Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 18H (Galois Theory)

A field homomorphism fixing $K$ is injective, since its kernel is an ideal of a field and cannot contain one. It is also $K$-linear. Equal finite field extension degrees then make it a bijection, hence a $K$-isomorphism.

Equal degrees alone do not ensure that any such homomorphism exists. For example $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ both have degree two, but a $\mathbb{Q}$-isomorphism would require a square root of two in $\mathbb{Q}(\sqrt{3})$. Squaring $a+b\sqrt{3}$ with rational $a,b$ would give $2ab=0$ and $a^2+3b^2=2$. Neither $a^2=2$ nor $b^2=2/3$ has a rational solution. Thus the two degree-two extensions are not isomorphic over $\mathbb{Q}$.

An isomorphism fixing $L$ fixes its subfield $K$, so an $L$-isomorphism is a $K$-isomorphism. The converse is false because an isomorphism over $K$ may act nontrivially on $L$.

Take $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2})$, $\alpha =2^{1/4}>0$, $F=L(\alpha )$ and $F^{\prime }=L(i\alpha )$. Both are degree-two extensions of $L$. The polynomial $X^4-2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion, and both $\alpha$ and $i\alpha$ are roots. Sending $\alpha$ to $i\alpha$ therefore gives a $\mathbb{Q}$-isomorphism $F\to F^{\prime }$; on $L$ it sends $\sqrt{2}$ to $-\sqrt{2}$.

There is no $L$-isomorphism. The two quadratic extensions adjoin square roots of $\sqrt{2}$ and $-\sqrt{2}$. Their square classes would agree only if $-1$ were a square in $L$. To check this criterion directly, a square root of $\sqrt{2}$ in $L(i\alpha )$ would have form $a+b(i\alpha )$; its square forces $ab=0$. The case $b=0$ would put $\alpha$ in $L$, while $a=0$ would require $b^2=-1$. Both are impossible, the latter since $L\subseteq \mathbb{R}$. Thus the extensions are isomorphic over $K$ but not over $L$.