$K[X,Y]$ is a PID and a primary ideal in it is not power of a maximal ideal?

Question

I wonder if $M$ is a maximal ideal, $Q$ is an ideal of $R$ and $\sqrt{Q}=M$ then $Q$ is an $M$-primary ideal.

The converse is not true (I know that). We also have that if $R$ is PID which is not field then the set of all primary ideals (maximal ideals) of $R$ is $\{0\}$ and $Rp^{n}$ with $p$ irreducible. It means that the primary ideals in PID are $\{0\}$ and powers of maximal ideals. But I see the example that is if $K$ is a field, take $(X,Y^{2})$ in $K[X,Y]$. It is an $M=RX+RY$-primary ideal of $R$, but it is not a power of maximal ideal (specifically $Rp^{n}$). I see it is contraction because $K[X,Y]$ is a PID and the primary ideal in it is not form of maximal ideal ? Can you explain it?

Answer 1

Yes, if $Q$ is an ideal and $\sqrt{Q}$ is maximal then $Q$ is primary. This is proposition 4.2 in Atiyah and MacDonald.

Answer 2

Theorem: Let $R$ be a commutative ring with identity. Then $R [X]$ is PID if and only if $R$ is a field.

Clearly, $X $ is not unite in $R [X]$, for every commutative ring. Thus, the above Theorem shows that $R [X,Y]=R [X][Y]$ is not a PID, for any commutative ring with identity.