I feel like I understand the theory of ideals (at least at the basic level required for this), but struggle with actual computations. I am trying to figure out an example of two ideals $I, A$ in $\mathbb{C}[x,y]$, $I\subseteq A$, such that there does not exist another ideal $J$ such that $I=AJ$.

I proved that in a commutative ring with identity, if an ideal $I$ is contained in a principal ideal $(a)$, then there exists an ideal $J$ such that $I=(a)J$.

So in $\mathbb{C}[x,y]$, I thus need to find a non-principal ideal and look at ideals contained within that.

My first instinct is to go with $A=(x,y)$, and then consider $I=(x)$ or $I=(x-y)$, but in computing I am not creating any contradictions taking $J$ to be some finitely generated ideal.

Is there some higher-level theorem I could use to point myself in a better direction?

Edit: I imagine it may have something to do with polynomials of minimal degree perhaps being impossible in the product of $A$ and $J$, however that is only a hunch.


1 Answer 1


$I=(x)$ and $A=(x,y)$ work. In fact, let $J$ be such that $JA=I$. We have $$JA=xJ+yJ=(x)$$ and therefore $x\mid yf$ for all $f\in J$. This means that $x\mid f$ for all $f\in J$ and, therefore, that $J\subseteq (x)$. If $J\subsetneq (x)$, then $JA\subseteq J\subsetneq (x)$: hence, $J=(x)$ is necessary. But $IA=(x^2,xy)\ne (x)$, because, for instance, if $f\in (x^2,xy)$ then either $f=0$ or the least total degree of a monomial of $f$ is at least $2$.

  • $\begingroup$ Thank you so much! I like the notation $xJ+yJ$--that makes it a lot more manageable. I was getting too turned around in messy computation and I wasn't able to see the forest for the trees. $\endgroup$
    – mathpanda
    Nov 23, 2020 at 23:04
  • 1
    $\begingroup$ @mathpanda Yeah, really $[AJ\subseteq(x)\Rightarrow J=(x)]$ holds as soon as $x$ is prime and $A\ne (x)$. $\endgroup$
    – user239203
    Nov 23, 2020 at 23:08

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