*Algebra*

## 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

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.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 \(F\).