Valuation rings of rank two

Let $A$ be a valuation ring of rank two. Then $A$ gives an example of a commutative ring such that $Spec(A)$ is a noetherian topological space, but $A$ is nonnoetherian. (Indeed, otherwise $A$ would be a discrete valuation ring.)

Is there a concrete example of such a ring $A$?

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Qiaochu's answer is sound in principle, but in practice one needs to be more careful with the definition of the ring. The quotient field $K$ of $A$ consists of formal Laurent series of the form

$f=\sum_{r=-r_0}^\infty x^r\sum_{s=-s_0(r)}^\infty a_{r,s}y^s.$

Here $r_0$ is an integer and for each integer $r$, $s_0(r)$ is an integer (depending on $r$). So these are the power series where the powers of $x$ are bounded below and for each integer $r$ the coefficient of $x^r y^s$ is zero for all $s$ below a bound depending on $r$. This complicated-looking condition ensures that the product of two elements of $K$ is also an element of $K$ (note that one cannot multiply two general Laurent series).

Then $A$ will consist of all such series with the additional conditions that $r_0=0$ and $s_0(0)=0$. The valuation of an element $f$ is the least $(r,s)$ under lexicographic ordering with $a_{r,s}\ne0$. Here the ordering is $(r,s)<(r',s')$ if $r < r'$ or $r=r'$ and $s < s'$.

A more high-brow interpretation of the condition for memebership of $K$ is that the support of $f$, the setof $(r,s)$ for which $a_{r,s}\ne0$, should be well-ordered, that is each subset of the support has a least element. (With repsct to this lexicographic ordering of course.)

By considering a version of this construction in $n$ variables one can construct explicitly a ring with a valuation of rank $n$.

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Thanks. Do we actually need to consider formal Laurent series though? As in, wouldn't it be sufficient to restrict to the intersection of this ring with $K(x,y)$? –  compact3dmanifold Jul 31 '10 at 22:53
Take the ring of formal power series (over $\mathbb{C}$, say) with exponents in $\mathbb{Z}^2$ under lex order.