# 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. – Hw Chu Mar 16 '18 at 0:39

## 1 Answer

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 '18 at 18:23
• In Section 7.3, the author proves that these two definitions are equivalent. – Kenny Wong Mar 16 '18 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 '18 at 21:54
• Thanks @kjo, good point. I have edited. – Kenny Wong Mar 23 '18 at 7:07