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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?

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

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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.

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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.

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