Is there a natural way to view the proof of Hilbert 90? I only know of one proof of Hilbert 90, which is very smart if not magical. See for example http://hilbertthm90.wordpress.com/2008/12/11/hilberts-theorem-90the-math/
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.
 A: 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)
A: Another way to conclude that $\tau$ has a fixed point using linear algebra is the following. David Speyer shows that $\tau$ is of order $n$. But, thanks to independence of characters, the minimal polynomial of $\tau$ is $X^n-1$ and thus $\tau$ is a cyclic endomorphism of $L$. Hence 1 is an eigenvalue of $\tau$ with multiplicity one.
A: 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.
A: 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.
A: 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.
