Modern Algebra: Groups, Rings, Modules, Fields

For the first assertion, consider the unique ring homomorphism \(\varphi:\Z \to F\text{.}\) Since \(F\) is a domain, the kernel of \(\varphi\) is a prime ideal (since \(\Z/\ker(\varphi) \cong \im(\varphi)\) and \(\im(\varphi)\) is a subring of \(F\)). The result holds since the only prime ideals of \(\Z\) are \((0)\) and \((p)\) for a prime integer \(p\text{.}\) (Note that this proof shows that, more generally, the characteristic of an integral domain must be either \(0\) or a prime.)

For the second assertion, observe that the smallest {} of \(F\) is the image of the ring map \(\varphi: \Z \to F\text{,}\) and by the first assertion, this image is isomorphic to either \(\Z\) or \(\Z/p\text{.}\) The latter is already a field and hence it is the prime field of \(F\text{.}\) In the former case, the prime subfield is isomorphic to the field of fractions of \(\Z\text{,}\) which is \(\Q\text{.}\)

Suppose \(F\) and \(E\) are fields and \(\s: F \to E\) is a ring homomorphism. Let \(\phi_F: \Z \to F\) and \(\phi_E: \Z \to E\) be the unique ring maps from \(\Z\) to \(F\) and \(E\text{.}\) Since \(\s \circ \phi_F\) is a ring map from \(\Z\) to \(E\text{,}\) we have \(\s \circ \phi_F = \phi_E\) by the uniqueness of \(\phi_E\text{.}\) Since \(F\) is a field and \(E\) is not the zero ring, the map \(\s\) is injective. Since \(\s\) is injective, it follows that \(\ker(\s \circ \phi_F) = \ker(\phi_F)\text{,}\) and hence we obtain that \(\ker(\phi_F) = \ker(\phi_E)\text{.}\) By definition of characteristic, we conclude \(\ker(E) = \ker(F)\text{.}\)

Let \(h = \gcd_{F[x]}(f, g)\text{.}\) To prove (1), we note that \(h\) is the unique monic polynomial such that - \(h = f \cdot p + g \cdot q\) for some \(p,q \in F[x]\text{,}\) - \(f = h \cdot \a\) for soem \(\a \in F[x]\text{,}\) and - \(g = h \cdot \beta\) for soem \(\beta \in F[x]\text{.}\) Since \(F[x]\) is a subring of \(L[x]\text{,}\) these three properties also hold when we regard \(h,f,g\) as belonging to \(L[x]\) and thus by the uniqueness property, we have \(h = \gcd_{L[x]}(f,g)\text{.}\)

For (1), suppose \(\a\) is a root of \(f(x)\) of multiplicity at least two. Then \(f(x) = (x-\a)^2g(x)\) in \(\ov{F}[x]\) and hence \(f'(x) = 2(x - \a) g(x) + (x-\a)^2 g(x)\text{,}\) by the Product Rule. It follows that \(f'(\a) = 0\text{.}\) Conversely, suppose \(f(\a) = f'(\a) =0\text{.}\) Since \(f(\a) = 0\text{,}\) we have \(f(x) = (x-\a)h(x)\) and hence \(f'(x) = h(x) + (x-\a)h'(x)\text{.}\) Since \(f'(\a) = 0\) it follows \(h(\a) = 0\) and thus \(\a\) has multiplicity at least two.

For the second assertion, by (1), we have that \(f\) is separable if and only if \(f\) and \(f'\) has no common roots in \(\ov{F}\text{.}\) The result thus follows from the Lemma [provisional cross-reference: cite]

For the final assertion, assume \(f(x)\) is irreducible. Since the degree of \(f'(x)\) is strictly less than the degree of \(f(x)\text{,}\) we have that \(\gcd(f(x),f'(x)) \ne 1\) if and only if \(f'(x) = 0\text{.}\)

Suppose \(f(x)\) has a repeated root \(\beta\) in some splitting field \(K\) of \(f(x)\text{,}\) i.e., \(f(x) = (x-\beta)^2 g(x)\) for some polynomial \(g(x) \in K[x]\text{.}\) Since \(\alpha\) is also a root of \(f(x)\text{,}\) we have \((\alpha - \beta)^2 g(\alpha) = 0\text{.}\) Since \(F\subseteq L\) is an algebraic extension, \(\alpha\) is algebraic over \(F\text{,}\) so \(g(\alpha)\neq 0\text{.}\) It follows that \((\alpha-\beta)^2=0\text{,}\) i.e., \(\alpha=\beta\text{,}\) which means that \(\alpha\) is a repeated root of \(f(x)\text{,}\) contradicting the assumption that \(f(x)\) has no repeated roots. Hence \(f(x)\) is separable, and since \(\alpha\) was an arbitrary algebraic element of \(L\text{,}\) we conclude that \(F\subseteq L\) is a separable extension.