TITLE: Are cluster variables prime elements?
QUESTION [10 upvotes]: Cluster algebras introduction
A cluster algebra is a subalgebra $A$ of $k[x_1^{\pm1},...,x_n^{\pm1}]$ generated by a set of cluster variables, which are elements which can be generated from the set $\{x_1,...,x_n\}$ by a certain recursion determined by a skew-symmetrizable matrix $M$ (mutation).
The mutation of the pair (called a seed)
$$((x_1,x_2,...,x_i,...,x_n),M)$$
at the $i$-th place is the seed
$$((x_1,x_2,...,\mu_i(x_i),...,x_n),\mu_i(M))$$
$$\mu_i(x_i)=\left[\prod_{j,M_{ij}>0} x_j^{M_{ij}}+\prod_{j,M_{ij}<0}x_j^{-M_{ij}}\right]x_i^{-1}$$
and $\mu_i(M)$ is a new skew-symmetrizable matrix I cannot get mathjax to display.
Mutation of seeds may be iterated arbitrarily.  It is a non-trivial theorem that the functions in the seed are always Laurent polynomials.  A cluster variable is any function appearing in a seed obtained by a sequence of mutations.
Ring-theoretic properties of cluster variables
As elements in $A$, each cluster variable $f$ is an irreducible element (it can't be factored).  This is not hard to show; it follows from the Laurent embedding for any cluster containing $f$, and the observation that $A$ does not contain any Laurent monomials with negative powers of (non-frozen) variables.
So then I ask, are cluster variables prime elements in $A$?
The analogous argument to the irreducible case doesn't seem to work, since one needs to consider more general Laurent polynomials in a given cluster, and there is no nice criterion for telling when a general Laurent polynomial is in the cluster algebra.  It also seems unlikely that $A$ is a UFD in general, which would be a standard trick for deducing primality for irreducibility.  Nonetheless, the examples I checked in Macaulay 2 were all prime.

REPLY [6 votes]: Boo.  The answer is no.  Consider the Markov cluster algebra, whose corresponding matrix is
$$\left[\begin{array}{ccc} 0 & -2 & 2 \\ 2 & 0 & -2 \\ -2 & 2 & 0 \end{array}\right]$$
For an initial cluster of $\{x_1,x_2,x_3\}$, the mutation relation at 2 is
$$ x_2x_2'=x_1^2+x_3^2$$
If the ground field has a square root $i$ of $-1$, then
$$ x_2x_2'=(x_1+ix_3)(x_1-ix_3)$$
Thus $x_2$ is not prime.