On a normal closure of a trivial (i.e. degree 1) extension

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?

Definitions

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.

• The normal closure of $L/K$ when $L = K$ is $L$ itself. If $f(x) \in K[x]$ is irreducible and have a root in $K$, then $f$ can only have degree 1, otherwise it will be reducible, so the $L/K$ is a normal extension, and hence the closure does not change. Mar 16, 2018 at 0:39

The normal closure of $K$ over $K$ is $K$.

To verify this, you just need to show that $K$ is a normal extension of $K$. (Clearly, it would follow that $K$ is the smallest normal extension.)

Howie's book gives you two characterisations of "normal extension" (which are equivalent for finite extensions), and we can verify either one:

• According to the first definition, $K$ is a normal extension of $K$ if $K$ is the splitting field over $K$ for some polynomial $f(X) \in K[X]$. Indeed it is: take $f(X) = X - 1$. (Or even $f(X) = 1$.)

• According to the second definition, $K$ is a normal extension of $K$ if every irreducible polynomial $f(X) \in K[X]$ with a root $\alpha$ in $K$ splits completely in $K[X]$. But the only irreducible polynomial with $\alpha$ as a root is the linear polynomial $f(X) = X - \alpha$, and this manifestly splits completely in $K[X]$.

• Thanks. I'm familiar with the second definition in you cite from Howie's book. What's the page for the first one?
– kjo
Mar 16, 2018 at 18:23
• In Section 7.3, the author proves that these two definitions are equivalent. Mar 16, 2018 at 19:50
• Actually, only one of those, the second one, is the definition of normal extension; the first one is equivalent to it, only when the extension is of finite degree (that's what Howie proves, at any rate), and therefore it is not as general a characterization of normality as that given by the definition.
– kjo
Mar 22, 2018 at 21:54
• Thanks @kjo, good point. I have edited. Mar 23, 2018 at 7:07