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I only know of one proof of Hilbert 90, which is very smart if not magical. See for example

Does anyone know of a more intuitive proof or know a good way to view the proof?

I have accepted the answer by Emerton, great thanks as well to David Speyer and Brian Conrad.

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It says that there's no obstruction to etale descent for 1-dimensional vector spaces: line bundle for the etale topology is "same" as a line bundle for the Zariski topology in case of spectrum of a field (true for any scheme) and so has a basis. Faithfully flat descent for quasi-coherent sheaves doesn't rely on Hilbert 90, and in the special case of line bundles and etale covers of spec of field it is precisely Hilbert 90. For vector bundles gives "Hilb 90 for GL_n". Grothendieck points out the link somewhere (intro of sga1?). That is a "good" way to view the meaning of the result. – BCnrd Apr 12 '10 at 15:04
Just one clarification: the weblink gives the special case of cyclic extensions in the language of norms. So to make the connection with Grothendieck, conceptually one applies degree-2 periodicity of Tate cohomology for cyclic groups which identifies the norm thing (which is Tate cohomology in degree -1) with usual degree-1 cohomology. The latter provides the framework for the "usual" formulation of Hilbert 90, which in turn is a super-special case of grothendieck's ff. descent theory. – BCnrd Apr 12 '10 at 15:07
Thanks a lot for the way pointing answer, it is a bit overwhelming though (T T). – abcdxyz Apr 12 '10 at 15:16
@Tran: Brian Conrad allows others to flesh out his comments in the form of an answer. Since I find this particular comment very interesting, I hope that someone eventually will... (See the comments here… ) – François G. Dorais Apr 12 '10 at 15:30
It might also be worth pointing out that faithfully flat descent is a special case of Beck's theorem, which is a result in pure category theory. In other words, its content is entirely category-theoretic, rather than geometric. – JBorger Apr 12 '10 at 23:58
up vote 22 down vote accepted

Here is a proof of Hilbert's Theorem 90 in the case of cyclic extensions which I think is fairly conceptual. The key point (which is also at the heart of Grothendieck's very general version in terms of flat descent) is that if we want to verify that a linear transformation has a certain eigenvalue (in our particular case, the eigenvalue of interest will be 1), we can do so after extending scalars.

The set-up: we have a cyclic extension $L/K$, with Galois group generated by $\sigma$, and an element $a \in L$ of norm 1. We want to find $b \in L$ such that $a = b/\sigma(b)$. As in David Speyer's answer, rewrite this as the equation $a\sigma(b) = b$.

The map $b \mapsto a\sigma(b)$ is a $K$-linear transformation of the $K$-vector space $L$, and we want to show that it has a fixed point, i.e. that it has $1$ as an eigenvector. Well, we can verify this after extending scalars (the eigenvectors of a matrix don't change if we enlarge the ground field), and so we tensor up with $L$ over $K$.

Now $L\otimes_K L \cong L\times\cdots \times L$, an isomorphism of $L$-algebras, and under this isomorphism the action of $\sigma$ on the left just becomes the cyclic permutation of factors on the right. (To see the isomorphism, write $L = K(\alpha),$ as we may by the primitive element theorem. If $f(X)$ is a minimal polynomial of $\alpha$ over $K$, then $L \cong K[X]/f(X),$ and so $L\otimes_K L \cong L[X]/f(X).$ But over $L$, the polynomial $f(X)$ splits as $f(X) = (X-\alpha_1)\cdots (X-\alpha_n),$ where the $\alpha_i$ are all the conjugates of $\alpha$. Choosing the labelling appropriately, we may assume that $\alpha_i = \sigma^{i-1}(\alpha)$. Then $L[X]/f(X) = L[X]/(X-\alpha_1)\cdots (X-\alpha_n) \cong L\times\cdots \times L,$ and $\sigma$ does indeed just permute the factors.)

Under the isomorphism $L\otimes_K L \cong L\times\cdots \times L,$ the base-change of our linear transformation $b \mapsto a \sigma(b)$ is given by $(b_1,\ldots,b_n) \mapsto (a b_n, \sigma(a) b_1, \ldots, \sigma^{n-1}(a) b_{n-1}).$ This transformation has the obvious non-zero fixed vector $(1,\sigma(a),\sigma(a)\sigma^2(a),\ldots,\sigma(a)\ldots\sigma^{n-1}(a)).$ (Remember that Norm$(a) = 1$, and so the last entry is also just $a^{-1}$.)

Thus our original linear transformation (before extending scalars) has a non-zero fixed vector as well, as required.

How does this relate to Brian Conrad's comment? Well, the preceding argument generalizes massively to Grothendieck's theory of faithfully flat descent, which in particular shows that any quasi-coherent sheaf in the flat topology in fact arises from a Zariski sheaf. That may sound quite complicated, but what the argument amounts to is precisely what we used in the preceding argument: If $A \rightarrow B$ is a faithfully flat map of rings, and we want to study the "spectral theory" of a linear operator on an $A$-module, we can do so after extending scalars to $B$. (Of course, one has to be precise about what "spectral theory" means when we are working over rings that aren't fields. This is where faithfully flat comes in: it is the condition that extending scalars from $A$ to $B$ is exact, and takes non-zero modules to non-zero modules; this turns out to be exactly the right generalization of the more naive notion we used above, that extending scalars preserves the eigenvalues of a matrix.)

Finally, here is an aside about the relation with Galois cohomology:

In cohomological language, Hilbert's Theorem 90 is the statement that $H^1(Gal(L/K), L^{\times}) = 0$ for any finite Galois extension of fields $L/K$. To recover the statement involving norms, one proceeds as follows: if $Gal(L/K)$ is cyclic, with generator $\sigma$, and the norm of $a \in L$ equals 1, then $\sigma \mapsto a$ determines a $1$-cocyle on $Gal(L/K)$ with values in $L^{\times}$. By the vanishing of $H^1$, this must be a coboundary, which means that there exists $b$ such that $a = \sigma(b)/b.$

The cohomological statement (which, as Brian Conrad pointed out, is still a very special case of Grothendieck's general theory) can be proved by the same extension of scalars argument as above.

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Thank you. It is really a through out and insightful answer. – abcdxyz Apr 12 '10 at 16:38
Beautiful ! – Qfwfq Apr 12 '10 at 17:23
Could you please add a few more details on why $\sigma$ permutes the factors of $L\times L\times ...\times L$ ? Because it is an $L$-left module algebra endomorphism of $L\otimes_K L$ of order $n$ ? – darij grinberg Aug 1 '10 at 15:11
Oh, I see. Maybe you should have written that the isomorphism $L\otimes_K L\to L\times L\times ...\times L$ that you are using is not some random isomorphism, but the one given by $a\otimes b\mapsto\left(a\sigma^{-0}\left(b\right),a\sigma^{-1}\left(b\right),...,a\sigma^‌​{-(n-1)}\left(b\right)\right)$. This makes clear the statements below. And you don't need the (ugly and non-canonical) primitive element theorem to prove that this is an isomorphism, because injectivity is simply the linear independence of characters and surjectivity follows from injectivity by the power of linear algebra. – darij grinberg Aug 1 '10 at 15:22
Dear darij, you are correct, but I have a personal phobia towards independence of characters, for some reason, and if you bring that result to the forefront, it pretty quickly leads to another proof of Hilbert's Thm. 90 (which is however not the one that I remember). Hence I prefer to think in terms of primitive elements (which I don't find as ugly as you do; but I suspect I'm in a minority there!). – Emerton Aug 1 '10 at 23:42

One argument I love is the following: let $L/K$ be a Galois extension with group $G$ and let $n\geq1$. One can show very straightforwardly that $H^1(G,\mathrm{GL}(n, L))$ classifies $K$-vector spaces $V$ such that $L\otimes_KV$ is isomorphic as an $L$-vector space to $L\otimes K^n$, up to $K$-linear isomorphisms; Serre does it in chapter X, §2, of his Corps Locaux. Now, linear algebra tells us that all such $V$'s are in fact isomorphic to $K^n$ as $K$-vector spaces, so we conclude that $H^1(G,\mathrm{GL}(n, L))$ is trivial.

This is, in fact, the same argument that Brian gave. Yet it is nice that the theorem becomes essentially a statement saying that all vector spaces of the same dimension are isomorphic :)

Also, other somewhat mystifying statements, like $«H^1(G,\mathrm{Sp}(n, L))=0»$ can be proved by exactly the same argument.

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"without group G"? Probably not. :) – Pete L. Clark Apr 12 '10 at 17:21
Heh. Fixed. Thanks! – Mariano Suárez-Alvarez Apr 12 '10 at 17:23
It's strange because Proposition 4 in chapter X, §2 Corps Locaux depends on that H1(G,GL(n,L)) is trivial. – user42690 Jan 26 at 7:20

Here is a good way to think of the standard proof:

Let $L/K$ be a cyclic extension of degree $n$, with $\sigma$ a generator of $Gal(L/K)$. Suppose that $N(a)=1$, for $a \in L$.

Define the operator $\tau: L \to L$ by $\tau(b) = a \sigma(b)$. We have $$\tau^n(b) = a \sigma(a) \sigma^2(a) \cdots \sigma^{n-1}(a) b = N(a) b =b$$ so $\tau^n$ is the identity. Also, $\tau$ is $K$-linear. So, considering $L$ as a $K$-vector space, we have a representation of $\mathbb{Z}/n$ on $L$.

We want to show that this representation has a trivial summand. If we can show this, we are done; if $\tau(c) = c$ for $c \neq 0$ then $a = c/\sigma(c)$. As you will learn in any course on representation theory, the operator $$\pi := (1/n) \left( 1+ \tau + \cdots + \tau^{n-1} \right)$$ is the projector onto the trivial summand of $L$. The standard proof is to verify that $\pi$ is nonzero.

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This is nice but it doesn't work in characteristic $p$ when $p|n$. Fortunately the standard proof actually verifies that regardless of characteristic it's nonzero without the factor $1/n$. – Noam D. Elkies Nov 17 '14 at 23:18

Suppose $E/K$ is a Galois extension with group $G=\operatorname{Gal}(E/K)$.

Construct the crossed product algebra $E\rtimes G$, which is the $E$-vector space with basis the elements of $G$ turned into a $K$-algebra in the unique way such that $eg\cdot e'h=eg(e')gh$ for all $e$, $e'\in E$ and all $g$, $h\in G$.

It is easy to check that the $K$-algebra $E\rtimes G$ is simple. Wedderburn's theorem then tells us that all simple $E\rtimes G$-modules are isomorphic. A corollary of this is:

Theorem. $H^1(G,E^\times)=0$.

Indeed, suppose $\phi:G\to E^\times$ is a $1$-cocycle. The $E$-vector space $V=E$ can be endowed in a unique way with an action of $E\rtimes G$ in such a way that $eg\cdot 1=e\phi(g)$ for all $e\in E$ and all $g\in G$. Since $V$ is one-dimensional as an $E$-vector space, it is a simple module over $E\rtimes G$ and it follows from Wedderburn's theorem that there is an $E\rtimes G$-linear isomorphism $f:E\to V$,for $E$ is also a simple $E\rtimes G$-module. If we set $y=f(1_E)$, then the coboundary of $y$ is $\phi$. $\Box$

More generally, from Wedderburn's theorem we get that all $E\rtimes G$-modules are direct sums of copies of $E$, and that implies in pretty much the same way that $H^1(G,GL(n,E))=0$ (and the infinite dimensional version that Serre leaves as an exercise in Corps locaux)

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