In the text by Dummit and Foote, there is the following proposition:

Proposition Let $E$ be the splitting field over $F$ of the polynomial $f(x) \in F[x]$. Then we have \begin{equation*} |Aut(E/F)| \leq [E:F] \end{equation*} with equality iff $f(x)$ is separable over $F$.

The proof of this proposition involves looking at the number of possible ways of extending an isomorphism $\phi: F \to F' $ to an isomorphism $\sigma: E \to E'$ between the splitting field $E$ of $F$ and the splitting field $E'$ of $F'$.

I am wondering if there is a proof of this proposition that uses linear algebra. This seems at least plausible to me, since the linear algebra analog of an automorphism of $K$ is just an invertible linear transformation. This suggests writing something like:

\begin{equation*} \Big(\text{number of invertible maps from V to V fixing F}\Big) \leq dim_F(V). \end{equation*}

But there are (at least) two problems with the above statement:

  1. The base field (scalar field) $F$ is not necessarily contained in $V$, and it is not clear (to me at least) how to embed $F$ into $V$.

  2. There is no linear algebra analog of a transformation "fixing" a space. (Note: I guess one could just say that $T$ fixes a subset $A$ of $T$ if $T$ is the identity on $A$, so maybe this is not really an issue...?)

If there is not a way to prove the above proposition using only linear algebra, is there a proof that at least uses a decent amount of linear algebra?

  • $\begingroup$ I think you wrote $K$ instead of $E$ in the first displayed formula in the body of your post $\endgroup$ Mar 17, 2017 at 0:43
  • $\begingroup$ Basically the standard proof uses only the fact that an element must be mapped to a root of its minimal polynomial. The rest ist combinatorics and technical stuff. There will not be a proof, which does not use this basic fact. So what is your hope here? $\endgroup$
    – MooS
    Mar 17, 2017 at 9:56
  • $\begingroup$ @MooS So, admittedly, I don't really have a super concrete idea in mind of what type of proof I am looking for. But I guess I would generally like to see how far one could go in phrasing the entire statement and its proof using only linear algebra. As an example of what I mean by this...would it be possible to explicitly write out a basis for $V$ and then use this to prove the inequality? $\endgroup$
    – Sam Y.
    Mar 17, 2017 at 23:34
  • $\begingroup$ This will not be possible. A basis for $V$ does not contain any information on the ring structure of $E$. $\endgroup$
    – MooS
    Mar 18, 2017 at 0:05


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