Part (a)

Definition: A field extension \(L/F\) is said to be a splitting field of a polynomial \(f(x)\) if \(L\) contains all roots of \(f\) and thus decomposes as \begin{align*} f(x) = \prod_{i=1}^n (x - \alpha_i)^{k_i} \in L[x] \end{align*} where \(\alpha_i\) are the distinct roots of \(f\) and \(k_i\) are the respective multiplicities.

Part (b)

Let \(F\) be a finite field with \(q\) elements, where \(q=p^k\) is necessarily a prime power, so \(F \cong { \mathbf{F} }_{p^k}\). Then any finite extension of \(E/F\) is an \(F{\hbox{-}}\)vector space, and contains \(q^n = (p^{k})^n = p^{kn}\) elements. Thus \(E \cong { \mathbf{F} }_{p^{kn}}\) Then if \(\alpha \in E\), we have \(\alpha^{p^{kn}} = \alpha\), so we can define \begin{align*} f(x) \coloneqq x^{p^{kn}} - x \in F[x]. \end{align*}

The roots of \(f\) are exactly the elements of \(E\), so \(f\) splits in \(E\).

Part (c)

The polynomial \(f\) is separable, since \(f'(x) = p^{kn}x^{p^{kn}-1} - 1 = -1\) since \(\mathrm{char}(E) = p\). Since \(E\) is a finite extension, \(E\) is thus a separable extension. Then, since \(E\) is a separable splitting field, it is a Galois extension by definition.

Problem 2

We can write \(I = \operatorname{Ann}_\mu\) for some \(\mu \in R\), so suppose \(xy \in I\) so \(xy\mu = 0\).

If \(y\mu = 0\), then \(y\in I\).

Otherwise, \(y\mu \neq 0\) and \(x\in \operatorname{Ann}_{y\mu}\). But by maximality, \(\operatorname{Ann}_{y\mu} \subseteq I\), so \(x\in I\).

Problem 3

Let \(I {~\trianglelefteq~}R\), then since \(R\) is a PID we have \(I = (b)\) for some \(b\in R\). We can write \((b) = Rb\); if \(a\in I\) is an irreducible element, we’d like to show that \(Rb = Ra\).

Note that since \(a \in (b)\), we have \((a) \subseteq (b)\) and thus \(Ra \subseteq Rb\).

Since \(a\in Rb\), we have \(a = rb\) for some \(r\in R\). Since \(a\) is irreducible, either \(r\) is a unit or \(b\) is a unit.

If \(r\) is a unit, then \(a = rb \implies r^{-1}a = b\). But then \(x\in Rb \implies x = r'b = r'r^{-1}a \in Ra\), so \(Rb \subseteq Ra\) and thus \(Ra = Rb = I\).

Otherwise, if \(b\) is a unit, \(a = rb \implies Ra = R\). But any ideal containing a unit is the entire ring, so \(Rb = (b) = R\) as well, so again \(Ra = I\).