) = φ(β)φ(α)(x). Since the basis if fixed, and each matrix is invertible, that implies φ(α)φ(β) = φ(β)φ(α).(2) (1) and (2) imply that the image of φ is a subfield of the ring of n × n matrices over F (the ring itself needs not be commutative). Furthermore, since the homomorphism is obviously nontrivial, by theorem 13.1.2, it is an isomorphism of fields. That is K ∼ = φ(K). QED 13.2.20. Consider the matrix A − Iα over K (not F) and its associated linear transformation in K induced by matrix multiplication. For any x in K we have: (A − Iα)x = Ax − Iαx = α(x) − α(x) = 0. Thus, det(A − Iα) = 0 in K (otherwise the formula for inverse matrix still holds since K is commutative and produces an inverse linear transformation, which is absurd). So α is a root of the characteristic polynomial of A. Another proof is that by the Cayley-Hamilton theorem, A satisfies its own characteristic polynomial in Mn (F ). By 13.2.19, φ is an isomorphism of K and a subfield of Mn (R) that sends α to A so α must also satisfies the characteristic polynomial of A in K. 1

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√ √ √ Let Q[ 3 2] be the field extension of Q by 3 2 then it is easy to see that [Q[ 3 2] : Q] = 3 as x3 − 2 is the minimal polynomial (irreducible √ in Q[x] by Eisenstein’s criterion and proposition 13.2.9). √ Furthermore,√since 3 2 does not √ satisfied any quadratic 3 3 3 polynomial, e1 = 1, e = 2 and e = 4 are a basis of Q[ 2] over Q. 2 √ 3 √ 3 3 Certainly, α = 2 ∈ Q[ 2] so we can apply the result above with the observation that α(e1 ) = e2 ,α(e2 ) = e3 α(e3 ) = 2e1 . So elementary algebra shows that the matrix  is:  0 0 2 A= 1 0 0  0 1 0 and its associated characteristic polynomial is −x3 + 2 so the monic minimal poly√ 3 3 nomial of 2 is x − √2 √ √ Similarly if β = 1 + 3 2 + 3 4 ∈ Q[ 3 2] and β(e1 ) = e1 + e2 + e3 ,β(e2 ) = 2e1 + e2 + e3 and α(e 2e2 + e3 so the assoicated matrix is:  3 ) = 2e1 + 1 2 2 A= 1 1 2  1 1 1 and its associated characteristic polynomial is (1 − x)3 − 6(1 − x) + 6 so the monic √ √ 3 3 3 minimal polynomial of 1 + 2 + 4 is x − 3x2 − 3x − 1 QED 13.2.21 √ √ Denote e1 = 1 and e2 = D and α = a + b D. We have α(e1 ) = ae1 + be2 and α(e2 )= bDe1 +ae2 so the associated matrix is: a bD A= b a Let the map: φ :K → M2 (Q)  be defined as p a bD φ(a + b (D) = b a We have  if  a bD A= b a   m nD and B = n m   m + a (n + b)D then A + B = n+b m+a   am + bDn (an + bm)D and AB = an + bm am + bDn So it is clear that φ is a ring  homomorphism from K onto the subring H consisting  a bD of matrices of the form A = b a   a −bD 1 Furthermore, since D is square-free: A−1 = a2 −b is well defined 2D −b a √ and A−1 = φ((a + B D)−1 ). Thus, H is a field and φ is a field isomorphism by theorem 13.1.2 Additional problems 1 Consider the field K = F2 (x)/(x3 + x + 1). Since x3 + x + 1 has no root in F2 , it

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is irreducible over F2 (x) and therefore, by theorem 13.2.11, K ∼ = F2 (α) with α is a root of x3 + x + 1. So 1, α, α2 is a basis of K over F2 and every element of K can be written in the form of a + bα + cα2 for a,b,c in F2 . Since F2 has exactly 2 elements, K has exactly 8 elements. The multiplication table for K is given as follows: k 0 1 α 1+α α2 α2 + 1 α2 + α 0 0 0 0 0 0 0 0 1 0 1 α 1+α α2 α2 + 1 α2 + α α 0 α α2 α2 + α 1+α 1 α2 + α + 1 2 2 2 2 α+1 0 1+α α +α α +1 α +α+1 α 1 α2 0 α2 1+α α2 + α + 1 α2 + α α α2 + 1 α2 + 1 0 α2 + 1 1 α2 α α2 + α + 1 1+α 2 2 2 2 α +α 0 α +α α +α+1 1 α +1 1+α α α2 + α + 1 0 α2 + α + 1 α2 + 1 α 1 α2 + α α2 QED 2 a. Statement: If φ is any F-algebra homomorphism from K to a F-algebra H then p(φ(α)) = 0 in H. Conversely, if a ∈ H such that p(a) = 0 in H then there is a unique F-algebra homomorphism that sends α to a. Proof: (⇒) If φ is any F-algebra homomorphism then obviously φ sends 0 to 0. Since p(α) = 0 and p is a polynomial with coefficients in F,p(φ(α)) = φ(p(α)) = 0 in H. (⇐) Suppose degree(p(x)) = n. By theorem 13.1.4 and 13.2.11, 1, α, ...., (α)n−1 is a basis for K over F. Since being a F-algebra homomorphism forces φ(1) = 1, once the value φ(α) is given, φ(x) = φ(a0 + ... + an−1 αn−1 ) = a0 + ... + an−1 φ(α)n−1 is uniquely determined. Furthermore, it is completely routine to check that φ as defined above is a F-algebra homomorphism (notice that ai ∈ F for i = 0, .., n − 1). So the homomorphism exists and is unique. b. Also suppose degree(p(x)) = n then by theorem 9.5.17, p(x) has no more than n roots counting multiplicity. Since p(x) splits completely into linear factors in K[x], it has exactly n roots in K counting multiplicity. By part (a), a F-algebra homomorphism to itself is corresponding to sending α to a root of p(x) (can be the identity map). So the number of distinct F-algebra homomorphism to itself is the number of distinct roots of p(x) in K and no more than n. c. Let φ be an F-algebra homomorphism from K to itself and φ(α) = β with β is a root of p(x) in K. Then φ(1) = 1 and for any a in F: φ(a) = φ(a.1) = aφ(1) = a with the 2nd equality comes from the fact that φ is a F-algebra homomorphism. So φ is fixing F pointwise. As p(x) is irreducible over F, β is not in F. By theorem 13.1.8, F (α) ∼ = F (β) as vector spaces over F so F (β) also has dimension n over F. So F (β) = K and 1, β, ..., β n−1 is also a basis of K over F. Therefore, φ is surjective. The injectivity of φ also follows immediately from the fact that any element in a vector space can be written uniquely as a linear combination of its basis. Thus φ is an automorphism from K to itself. QED

α2 + α + 1 0 α2 + α + 1 α2 + 1 α 1 α2 + α α2 1+α