Today I Learned

April 19th

Today I learned (finally) the proof of the fundamental correspondence in Galois theory, from Marcus (say). We assume that a finite extension $L/K$ has $[L:K]$ embeddings into $\CC,$ and we take the primitive element theorem (for separable extensions) on faith. The main theorem is the following.

We can by showing the map is well-defined. The set of elements of $M$ fixed by a particular subgroup does indeed make a field: at least $\alpha,\beta\in M^H$ implies\[\begin{cases} \sigma(\alpha+\beta)=\sigma(\alpha)+\sigma(\beta)=\alpha+\beta, \\ \sigma(\alpha\cdot\beta)=\sigma(\alpha)\cdot\sigma(\beta)=\alpha\cdot\beta,\end{cases}\]and a subring of a field (say, containing $1$) is a subfield.

To show surjectivity, pick up any intermediate field $L,$ and we would like to take it to $\op{Gal}(M/L)\subseteq\op{Gal}(M/K).$ Note that any embedding of $M$ fixing $L$ is also an embedding of $M$ fixing $K$ and therefore lives in $\op{Gal}(M/K).$ Further, this subset of $\op{Gal}(M/K)$ is closed under composition and inversion (which we won't show explicitly), so it is a subgroup. We let this subgroup be $H:=\op{Gal}(M/L)\subseteq\op{Gal}(M/K).$

It remains to show that $L$ is actually the fixed field of $H.$ This is a size problem. Focusing on $M/L,$ we see that it is a finite, separable (inherited from $M/K$), normal (above) extension, so we have\[[M:L]=\#\op{Gal}(M/L)\]total embeddings of $M$ into $\CC$ fixing $L.$ However, we must also have $\op{Gal}(M/L)\subseteq\op{Gal}(M/M^H),$ implying $[M:L]\le[M:M^H],$ and certainly $L\subseteq M^H,$ so we conclude $L=M^H.$

For injectivity, we have to show that two subgroups $H_1$ and $H_2$ cannot have the same fixed field $L\subseteq M.$ For this, we claim that, actually,\[\op{Gal}(M/M^H)=H,\]implying that we can read the subgroup off of the fixed field. (The Galois group is defined because $M^H$ is an intermediate field of the Galois extension $M/K$; see above.) In particular, this will be enough because, if $M^{H_1}=M^{H_2},$ then $\op{Gal}(M/M^{H_1})=\op{Gal}(M/M^{H_2}),$ so $H_1=H_2$ as we need.

Showing that $\op{Gal}(M/M^H)=H$ comes down to a size condition. Certainly all automorphisms in $H$ belong in $\op{Gal}(M/M^H),$ but the issue is that we might have more automorphisms. To control this, we let $L=M^H,$ and we're showing that $L$ is not the fixed field of any proper subgroup of $\op{Gal}(M/L)$ (which $H$ would be if there were extra automorphisms).

This, again, comes down to a size condition. Fix $H\subseteq\op{Gal}(M/L)$ whose fixed field is $L.$ Using the primitive element theorem, we can fix $M=L[\alpha],$ and then we focus on the polynomial\[f(x):=\prod_{\sigma\in H}(x-\sigma\alpha).\]The coefficients of $f(x)$ are fixed by $H$—they are symmetric sums of products of $\sigma_\bullet$s, so applying anything in $H$ merely permutes the symmetric sum—meaning that the coefficients live in $L,$ or $f\in L[x].$ Expanding $M$ as $L$ with a power basis $\alpha^\bullet,$ we see that we are forced to have\[\deg f=[M:L]=\#\op{Gal}(M/L).\]In particular, $\#H=\#\op{Gal}(M/L),$ so we had better have $H=\op{Gal}(M/L).$ This completes the proof. $\blacksquare$

We remark that the proof also gives the inclusion reversing cleanly. In particular, subgroups $A,B\subseteq\op{Gal}(M/K)$ correspond (one-to-one) with their fixed fields $M^A$ and $M^B.$ But $A\subseteq B$ if and only if $M^B\subseteq M^A$ ("the set of elements fixed by $B$ is also fixed by $A$''), which is the inclusion-reversing property.

We also prove the following refinement in the context of the previous theorem.

The main idea is that, for an intermediate field extension $L$ between $M/K$ and $\sigma\in\op{Gal}(M/K),$ we have\[\op{Gal}(M/\sigma L)=\sigma\op{Gal}(M/L)\sigma^{-1}.\]Indeed, $\tau\in\op{Gal}(M/\sigma L)$ if and only if $\tau$ fixes $\sigma L$ if and only if $\sigma^{-1}\tau\sigma$ fixes $L$ if and only if $\tau\in\sigma\op{Gal}(M/L)\sigma^{-1}.$

Now, the statement that $L$ is normal over $K$ is that all embeddings of $L/K$ are in fact automorphisms. Well, any of the embeddings of $L/K$ can be extended to embeddings (and therefore automorphisms) of $M/K,$ so we only have to check restrictions of $\op{Gal}(M/K).$ In particular, $L$ is normal over $K$ if and only if\[\sigma L=L\]for each $\sigma\in\op{Gal}(M/K).$ Well, the subgroup fixing $\sigma L$ is $\sigma\op{Gal}(M/L)\sigma^{-1},$ so the Galois correspondence asserts that this holds if and only if\[\sigma\op{Gal}(M/L)\sigma^{-1}=\op{Gal}(M/K)\]for each $\sigma\in\op{Gal}(M/K).$ In particular, $L$ is normal over $K$ if and only if $\op{Gal}(M/L)$ is a normal subgroup, which is what we wanted. $\blacksquare$