There's a step at the beginning of a proof in my textbook that I can't figure out.
Theorem 7.28 1
Let $L:K$ be a separable extension of finite degree $n$. Then there are precisely $n$ distinct $K$-monomorphisms of $L$ into a normal closure $N$ of $L$ over $K$.
The book's proof begins like this:
The proof is by induction on the degree $[L:K]$. If $[L:K] = 1$, then $L = K = N$ [sic!], and the only $K$-monomorphism of $K$ into $N$ is the identity mapping.
I get stuck at the second equality of $L = K = N$. What justifies it?
I understand that $[L:K] = 1$ iff $L = K$, and therefore that $K = L$ is separable (by assumption). On the other hand, $K = N$ would mean that $K$ is a normal closure of itself over itself, which would require that $K$ be normal, but I don't see why we can conclude this.
Is the second equality in "$L = K = N$" a typo?
A field extension $L : K$ is said to be normal if every irreducible polynomial in $K[X]$ having at least one root in $L$ splits completely over $L$.
If $L$ is a finite extension of a field $K$, a field $N$ containing $L$ is said to be a normal closure of $L$ over $K$ if (1) it is a normal extension of $K$; and (2) if $E$ is a proper subfield of $N$ containing $L$, then $E$ is not a normal extension of $K$.
An irreducible polynomial with coefficients in $K[X]$ is said to be separable over $K$ if it has no repeated roots in a splitting field. An arbitrary polynomial in $K[X]$ is said to be separable over $K$ if all its irreducible factors are. An algebraic extension $L$ of $K$ is called separable if every $\alpha$ in $L$ is separable over $K$.
1 John M. Howie, Fields and Galois Theory, p. 116.