Whenever I read anything about valuations or things related to them (such as local fields) extensions always occupy a prominent position and a huge amount of effort is expended to derive results about such extensions. I have a little understanding of valuations and I think I have some understanding about how they can be helpful, but I don't understand why so soon after valuations are introduced everything is ignored except for extensions.

6$\begingroup$ What, for example, is being ignored? $\endgroup$– Tom GoodwillieJan 26, 2011 at 4:36
2 Answers
I don't find your claim that "so soon after valuations are introduced everything is ignored except for extensions" to be accurate. See for instance my notes on valuation theory (as part of a second graduate course on number theory) here. Extensions of valuations play a prominent role there, but it is not as if they are the only things discussed. (I think by the way that the same is more or less true for every text on valuation theory I've seen. I picked my notes because I know what's in them and it's easy for me to point to them, not because they are especially exemplary in any way.)
But if I may adjust your question to "Why is it so important to study extensions of valuations?" I can try to give an answer. I think it is helpful to think of things in terms of number theory. (Well, I would, wouldn't I?) If $L/K$ is a (finite!) extension of number fields, then one of the fundamental problems in algebraic number theory is to understand how the prime ideals in the ring of integers $\mathbb{Z}_K$ of $K$ decompose when pushed forward to ideals in the ring $\mathbb{Z}_L$. This theme is relentlessly pursued: Hilbert's ramification theory, Cebotarev density, class field theory...Let's accept that this is really important!
The question of extension of valuations is a precise analogue and generalization of this question. Namely, if $R$ is a Dedekind domain with fraction field $K$, $L$ is a finite field extension of $K$ and $S$ is the integral closure of $R$ in $L$ (again a Dedekind domain, by the KrullAkizuki Theorem), then to each nonzero prime ideal $\mathfrak{p}$ of $R$ we can attach the $\mathfrak{p}$adic valuation $v_{\mathfrak{p}}$ which assigns to $x \in K^{\times}$ the exponent to which $\mathfrak{p}$ appears in the factorization of the fractional ideal $xR$ into primes. Then there is a bijective correspondence between the valuations on $L$ extending $v_{\mathfrak{p}}$ and the primes of $S$ lying over $\mathfrak{p}$. This take on the classical number theory is useful even in studying the classical case: for instance, an especially clean version of "Kummer's criterion" for how a prime factors is to look at the direct factor decomposition of the algebra $L \otimes_K K_{\mathfrak{p}}$. In other words, the valuationtheoretic perspective (and the business with completions, which is certainly another really important part of valuation theory!) allows us to get the splitting information we want not from the ring $S$ but from the field $L$. This is nice because the classical Kummer criterion has problems with a finite number of primes, owing to the fact that $S$ need not be monogenic as an $R$module. There are no exceptions to the valuationtheoretic analogue.
And from here one gets interested in the problem of extension of valuations for its own sake. One could equally well give an answer in terms of algebraic geometry: if $K$ is a $k$algebra, then the set of valuations on $K$ which are trivial on $k$ is the Zariski Riemann space of $K$. When $K/k$ is finitely generated, this space has a lot of information about the various models of $K$, i.e., $k$varieties with function field $K$. When $K$ has transcendence degree $1$ the valuations are precisely the closed points on the nonsingular model. In higher dimensions things are more complicated but also more interesting. So then if I have a finite field extension $L/K$, then looking at the extension problem is asking for information about a corresponding dominant rational map on algebraic varieties.
It is certainly useful to know that any nonArchimedean valuation can be extended to any field extension. For instance, Paul Monsky famously and brilliantly proved that it is not possible to dissect a square into an odd number of triangles all of the same area: see here. The "technology" of his argument is the fact that the $2$adic valuation on $\mathbb{Q}$ extends to a valuation on $\mathbb{R}$. To the best of my knowledge, no one has ever given a proof of Monsky's theorem that does not use some form of this fact on extensions of valuations!

$\begingroup$ Pete, can you recommend a reference that develops the stuff you talked about in your second to last paragraph? $\endgroup$ Jan 26, 2011 at 15:42

$\begingroup$ Dear Phillip: Try Section 17 of Commutative Algebra: Volume II by Zariski and Samuel and also the papers of Zariski referred to at the end of the section. (I have not thoroughly read this material myself.) $\endgroup$ Jan 27, 2011 at 1:12

1$\begingroup$ @LSpice: I fixed it in my last edit. (This is the broken link problem described on my profile page.) $\endgroup$ Feb 22, 2016 at 1:26

$\begingroup$ @PeteL.Clark, I love such "for your convenience" changes. (What steenking redirects?) My own university recently renamed its so called TR department to the more usual IT department, but (years later) lots of their links still point to URLs with 'TR' in the name. $\endgroup$– LSpiceFeb 22, 2016 at 2:13

$\begingroup$ Oops, I should have looked at the profile! $\endgroup$ Feb 22, 2016 at 2:42
This is not properly an answer to the title question, but rather a response to the premise underlying the question, which I believe is exaggerated.
Valuation theory is actually a huge field; I am guessing you are being exposed to a part of it that has to do with applications to algebraic number theory, where of course one studies extensions all the time (whether local or global, whether of fields or of domains, etc.). Here there is a strong tendency for valuations to be discrete valuations, but this scratches the barest surface of what valuation rings can be like.
One good illustration of what valuation rings can be like comes about by contemplating orders of growth of functions. Consider all the functions that you can build starting with polynomials (over the reals), the exponential and logarithmic functions, and closing up under the four arithmetic operations and composition. (These are the types of functions one often sees in computer science  Knuth dwells for many pages on them in The Art of Computer Programming  and in analytic number theory whenever one is interested in orders of growth.) This class of functions turns out to be totally ordered by $f < g$ if $f(x) < g(x)$ for all sufficiently large $x$. Thus, one is considering germs of such functions at infinity, and these form a (highly nonarchimedean) totally ordered field.
Now, this field $K$ contains a valuation ring $O$ consisting of those functions that are bounded; one sees that for any $f \in K$, that either $f \in O$ or $1/f \in O$. The corresponding valuation group $K^\ast/O^\ast$ is huge! It is an ordered group: letting $[f] = fO^\ast$ be a coset, we have $[f] \leq [g]$ if $f/g \in O$, or if $f/g$ is bounded. In other words, punning a little, $fO^\ast \subseteq gO^\ast$ iff $f = O(g)$ (bigO notation). In other words, the valuation group is the group of orders of growth of such functions. It has received a lot of attention (try googling "Hardy field").
Other examples along these lines: nonstandard fields of real numbers, surreal numbers.
Valuation theory enters into the study of algebraically closed fields. You may have heard of Puiseux series (like power series but allowing fractional exponents), and how Newton essentially used Puiseux series expansions of algebraic functions (using his famous technique of Newton polygons). In this case, we have a valuation ring whose valuation group is the additive group of rational numbers. Here is a terrific theorem which allows the (explicit) construction of exotic algebraically closed fields:
 Let $k$ be algebraically closed, and for an ordered group $G$, let $k[x^G]$ denote the ring of functions $f: G \to k$ such that $\{x \in G: f(x) \neq 0\}$ is wellordered. Then $k[x^G]$ is algebraically closed if and only if $G$ is a divisible group.
This $k[x^G]$ is a valuation field whose valuation group is $G$; the valuation $v$ is defined by taking $v(f)$ to be the least $g$ such that $f(g) \neq 0$. Similarly, if $k$ is real closed, the valuation field $k[x^G]$ is real closed iff $G$ is divisible.
Of course, extensions are also important for the examples I just mentioned. I'm only saying that not "everything is ignored except for extensions"  there is a lot to say, depending on the sources you consult.