Brian Bi
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Section 16.11. Kummer Extensions

Exercise 16.11.1 If the discriminant is not a square, Theorem 16.8.5 tells us that the splitting field has degree 6 over $$F$$, therefore it cannot be obtained by adjoining a cube root.

Exercise 16.11.2

1. If $$b$$ has a $$p$$th root $$\beta$$ that lies in $$F$$, then $$\beta, \zeta\beta, \ldots, \zeta^{p-1}\beta$$ are $$p$$ distinct roots of $$g$$ that lie in $$F$$, therefore $$g$$ splits completely in $$f$$.

If not, let $$\beta$$ denote a $$p$$th root of $$F$$ in a splitting field $$K$$. Again, the roots of $$g$$ are $$\beta, \zeta\beta, \ldots, \zeta^{p-1}\beta$$. Suppose $$g = PQ$$ where $$P$$ and $$Q$$ are monic polynomials with coefficients in $$F$$. One of $$P$$ and $$Q$$ must have $$\zeta^{p-1} \beta$$ as a root; let us say it is $$P$$. Let $$d = \deg Q$$; the roots of $$Q$$ are $$\zeta^{a_1} \beta, \ldots, \zeta^{a_d} \beta$$ where $$a_1, \ldots, a_d$$ are some distinct integers between 0 and $$p-2$$, inclusive. The coefficient $$c$$ of $$x^{d-1}$$ in $$Q$$ lies in $$F$$ and is equal to $$-\beta\sum_{i=1}^d \zeta^{a_i}$$. Since $$-\beta \notin F$$, this equality can only be satisfied if $$\sum_{i=1}^d \zeta^{a_i} = 0$$. However, since $$\{1, \zeta, \ldots, \zeta^{p-2}\}$$ are linearly independent over $$\Q$$, it follows that there are no terms on the left-hand side; that is, $$Q = 1$$. Therefore, $$g$$ is irreducible.

2. A solution is given here.

Exercise 16.11.3 The conjecture is true; indeed, a fundamental result in Kummer theory is the generalization of Theorem 16.11.1 to non-prime powers. However, we will only consider the case $$n = 4$$ here, as the problem asks.

By the fundamental theorem of Galois theory, the extension $$K/F$$ has exactly one proper intermediate field $$L$$, of which $$K$$ is a quadratic extension. Thus, there is an element $$\beta \in K$$ such that $$K = L(\beta)$$ and $$\beta^2 \in L$$. Let $$\sigma$$ be a generator of $$\Gal(K/F)$$. Then $$\sigma^2$$ fixes $$L$$, so $$\sigma^2(\beta)^2 = \sigma^2(\beta^2) = \beta^2$$, thus $$\sigma^2(\beta) = -\beta$$. Write $$\beta' = \sigma(\beta)$$, so the $$\sigma$$-orbit of $$\beta$$ is $$\beta \to \beta' \to -\beta \to -\beta'$$.

If $$\beta' = i\beta$$, then let $$\alpha = \beta$$. Otherwise, if $$\beta' = -i\beta$$, then replace $$\sigma$$ by $$\sigma^{-1}$$ and let $$\alpha = \beta$$. Otherwise, let $$\alpha = \beta - i\beta'$$, so that $$\sigma(\alpha) = \beta' - i(-\beta) = i(\beta - i\beta') = i\alpha$$, and $$\alpha \ne 0$$. Thus, the orbit of $$\alpha$$ under $$\sigma$$ is $$\alpha \to i\alpha \to -\alpha \to -i\alpha$$.

Since $$\alpha$$ isn't fixed by $$\sigma^2$$, $$\alpha \notin L$$, so the intermediate field $$F(\alpha)$$ must be all of $$K$$. However, $$\alpha^4 = (i\alpha)^4 = \sigma(\alpha)^4 = \sigma(\alpha^4)$$, therefore $$\alpha^4$$ is in the fixed field of $$\sigma$$, which is $$F$$.
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