# Proof of Laurent Phenomenon for Cluster Algebras

I went through the proof of the Laurent phenomenon for Cluster Algebras in Fomin and Zelevinsky's initial paper: Cluster Algebras I: Foundations. I am stuck at their claim that the gcd of two exchange polynomials is 1.

I am working in a setup, where all coefficients are set to 1, thus the coefficient group is just the group of integers (therefore the cluster algebras are of geometric type). But why do the exchange polynomials have to be different? In my opinion their gcd could be a binomial in the cluster variables. This would not be a unit in the ring of Laurent polynomials. What am I missing?

Cheers

• There are some partial results about coprime exchange polynomials, if they are stemming from a sign-skew symmetric matrix with full rank as stated in Bernsteins, Fomin and Zelevinskys paper Cluster Algebras III: Upper Bounds and double Bruhat cells. However not all Cluster Algebras (of geometric type) have such matrices. Nonetheless the authors claim that cluster Algebras of geometric type are coprime. Does anyone have an idea why they ignore for instance the possibility of equal exchange polynomials? – Confused Jul 8 '16 at 9:06

See the discussion on page 11 within the proof of Theorem 3.1 for the universal coefficients set up. Essentially what is done is a polynomial ring with an indeterminate for each for each coefficient in the exchange polynomials. That is $\mathbb P$ is the free abelian group generated by a certain collection of coefficients, and then we look at $\mathcal{L}_0$ which is a Laurent polynomials ring over $\mathbb Z \mathbb P$. It is only in the ring of Laurent polynomials with these coefficients the coprime condition happens. Note with this set up exchange polynomials which were the same before become distinct.