3
$\begingroup$

I am reading Lang's book algebra. I don't understand his proof of the first part of theorem 5.1 on page 285.

His definition of norm:

Definition. Let $E$ be a finite extension of $k$. Let $[E:k]_{s}=r$, and let $p^{\mu}=[E:k]_{i}$ if the characteristic is $p>0$, and $1$ otherwise. Let $\sigma_{1},\cdots,\sigma_{r}$ be the distinct embeddings of $E$ in an algebraic closure $k^{a}$ of $k$. If $\alpha\in E$, define its norm from $E$ to $k$ to be $$ N_{E/k}(\alpha)=N_{k}^{E}(\alpha)=\prod_{\nu=1}^{r} \sigma_{\nu}\alpha^{p^{\mu}}. $$

Then he gives the following theorem:

Theorem 5.1. Let $E/k$ be a finite extension. Then the norm $N_{k}^{E}$ is a multiplicative homomorphism of $E^{*}$ into $k^{*}$ and ...

In the proof, he says:

Proof. For the first assertion, we note that $\alpha^{p^{\mu}}$ is separable over $k$ if $p^{\mu}=[E:k]_{i}$. The product $\prod_{\nu=1}^{r}\sigma_{\nu}\alpha^{p^{\mu}}$ is left fixed under any isomorphism into $k^{a}$ because applying such an isomorphism simply permutes the factors. Hence this product must lie in $k$ since $\alpha^{p^{\mu}}$ is separable over $k$. ...

(1) Why is $\alpha^{p^{\mu}}$ separable over $k$?

(2) What isomorphisms is he talking about? I know that if $K/k$ is Galois and if $\beta\in K$ is fixed by all $\sigma\in Gal(K/k)$, then $\beta\in k$. But here we don't have a Galois extension, and the product $\prod_{\nu=1}^{r}\sigma_{\nu}\alpha^{p^{\mu}}$ is only in $k^{a}$.

Thank you for your help!

$\endgroup$
0

1 Answer 1

4
$\begingroup$

For (1), see Proposition V.6.1 on page 247. The proposition says that the minimal polynomial of $\alpha$ has roots $\alpha_1,\dots,\alpha_r$ where $r = [k(\alpha) : k]_s$ and each root has the same multiplicity $p^m$ for some $m \ge 0$. Furthermore, $\alpha^{p^m}$ is separable. Since we necessarily have $p^m = [k(\alpha) : k]_i \mid [E : k]_i$, it follows that $\alpha^{p^\mu}$ is also separable.

For (2), note that if $K$ is normal over $k$ (in particular $K = k^a$), then the fixed field of $\operatorname{Aut}(K/k)$ is a purely inseparable extension of $k$. Since $N_{E/k}(\alpha)$ is separable and lies in a purely inseparable extension of $k$ it must be that $N_{E/k}(\alpha) \in k$ (c.f. Proposition V.6.11 on page 251).

$\endgroup$
4
  • $\begingroup$ Dear, Trevor! Let me ask you the following question: Could you explain why $N_{E/k}(\alpha)$ lies in a purely inseparable extension? I guess in order to show it we've to show that it's fixed any automorphism of $k^a$ over $k$, right? But why is it fixed under such automorphisms? $\endgroup$
    – RFZ
    Apr 22, 2019 at 1:24
  • $\begingroup$ @ZFR Because $N_{E/k}$ is defined by taking a product over such automorphisms. Any automorphism will just permute the values $\{\sigma_{\nu}\alpha^{p^{\mu}} : 1 \le \nu \le r\}$ and thus the product is fixed. $\endgroup$ Apr 22, 2019 at 1:59
  • $\begingroup$ Sorry but the norm is defined as the product of embeddings of $E$ into $k^a$ over $k$ NOT automorphism of $k^a$ over $k$. An also could you give more details why it just permutes these values? It is not so obvious to me. $\endgroup$
    – RFZ
    Apr 22, 2019 at 2:03
  • 1
    $\begingroup$ @ZFR Given one embedding of $E$ into $k^a$, any automorphism will take this embedding to a different embedding. So the effect on $\{\sigma_{\nu}\alpha^{p^{\mu}} : 1 \le \nu \le r\}$ is just a permutation. If you need more information, open a new question. $\endgroup$ Apr 22, 2019 at 2:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .