Concrete examples of 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 $\mathrm{Spec}(A)$ is a noetherian topological space, but $A$ is non-noetherian. (Indeed, otherwise $A$ would be a discrete valuation ring.)
Is there a concrete example of such a ring $A$? 
 A: I'm late to this game, and Robin Chapman's elaboration of Qiaochu's example is good -- so good I just used it as a (counter)-example in my answer here. Whilst thinking about that, I realised there is another example which feels slightly more natural to me as someone who works with $p$-adics a lot:
Let $K$ be the field $\mathbb Q_p((X))$ of Laurent series over the $p$-adics. We have the obvious rank-one discrete valuation
$$w_1(\sum a_i X^i) = \min\{i: a_i \neq 0\}$$
like over any other field of Laurent series, just treating the coefficient field as being of zero valuation and identifying it with the residue field of the valuation ring $R_1 = \{x \in K: w_1(x) \ge 0\} = \mathbb Q_p[[X]]$ with respect to its maximal ideal $\mathfrak{m}_1 = \{x \in K: w_1(x) \ge 1\}= X \cdot \mathbb Q_p[[X]]$.
So far we have not used that $\mathbb Q_p$ has a valuation $v_p$ as well. Let's do it. Define a rank-two valuation $w_2 : K \rightarrow \mathbb Z \times \mathbb Z \cup \{\infty\}$ as follows: $w_2(0) := \infty$ and for $0 \neq x = \sum a_i X^i$ set
$$w_2(x) := \left(w_1(x), v_p(a_{w_1(x)})\right)$$
i.e. we "refine" the rank-one valuation $w_1$ by also keeping track of the $p$-adic valuation of the leading coefficient. Of course we give the value group the lexicographic order here. Check that the valuation ring to this is
$$R_2 = \{x \in K: w_2(x) \ge (0,0)\} = \{\sum_{i \ge 0} a_i X^i \in \mathbb Q_p[[X]]: a_0 \in \mathbb Z_p\};$$
and the maximal ideal of $R_2$ is
$$\mathfrak m_2= (p) = \{x \in K: w_2(x) \ge (0,1)\} = \{\sum_{i \ge 0} a_i X^i \in \mathbb Q_p[[X]]: a_0 \in p\mathbb Z_p\}.$$
Note that we have proper inclusions $R_2 \subsetneq R_1$ and $\mathfrak m_1 \subsetneq \mathfrak m_2$ (indeed, "the other way around"!), exactly as in the answer linked above. I found it worthwhile to think about how the valuation rings and their maximal ideals relate for this kind of "refinement" of a given valuation. One could obviously iterate the procedure.
It is also worthwhile to think how this example and Robin Chapman's / Qiaochu's are "almost the same". Actually, these are standard first examples of higher(-dimensional) local fields which have been studied in recent decades for their class field theory (Kato, Fesenko), and in connection to $p$-adic Langlands.
A: 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 $n$.
A: Take the ring of formal power series (over $\mathbb{C}$, say) with exponents in $\mathbb{Z}^2$ under lex order.
Edit:  As Robin Chapman mentions, one must be careful about what this means.  The precise construction for any totally ordered abelian group is described in the Wikipedia article.
