# A question about Hilbert Theorem 90 and Artin-Schreier Theorem

I'm reading Lang's "Algebra" and there's a passage in the proof of Theorem 6.3 pg.290 (namely Hilbert's Theorem 90 additive form) for which I can't find a justification, if anyone could provide an explanation, I would appreciate it.

This theorem is used for recovering Artin-Schreier Theorem for cyclic extensions in characteristic p (so, even if not explicitly stated in the theorem, the hypothesis of characteristic p can be assumed if necessary). The precise statement of the theorem is the following:

"Let $$K/F$$ be a cyclic field extension of degree $$n$$, $$\sigma$$ be a generator of the Galois group $$\text{Gal}(K/F)$$ and $$\beta\in K$$. Then, the trace $$\text{Tr}_{K/F}(\beta)=0$$ if and only if there is an element $$\alpha\in K$$ s.t. $$\beta=\alpha-\sigma(\alpha)$$."

Here, by $$\text{Tr}_{K/F}(\beta)$$ we mean the trace of the matrix representation of the $$F$$-linear map given by the multiplication by $$\beta$$ in $$K$$, seen as $$F$$-vector space (clearly, with respect to some fixed basis).

The unclear passage of the proof is direction $$[\Rightarrow]$$, in which it is claimed without any further explanation the existence of some $$\theta\in K$$ s.t. $$\text{Tr}_{K/F}(\theta)\neq 0$$. Do you see why this is the case?

• You might want to check out Grillet's book Abstract Algebra for details like this that Lang sometimes annoyingly omits. Commented Jun 19 at 22:39
• Thank you for the nice suggestion ;) Commented Jun 20 at 11:39

That is because, since $$K/F$$ is Galois, there exists $$\vartheta\in K$$ such that the family $$(\sigma(\vartheta))_{\sigma\in{\rm Gal}(K/F)}$$ is an $$F$$-basis of $$K$$. Now, $${\rm Tr}_{F/K}(\vartheta)=\sum_{\sigma\in{\rm Gal}(K/F)}\sigma(\vartheta)$$ can't be $$0$$ because this is a non-trivial linear combination composed of vectors of a free family. Note that the fact that $$K/F$$ is cyclic is not used here.

Here is a more hands-on argument avoiding the normal basis theorem. I assume you know this, but just to state the difficulty clearly for a general audience: we of course always have $$\text{tr}_{K/F}(1) = n$$, which settles the question in any characteristic not dividing $$n$$. However, for Artin-Schreier theory we need to apply this result in characteristic $$p$$ and when $$n = p$$, so we can't get away with this and we need to find some other element.

We can instead use primitive elements. Let $$\alpha \in K$$ be a primitive element, so that $$\{ 1, \alpha, \dots \alpha^{n-1} \}$$ is a basis for $$K$$ over $$F$$. Then there is some element of $$K$$ with nonzero trace iff $$\text{tr}_{K/F}(\alpha^k) \neq 0$$ for some $$0 \le k \le n-1$$, so let's try to prove this. These traces are the sums of the powers of the conjugates of $$\alpha$$, namely

$$\text{tr}_{K/F}(\alpha^k) = \sum_{i=0}^{n-1} (\sigma^i \alpha)^k$$

and we have the nice result that if $$x_1, \dots x_n$$ are elements of any field of characteristic $$0$$ such that the power sum symmetric functions $$p_k = \sum x_i^k$$ are all zero, then the $$x_i$$ are all zero, using Newton's identities to express the elementary symmetric functions $$e_k$$ of the $$x_i$$ in terms of the power sums.

But Newton's identities express $$ke_k$$ as an integer polynomial in the smaller $$e_i$$ and $$p_i$$, so in characteristic $$p$$ what we get is that $$e_k = 0$$ for all $$k$$ not divisible by $$p$$ but not necessarily otherwise. If $$p \mid n$$ (which as above is the interesting case) it follows that if the traces $$\text{tr}_{K/F}(\alpha^k)$$ are all zero then the characteristic polynomial $$\chi_{\alpha}(t)$$ of $$\alpha$$ is a polynomial in $$t^p$$, and such a polynomial cannot be irreducible. Taking the contrapositive, some trace does not vanish.

This argument does not require $$K/F$$ to be cyclic or even Galois; we actually get the result assuming separability only (in which case we replace the $$\sigma^i \alpha$$ above with the roots of the characteristic polynomial of $$\alpha$$ in a normal closure of $$K$$). In fact more is true: a finite extension is separable iff the trace form is nondegenerate.