What are the prime ideals of k[[x,y]]? Let $k$ be a field. Then $k[[x,y]]$ is a complete local noetherian regular domain of dimension $2$. What are the prime ideals?
I've browsed through the paper "Prime ideals in power series rings" (Jimmy T. Arnold), but it does not give a satisfactory answer. Perhaps there is none. Of course you might think it is more natural to consider only certain prime ideals (for example open/closed ones w.r.t. the adic topology), but I'm interested in the whole spectrum.
A first approximation is the subring $k[[x]] \otimes_k k[[y]]$. If we know its spectrum, perhaps we can compute the fibers of $\text{Spec } k[[x,y]] \to \text{Spec } k[[x]] \otimes_k k[[y]]$. Now the spectrum of the tensor product consists of $(x),(y),(x,y)$ and $\text{Spec } k((x))  \otimes_k k((y))$. The latter one is still very complicated, I think. For example we have the kernel of $k((x)) \otimes_k k((y)) \to k((x))$. Also, for every $p \in k[[x]]$, we have the prime ideal $(y - p)$.
 A: You will find the proof of the fact that $k[[x_1,\dots,x_n]]$ is a UFD in Shafarevish, Basic algebraic geometry, Chapter II, 3.1. The proof uses the Weierstrass preparation lemma. The same section contains the proof that a codimension 1 prime ideal is principal. The only codimension prime ideal of codimension 2 is $(x,y)$ of course, since it has to be contained in $(x,y)$. Miles Reid's Undergraduate commutative algebra contains the proof as well, I think.
A: The ring $k[[x,y]]$ is a local UFD of dimension 2; so its prime ideals are the zero ideal, the maximal ideal, and all the ideals generated by an irreducible element. They are all closed (all ideals in a noetherian local ring are closed). Classifying them is an extremely complicated business, already when $k = \mathbb C$.
The ring $k[[x]] \otimes_k k[[y]]$ is truly nasty, it is not even noetherian, and I doubt it would help.
A: I really come long after the battle on this one, but I have something to add to Angelo's answer. To rephrase what Angelo said, there is only one prime ideal of height 2, namely the maximal ideal $(x,y)$, and one of height $0$. The problem is to classify the prime ideals of height one which (as in any UFD) are exactly the principal ideals generated by an irreducible element $f(x,y)$.
Now there are several possible meaning for "classify", but one reasonable sense is
to describe all such prime ideal up to an automorphism of $k[[x,y]]$ (automorphism of $k$-algebra, I mean, of course.) 

Theorem: if $k=\mathbb C$, then up to a an automorphism of $k[[x,y]]$ any prime ideal of height one is of the form $(f(x,y))$ where $f(x,y)$ is an irreducible polynomial in two variables.

A proof for this theorem (that is, a combination of two references that do the job) can be found in my answer in the following question. I think that the theorem is true for any field $k$, but one of the references used there use $k=\mathbb C$.
EDIT: I give a little bit more explanation as requested. Let start with $f(x,y)$ an irreducible element. One has $f(0,0)=0$ otherwise $f$ would be a unit. Then a theorem
of Artin (references in the question linked above) tells that up to a change of variable
$x \rightarrow x'$, $y \rightarrow y'$, one can assume that $f(x,y)=g(x',y')$ will be analytical, that is the power series $g$ converges on some neighborhood of $(0,0)$. 
In algebraic terms, the change of variables defines an automorphism $\psi$ if $k[[x,y]]$ sending $x$ to $x'$ and $y$ to $y'$, and $g = f \circ \psi^{-1}$. This theorem is actually much more general: it works in any dimension and over any field, with analytic replaced then by "in the Henselization of the subring of polynomials". In the case of dimension $2$ over $\mathbb C$, i would not be surprised if it was much older (Puiseux? Newton?). 
The second step is proving that analytic $g(x,y)$ can be changed, up to a second change of variable to an $h(x,y)$ which is simply a polynomial. The reference given in the linked question is to a book of complex geometry, hence is over $\mathbb C$, but the result 
may be true over any field. However it is not true in three and more variable, as explained in Moret-Bailly's answer to this question. 
So we have an automorphism $\psi$ such that $h = f \circ \psi^{-1}$. Since $(f)$ is prime, so is $h$, hence $h$ is irreducible as a power series, hence as a polynomial.
To make this more concrete, let me gives two examples. if $f(x,y) =a x + b y$ + higher terms, with $a$, or $b$, or both, non zeros (This condition is equivalent to $A/(f)$ being regular, or a discrete valuation ring.) Say $a \neq 0$, then one can take for new variables $x'=f(x,y)$, $y'=y$, and $h(x,y)$ is just $x$.
If on the contrary $a=b=0$, looking at the quadratic terms of $f$ gives a quadratic forms over $k$. If this quadratic form $h$ is non-degenerate, then by the theory of Morse function $f$ can be transformed to $h$ (I guess $\operatorname{char} k \neq 2$ may be necessary).      
