# Ideals in $\mathbb{C}[x,y]$ which are not the product of another ideal

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.

$$I=(x)$$ 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 $$2$$.
• 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. Commented Nov 23, 2020 at 23:04
• @mathpanda Yeah, really $[AJ\subseteq(x)\Rightarrow J=(x)]$ holds as soon as $x$ is prime and $A\ne (x)$.