Is there a natural way to view the proof of Hilbert 90?

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.

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.

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.