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This question already has an answer here:

How to show that when $K$ is a field, $K[X^2,X^3]$ is not factorial? What in fact is the ring $K[X^2,X^3]$, is this all the polynomials with exponents $2,3,6,8,9,...$?

We need to show that there is some $f\in K[X^2,X^3]$ such that:

$$f\neq uf_1f_2f_3....f_n$$

Where $u$ is a unit.

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marked as duplicate by rschwieb abstract-algebra Jan 29 at 20:56

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  • $\begingroup$ $K[X^2, X^3]$ is the subring of $K[X]$ consisting of all polynomials which can by "built" using $X^2$ and $X^3$. In particular, this also involves $X^5 = X^2 \cdot X^3$, which is missing in your list. $\endgroup$ – red_trumpet Jan 29 at 20:45
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    $\begingroup$ @red_trumpet Taking the thought one step further, you will get every $x^n$ where $n>1$. $\endgroup$ – Aaron Jan 29 at 20:48
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    $\begingroup$ $K[X^2,X^3]$ is not integrally closed ($X\in \text{Frac}(K[X^2,X^3])$ is integral over $K[X^2,X^3]$ but doesn't belong to it). In particular it can't be factorial. $\endgroup$ – yamete kudasai Jan 29 at 20:51
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Hint: Consider $(X^2)^3=X^6=(X^3)^2$.

It seems that you're trying to find an element that cannot be decomposed into irreducibles. You won't find any because $K[X]$ is factorial. You'll need to disprove uniqueness of factorization. Hence the hint.

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