How to show that this ideal is not primary? I am stuck at the following exercise:

Let $R$ be the subring of $\mathbb{Z}[X]$ consisting of the polynomials whose
coefficient of $x$ is even. Let $P$ the ideal of those polynomials in $R$ whose constant
coefficient is zero. Then $P$ is prime, and $P^2$ is not primary.

I understand that $P$ is a prime ideal, but I do not see how I can show that $P^2$ is not be primary. I think one way to do this might be to show that $\sqrt{P^2}$ is not prime, but again I do not see how. Could you help me?
 A: Your proof approach is not quite right; in fact, for an arbitrary commutative ring $R$ and an arbitrary prime ideal $P<R$, we have $\sqrt{P^n}=P$ for any $n\in\mathbb{N}$. Indeed, certainly $P\subseteq\sqrt{P^n}$, since $p^n\in P^n$ for any $p\in P$. On the other hand, suppose $q\in \sqrt{P^n}$. Then $q^k\in P^n$ for some $k\in\mathbb{N}$, and hence, since $P^n\subseteq P$, in particular $q^k\in P$. Since $P$ is prime, this forces $q\in P$.
Now let's restrict to the $R$ and $P$ of your question. Consider the element $4x^2=(2x)^2\in P^2$, and decompose this element as a product $ab$, where $a=2x^2$ and $b=2$. We have $a\notin P^2$, since the coefficient of the $x^2$ term in any element of $P^2$ must be divisible by $4$. (Indeed, an arbitrary element of $P$ looks like $2m_1x+m_2x^2+m_3x^3+\dots$, where the $m_i$ lie in $\mathbb{Z}$. So, the product of two arbitrary elements of $P$ is of the form
\begin{align*}
&(2m_1x+m_2x^2+m_3x^3+\dots)(2n_1x+n_2x^2+n_3x^3+\dots) \\
=\text{ }&4m_1n_1x^2+2(m_1n_2+m_2n_1)x^3+(2m_1n_3+m_2n_2+2m_3n_1)x^3+\dots.
\end{align*} In particular, the coefficient of the $x^2$ term of such an element is divisible by $4$.) But also $b\notin\sqrt{P^2}=P$, since $b$ has non-zero constant term. Thus $P^2$ is not primary, as desired.
