Showing that $X^2$ and $X^3$ are irreducible but not prime in $K[X^2,X^3]$

Question

Show that $X^2$ and $X^3$ are irreducible but not prime in $K[X^2,X^3]$.

My reasoning is as follows:

Since the only ways $X^2$ and $X^3$ can be factored in $K[X^2,X^3]$ are $X^2 = X^2 * 1$ and $X^3=X^3*1$ it follows that $X^2$ and $X^3$ are irreducible.

To show that they are not prime in $K[X^2,X^3]$ I argue that $X^2\mid X^3*X^3$ but $X^2\nmid X^3$ and $X^3\mid X^4*X^2$ but $X^3\nmid X^4$ or $X^3\nmid X^2$.

Is that correct?

Use that $K[X]$ is a unique factorization domain, and $K[X^2,X^3]$ is a subring, to prove they are irreducible. Writing "can be factored ..." is purely assertion. — Thomas Andrews, Nov 20 '15 at 18:15
could i enquire about the reference Where did you come across this problem? ... I am preparing for grad exams and this might be helpful. — spaceman_spiff, Jul 21 '16 at 18:35

score 0 · Accepted Answer · answered Dec 07 '15 at 06:24

You can do as Thomas said, by appealing to unique factorization in the larger ring, but here it is simple enough to do it directly. By definition of product, the degrees of the factors must sum to the degree of the product. Now you can prove by induction that $K[X^2,X^3]$ does not have any element with a nonzero coefficient for $X^1$, or alternatively you can prove that $K + X^2 K[X]$ is a ring and hence contains $K[X^2,X^3]$ since it already contains $X^2,X^3$. Then clearly $X^2,X^3$ cannot have any factor of degree $1$, and so you are quite done. Similarly, you can justify that $X^2 \nmid X^3$ and $X^3 \nmid X^4$ because the other factor must have degree $1$.

@user26857: That's true, but I was using the bare minimum necessary. In general the polynomials in $K[X^a,X^b]$ are exactly those with zero $X^c$ coefficient for every $c$ that is not a non-negative integer linear combination of $a,b$. — user21820, Mar 23 '16 at 00:59

Showing that $X^2$ and $X^3$ are irreducible but not prime in $K[X^2,X^3]$

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