Brian Bi
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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

1. It is clear that $$x^3 - 2$$ is irreducible, so that adjoining the root $$\sqrt{2}$$ to $$\Q$$ produces a field extension of degree 3. But $$\Q(\sqrt{2})$$ 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.
2. 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.
3. 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.