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I've been doing the exercises from Section 9.1 of Dummit and Foote and got stuck on the following problem:

Show that the radical of the ideal $I=(x, y^2)$ in $\mathbb{Q}[x, y]$ is $(x, y)$. Deduce that $I$ is a primary ideal that is not a power of a prime ideal.

I've done the first part of this exercise, including proving that $I$ is a primary ideal, but I'm not sure why I can deduce from that that $I$ is not a power of a prime ideal. I'm aware that this same exercise has been posted here before, but the OP was stuck on the first part of the problem, which I think I was able to solve.

My work so far can be seen below (which I would appreciate if someone could verify):
Any element in $(x, y)$ is of the form $f(x, y)=xg(x, y)+yh(x, y)$, so that $$f^2(x, y)=x^2g^2(x, y)+y^2h^2(x, y)+2xyg(x, y)h(x, y)=x[xg^2(x, y)+2yg(x, y)h(x, y)]+y^2[h^2(x, y)]$$ and, in particular, $f \in \sqrt{(x, y^2)}$. Conversely, if $f \in \sqrt{(x, y^2)}$, then $f^n \in (x, y^2)$ for some $n$. If $f$ had a constant term $c \neq 0$, then $f^n$ would have $c^n \neq 0$ as its constant term, a contradiction. Hence, $f$ has no constant term, so $f \in (x, y)$. Thus, $\sqrt{(x, y^2)}=(x, y)$. To show that $I$ is primary, let $fg \in (x, y^2)$ with $f \not\in (x, y^2)$. The polynomial $g(x, y)$ then can't have constant term different than zero, because that would mean $fg$ has either a constant term different than zero (in the case $f$ also has such a term) or a term of the form $ay$ for $a \in \mathbb{Q}$. Hence, $g \in (x, y)$, so $g^2 \in (x, y^2)$.

With that, I'm not sure why I can conclude that $I$ isn't a power of a prime ideal. I know that the fact that $I$ is primary implies $\sqrt{(x, y^2)}=(x, y)$ is prime, but not sure how to proceed from here. I would particularly appreciate if someone could give me a hint or provide an answer that doesn't use results that come after Section 9.1 of D&F.

Will199
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  • Your ideal $I$ sits properly between $(x,y)$ and $(x,y)^2.$ Note that $(x,y)$ is the smallest prime containing $I.$ – Chris Leary Aug 15 '21 at 14:48

1 Answers1

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Your proof looks correct, however there is a faster way to prove that $I$ is a primary ideal. Notice that

Let $R$ be a commutative ring. An ideal $I\subseteq R$ is primary $\iff$ every zero-divisor of $R/I$ is nilpotent

Proof: ($\Rightarrow$) Let $I$ be primary, and suppose that $x\in R/I$ is a zero divisor. Then $xy=0$, for some $y\in R/I$. We therefore have that either $x\in I$ or $y^n\in I$ for some $n$. That is, they are nilpotent.

($\Leftarrow$) Suppose every zero divisor of $R/I$ is nilpotent. Then if $xy=0$ either $x=0$ or $y^n=0$ for some $n$. That is, either $x\in I$ or $y^n\in I$ for some $n$.$\blacksquare$

Then it follows that every zero divisor of $\Bbb Q[x,y]/I\cong \Bbb Q[y]/(y^2)$ is inside $(y)$.

To prove that $I\neq \mathfrak p^n$, with $\mathfrak p\subseteq\Bbb Q[x,y]$ a prime ideal and $n\geq 2$, assume that $$I=\mathfrak p^n$$

Then notice that $I\subseteq\mathfrak p$, what can you then say about $\mathfrak p$?

cansomeonehelpmeout
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  • Since $y^2 \in \mathfrak{p}$, we have $y \in \mathfrak{p}$, so $(x, y) \subseteq \mathfrak{p}$. But $\mathfrak{p}$ cannot be the prime ideal $(x, y)$ because then $\mathfrak{p}^2 \subset I=\mathfrak{p}^n$, forcing $n=2$, a contradiction ($x \in I$ but $x \not\in (x, y)^2$). This tell us $(x, y)$ is properly in $\mathfrak{p}$. We can then take some polynomial $f \in \mathfrak{p} \setminus (x, y)$, meaning $f$ has a nonzero constant term $c$, which must hence be in $\mathfrak{p}$, implying $1=\frac{1}{c} c \in \mathfrak{p}$, which contradicts the primality of $\mathfrak{p}$. Would that be right? – Will199 Aug 15 '21 at 16:57
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    @Will199 Yes, this seems right! You are right that $(x,y)\subseteq\mathfrak p$. Notice that we also have $\mathfrak p\subseteq (x,y)$, since $(x,y)$ is a maximal ideal. Therefore $(x,y)=\mathfrak p=I$, which is a contradiction. – cansomeonehelpmeout Aug 15 '21 at 18:20
  • This makes total sense. Thank you very much for your help. I really appreciate it! – Will199 Aug 15 '21 at 18:49