Missing step in proving that $[L:K] = n \Rightarrow$ there are precisely $n$ distinct $K$-monomorphisms $L \rightarrow N$.

Question

The following is a proof that $[L:K] = n \Rightarrow$ there are precisely $n$ distinct $K$-monomorphisms $L \rightarrow N$, where $N$ is the normal closure of $L:K$.

My concern is that the author states "by induction there are precisely $s$ distinct $K( \alpha )$-monomorphisms $p_1, ..., p_s : L \rightarrow N$". Shouldn't the induction step allow us to state only that "by induction there are precisely $s$ distinct $K( \alpha )$-monomorphisms $p_1, ..., p_s : L \rightarrow N'$, where $N'$ is the normal closure of $L:K( \alpha )$"?

Such a problem could be solved by noticing that $N' \subseteq N$, though the only proof I know of the latter fact is quite long and messy. Is there an easier way to understand how the statement made by the author is correct?

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I would really appreciate any help.

Answer 1

By definition, a field $R$ is the "normal closure of $M$ over $F$" (or in your notation, "normal closure of $M\colon F$") if and only if:

1. $R$ is an extension of $F$ that contains $M$ ($F\subseteq M\subseteq R$);
2. $R$ is a normal extension of $F$ ($R\colon F$ is normal); and
3. $R$ is the smallest normal extension of $F$ that contains $M$ (if $S$ is any field with $F\subseteq M\subseteq S$ and $S\colon F$ is normal, then $R\subseteq S$).

Added: The above assumes we are working over a particular algebraic closure; probably better would be to replace the inclusions with $F$-embeddings, to allow for “up to isomorphism” situations.

In your set-up, $N'$ is the normal closure of $L$ over $K(\alpha)$, and $N$ is the normal closure of $L$ over $K$. You want to show that $N'\subseteq N$. To that end, it is enough (by part 3 of the definition) to show that $N$ is a normal extension of $K(\alpha)$ that contains $L$.

Indeed, recall that if $F\subseteq F'\subseteq M$, and $M$ is normal over $F$, then it is also normal over $F'$ (the converse does not hold, even if you assume $F'$ is normal over $F$). In particular, since $N$ contains $K(\alpha)$ and is normal over $K$, $N$ is also normal over $K(\alpha)$. Since $K(\alpha)\subseteq L\subseteq N$, then $N$ is a normal extension of $K(\alpha)$ that contains $L$. By definition, $N$ contains the normal closure of $L$ over $K(\alpha)$, namely $N'$. Thus, $N'\subseteq N$, as desired.

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In comments, the OP provides a different definition of “normal closure”, where we have

3’. $R$ is the smallest normal extension of $F$ that contains $M$: if $M\subseteq S\subseteq R$, and $S$ is normal over $M$, then $S=R$. In other words, no proper subextension of $R:M$ can be normal over $F$.

That would account for the larger difficulty alluded to by the OP.