$K[X^2,X^3]$ is a non-unique factorization ring

Let $$K$$ be any field. I would like to prove that any element of $$K[X^2,X^3]$$ can be written as a product of irreducible elements, in a possibly non-unique fashion. The non-unique part can be easily proven when noticing that $$X^6 = (X^2)^3=(X^3)^2$$ and given that $$X^2$$ and $$X^3$$ are both irreducible in $$K[X^2,X^3]$$ (writing any of them as a product of two non-invertible elements would lie a factor of degree $$1$$, which can not be an element of $$K[X^2,X^3]$$).

However, I can't find how to prove the existence of such a decomposition. Could someone please give a hand for this exercise ?

• Isn't that ring Noetherian? Then you can use the fact that every element in a Noetherian ring is product of irreducibles. – Bias of Priene Nov 17 '18 at 14:11
• I was not aware of this result - or rather, I totally forgot it. Thank you very much ! – Suzet Nov 17 '18 at 14:25

The $$K$$-algebra $$K[X^2,X^3]$$ is finitely generated over $$K$$, so by Hilbert's basis theorem, it is noetherian. In particular, it satisfies ACC on principal ideals, hence every element has a factorization into irreducibles.
Another approach: unique factorization domains are normal, that is, integrally closed in their field of fractions. But the field of fractions of your domain is $$k(x)$$, and the closure of your domain in this is $$k[x]$$.