Download Minimal Polynomial of √2+√3: Proofs of Irreducibility and more Assignments Linear Algebra in PDF only on Docsity! MATH 731, FALL 2008 HOMEWORK SET 9 sample solutions A. Find the minimal polynomial of √ 2+ √ 3 over Q (and verify it is the minimal polynomial). Show that Q[ √ 2, √ 3] = Q[ √ 2 + √ 3]. Answer & Proof. Let α = √ 2 + √ 3. Then α2 = 5 + 2 √ 6, so (α2 − 5)2 = 24. This means α is a root of f = x4 − 10x2 + 1. We believe this is the minimal polynomial, but all we know immediately is that it’s a multiple of the minimal polynomial. We give four proofs of this fact, but only the first one is complete. Proof 1 that f is the minimal polynomial. If we show f is irreducible, it’s the minimal polynomial. It would be nice to apply Eisenstein’s criterion — perhaps to f(x ± a) — but that doesn’t seem to work. The rational root test tells us ±1 are the only possible roots and they are not roots, so if f factors, it factors as f = gh with g, h of degree 2. By Gauss’s Lemma, if there is such a factorization in Q[x] , there must be one in Z[x] . Hence we may assume g = x2 + ax+ b , h = x2 + cx+ d , c, d ∈ Z . Thus x4− 10x2 + 1 = x4 + (a+ c)x3 + (b+ d+ ac)x2 + (ad+ bc)x+ bd . This immediately tells us c = −a , and equating the other coefficients, we see b+ d− a2 = −10, a(d− b) = 0, bd = 1. The last equality tells us b = d = ±1, and the first equality then becomes ±2− a2 = −10, which has no integer solutions. Thus it is impossible for such g, h to exist. Proof 2 that f is the minimal polynomial. Again, we show f is irreducible. One can show that if α, β, γ, δ are the 4 numbers ± √ 2± √ 3, then α, β, γ, δ are roots of f . Thus f = (x − α)(x − β)(x − γ)(x − δ). You can check(!) that none of these numbers are rational, so f has no linear or cubic factors in Q[x] . Thus if f is reducible, it has two quadratic factors. That means some pair of α, β, γ, δ has rational sum and rational product. You can check(!) that this is not the case. For example, the sum is either ±2 √ 2 or ±2 √ 3 or 0, and if it is 0, the product is −( √ 2± √ 3)2 , which is not rational. Thus f is irreducible. Proof 3 that f is the minimal polynomial. As we noted above, the rational root test shows that f has no linear, and hence no cubic, factors. Thus if f is not the minimal polynomial, the minimal polynomial must have the form g = x2 + ax + b . Now g(α) = 2 √ 6 + a √ 2 + a √ 3 + (b+ 5). This can never equal zero, since 1, √ 2, √ 3, √ 6 are linearly independent over Q . (But is this linear independence obvious?) Proof 4 that f is the minimal polynomial. α = √ 2 + √ 3 lies in E = Q[ √ 2, √ 3]. Note that [E : Q] = 4, since √ 3 /∈ Q[ √ 2]. (Why not?) Thus the splitting field for the minimal polynomial of α must be a subfield of E . If it is E , then x4 − 10x2 + 1 is the minimal polynomial of α . If the splitting field is smaller than E , then it has dimension 2 over Q . Since the automorphism group of E is isomorphic to Z2 ⊕ Z2 , there are only three quadratic extensions of Q inside E , namely Q[ √ 2], Q[ √ 3], Q[ √ 6]. To finish the proof, show that√ 2 + √ 3 is not in any of those fields. Once we know f is the minimal polynomial, we know [Q[α] : Q] = 4. Since α = √ 2+ √ 3, we clearly have Q[α] ⊆ Q[ √ 2, √ 3]. But[ Q[ √ 2, √ 3] : Q ] = [ Q[ √ 2, √ 3] : Q[ √ 2] ] [ Q[ √ 2] : Q ] ≤ 2 · 2 = 4. Thus the dimensions are both 4 and Q[α] = Q[ √ 2, √ 3]. 1
