# Uniqueness of splitting fields

I am reading a set of Galois theory notes, and I don't understand the proof of the following theorem:

Let $f(X)$ be in $K[X]$, let $L/K$ be a splitting field for $f(X)$. Let $s\colon K\to M$ be any embedding such that $sf$ splits into linear factors in $M[X]$. Then the following hold:

1. There exists at least one embedding $t\colon L\to M$ extending $s$.
2. The number of embeddings $t$ as in (1) is at most $[L:K]$, with equality holding if $f(X)$ has no repeated roots in $L$, i.e., if it splits into distinct linear factors.

For now I am only asking for a proof of the first part of (2). The more intuitive it is, the better.

Thanks!

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Perhaps you could give the proof that is troubling you? –  Dylan Moreland Nov 9 '11 at 20:14
OP's taking this from Theorem 6.6, pages 23 & 24, from these notes. –  anon Nov 9 '11 at 20:19
@anon You're psychic! (Or you searched for some phrases, but that would be less exciting). To Martin: What, in particular, troubles you about the proof? Is it the lemmas used, the induction, ...? –  Dylan Moreland Nov 9 '11 at 20:33

The important calculation is the following. Let $\alpha$ be a root of $f$ in $L - K$, and let $g$ be the irreducible polynomial of $\alpha$ over $K$. You can easily check that embeddings of $K(\alpha)$ in $M$ over $s$ correspond precisely to roots of $g^s$ in $M$. There are at most $\deg g = [K(\alpha) : K]$ such roots, and there is at least one because $g^s$ divides $f^s$ and the latter polynomial splits in $M$.
Now, since $L$ is still a splitting field of $f$ over $K(\alpha)$ we can repeat this process until we've adjoined so many roots of $f$ that we hit $L$. This is formalized by induction. Thankfully, degrees of field extensions are multiplicative in towers, e.g. $[L : K] = [L : K(\alpha)][K(\alpha) : K]$, and an embedding of $L$ into $M$ over $s$ corresponds to a choice of embedding $t\colon K(\alpha) \to M$ over $s$ and then the choice of an $L \to M$ over $t$. In other words, counting the embeddings is multiplicative as well.