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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 ?

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

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  • $\begingroup$ Thank you very much ! I absolutely forgot about this result. $\endgroup$ – Suzet Nov 17 '18 at 14:26
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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 $k[x]$.

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