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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Hw Chu Mar 16 '18 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 $K[X]$.

  • $\begingroup$ Thanks. I'm familiar with the second definition in you cite from Howie's book. What's the page for the first one? $\endgroup$ – kjo Mar 16 '18 at 18:23
  • 1
    $\begingroup$ In Section 7.3, the author proves that these two definitions are equivalent. $\endgroup$ – Kenny Wong Mar 16 '18 at 19:50
  • 1
    $\begingroup$ 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. $\endgroup$ – kjo Mar 22 '18 at 21:54
  • 1
    $\begingroup$ Thanks @kjo, good point. I have edited. $\endgroup$ – Kenny Wong Mar 23 '18 at 7:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.