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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?

3nondatur
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    I suspect the question is incorrect. I tried to work it out, and seem to have a proof that $P^2$ is primary [I'm not yet ruling out the small possibility that I made some error]. If it were true, this would be a much simpler example than the multivariate polynomial ring ones that come up when you search for an example of a prime ideal whose square is not primary: e.g., https://math.stackexchange.com/questions/292362/elements-of-the-square-of-a-prime-ideal and https://math.stackexchange.com/questions/2694646/is-there-an-example-of-a-power-of-prime-ideal-in-a-polynomial-ring-that-is-not-p. – M. Vinay Mar 26 '21 at 03:55
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    dear @M.Vinay, I believe the key thing to note (at least in the second question you link to) is that that post asks about polynomial rings, and indeed the only examples I know of in polynomial rings are quite involved. however, the ring $R$ given by OP is not a polynomial ring, and indeed the $P^2$ in their example is not primary. (see the argument below, for example) – Atticus Stonestrom Mar 26 '21 at 06:16
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    @AtticusStonestrom Ah, I see. I must've gone wrong in my computation somewhere then. Thank you. – M. Vinay Mar 26 '21 at 08:08

1 Answers1

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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.

Atticus Stonestrom
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    Thanks for your answer, but there is a part that I do not get: You say that $a = 2x^2 \not \in P$, since the coefficient for $x$ in $a$ is not divisible by $4$. But it is, since it is $0$.

    Another point I do not get is: Why do the coefficients of $P^2$ need to be divisible by $4$ in the first place? If $f := a_nx^n+\ldots + a_0 ,g := b_mx^m+\ldots + b_0 \in P$ and $fg := c_{n+m}x^{n+m}+\ldots + c_0$ we have $c_1 = a_0b_1 + a_1b_0$, so $c_1$ should always be zero.

     – 3nondatur Mar 26 '21 at 10:38
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    I think you meant to write "the coefficient of $x^2$ in any element of $P^2$ must be divisible by $4$" (and the coefficient of $x$ should be zero, of course). – M. Vinay Mar 26 '21 at 13:02
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    @3nondatur Looks like a typo. See my comment above. And if you ask why, just multiply two arbitrary elements of $P$, and observe that the product is of the form $4a x^2 + 2 x^3 p(x) + x^4 q(x)$ ($a \in \mathbb Z$, $p(x), q(x) \in \mathbb Z[x]$), and even finite integer-linear combinations of such polynomials will be of the same form. Thus, all elements of $P^2$ are of this form. – M. Vinay Mar 26 '21 at 13:06
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    @3nondatur oops, apologies for the typo! M. Vinay is absolutely right, I indeed meant to say the $x^2$ term rather than the $x$ term. I've added some explanation to the answer and expanded on M. Vinay's comment, hopefully it helps :) – Atticus Stonestrom Mar 26 '21 at 14:46
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    (@M.Vinay you are of course absolutely right, thank you :) I have edited to correct!!) – Atticus Stonestrom Mar 26 '21 at 14:47