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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.

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Only if it is short, can you describe 'mutation' and how cluster variables are generated? –  Mahdi Majidi-Zolbanin Aug 16 '11 at 22:08
    
I tried to add an explanation, Mahdi, but I had difficulty getting the matrix mutation to display. A reasonably compact definition can be found in Section 2.1. of arxiv.org/abs/math/0309138 –  Greg Muller Aug 16 '11 at 22:37

1 Answer 1

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.

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If you enclose the expression by backticks the matrix will display properly. –  Gjergji Zaimi Aug 17 '11 at 7:19
    
Fixed; thank you. –  Greg Muller Aug 17 '11 at 16:44

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