# showing that there are $n$ distinct $K$-monomorphisms

I am trying to understand this proof from the book Galois Theory by Ian Stewart. I am stuck on understanding the motivation of using $\alpha$ in the field $L$ but not in the field $K$. I don't see where he uses this result for the remainder of the proof, by this I mean why is $\alpha$ in $L$ and not in $K$ useful?

• If $\alpha$ were an element of $K$, then its minimal polynomial would have degree $1$. Consequently $[L:K(\alpha)]=[L:K]$ meaning that the induction hypothesis could not be applied. – Jyrki Lahtonen Apr 11 '16 at 5:18

If $\alpha\in K$, then $K(\alpha)=K$ and so $[L:K(\alpha)]=[L:K]$. This means you cannot use the induction hypothesis on the extension $L:K(\alpha)$. Specifically, the sentence that breaks down is:
By induction there are precisely $s$ distinct $K(\alpha)$-monomorphisms $\rho_1,\dots,\rho_s:L\to N$, where $s=[L:K(\alpha)]=k/r$.
This sentence requires you to already know that the theorem is true for the extension $L:K(\alpha)$. If $\alpha\not\in K$, then $[L:K(\alpha)]<[L:K]$ so you know this by the induction hypothesis. But if $\alpha\in K$, then you don't know anything.