*Algebra*

## Section 16.3. Splitting Fields

Exercise 16.3.1 We prove the claim by induction on degree. In the base case, the polynomial has degree 1 and the claim is obvious. Now assume the claim is proved for all degrees smaller than \(n\) and suppose \(f\) has degree \(n\).

Suppose first that \(f\) is irreducible. Let \(\alpha \in K\) be one of the roots of \(f\), so that \(f(x) = (x - \alpha)g(x)\). Clearly, \([F(\alpha) : F] = n\) and \(g(x) \in F(\alpha)[x]\). There exists a field \(L\) with \(F(\alpha) \subseteq L \subseteq K\) that is a splitting field of \(g\), since \(g\) splits completely in \(K\), and by the inductive hypothesis, \([L : F(\alpha)] \divides (n-1)!\). In fact, since \(L\) contains all the roots of \(g\) as well as \(\alpha\), it contains all the roots of \(f\), therefore \(K \subseteq L\). This implies \(K = L\). Thus, \([K : F] = [K : F(\alpha)][F(\alpha) : F] = n[K : F(\alpha)] \divides n(n-1)! = n!\).

Now consider the case where \(f\) isn't irreducible. Write \(f = gh\), where \(g\) and \(h\) both have positive degree. Let \(F'\) be the subfield of \(K\) obtained by adjoining the roots of \(g\) to \(F\); then \(F'\) is a splitting field for \(g\), and by the induction hypothesis, \([F' : F] \divides (\deg g)!\). Furthermore, adjoining the roots of \(h\) to \(F'\) must yield \(K\), so \(K\) is a splitting field for \(h\) when considered as a polynomial over \(F'\); the induction hypothesis implies \([K : F'] \divides (\deg h)!\). Thus, \([K : F] = [K : F'][F' : F] \divides (\deg g)!(\deg h)! \divides (\deg g + \deg h)! = n!\).

Exercise 16.3.2

- It is clear that \(x^3 - 2\) is irreducible, so that adjoining the root \(\sqrt[3]{2}\) to \(\Q\) produces a field extension of degree 3. But \(\Q(\sqrt[3]{2})\) is still a subfield of \(\R\), so it doesn't contain the two complex roots of \(x^3 - 2\). Therefore 3 is a proper divisor of the degree of the splitting field of \(x^3 - 2\). But Exercise 16.3.1 implies that the degree of the splitting field divides 6. Therefore, it must be equal to 6.
- We have the factorization \(x^4 - 1 = (x^2 + 1)(x + 1)(x - 1)\) and clearly \(x^2 + 1\) is irreducible over \(\Q\). It is easy to see that \(\Q(i)\) is a splitting field for \(x^2 + 1\) and thus for \(x^4 - 1\). Therefore the degree of the splitting field is 2.
- The roots of \(x^4 + 1\) are \(\omega, \omega^3, \omega^5, \omega^7\) where \(\omega\) is a primitive eighth root of unity. Any field extension of \(\Q\) that contains one of the four roots also contains the other three; therefore, \(\Q(\omega)\) is a splitting field for \(x^4 + 1\). We showed in Exercise 15.3.4(c) that the minimal polynomial for \(\omega\) over \(\Q\) has degree 4, therefore \([\Q(\omega) : \Q] = 4\).

Exercise 16.3.3 Let \(K\) be a splitting field for \(x^2 - u\). In \(K\), this polynomial must factor as \((x - f)(x - g)\), so \(f + g = 0\) and \(fg = -u\), thus, \(f^2 = u\). Suppose \(f \in F\). Then write \(f = g/h\) where \(g, h \in \mathbb{F}_2(u)\). Then \(f^2 h^2 = g^2\), but \(f^2 = u\), so \(uh^2 = g^2\). Obviously both sides are nonzero, so their degrees must be equal, which is a contradiction since the degree of \(uh^2\) is odd and that of \(g^2\) is even. Our assumption that \(f \in F\) must therefore be wrong, and \(x^2 - u\) is irreducible over \(F\). Note that since \(K\) has characteristic 2, we have \(x^2 - u = (x + f)(x - f) = (x - f)(x - f)\), so \(f\) is a double root.