# Product and quotient of ideals

I am trying to show the equality of two complex space germs related to double points of singular map germs $f:(\mathbb C^n,0)\to (\mathbb C^p,0)$. This two spaces are given by two (possibly non reduced) ideals, let's say $I,J$, in $\mathcal O_{2n+s}$. I can show $J\subseteq I$ and, if we call $K$ the ideal wich defines the diagonal of $\mathbb C^n\times \mathbb C^n$ (seen into $\mathcal O_{2n+s}$), I can show that $IK\subseteq JK$. Thus, my questions are:

(1) When does $IK\subseteq JK$ imply $I\subseteq J$?

(2) When does the equality $(JK):K=J$ hold?

(3) Would the statements in (1) and/or (2) be true, provided that $\mathcal O_{2n+s}/K$ is regular?

Thanks!

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Welcome to MO! This question is not phrased in a way as it is recommended here. (For detailed suggestions please see 'how to ask'.) More specifically, this question feels a bit formulated like an excercises. It is best to avoid doing this. Homework in a narrow sense is considered as off-topic; things formally similar to it can face (sometimes unjustified) problems. Thus, I would like to ask you to include some context and motivation for these questions. – user9072 Mar 6 '13 at 12:36
And to avoid a potential misunderstanding, let me add that I do not mean a general mathematical context or motivation (say definitions of the involved notions) but by contrast an explanation why you would like to know this. – user9072 Mar 6 '13 at 12:39
I'm going to refrain from answering this unless/until it's rewritten in response to quid's comments. – Steven Landsburg Mar 6 '13 at 12:46
Thank you for including the motivation! – user9072 Mar 6 '13 at 14:08
Thank you for the advice – Guillermo Peñafort Mar 6 '13 at 14:18

This is a partial answer. An ideal $K$ is a cancellation ideal if for any ideals $I$ and $J$, $IK=JK$ implies that $I=J$. Your question (1) says that $K$ is a cancellation ideal. It is known that a principal ideal $K=(a)$ in a commutative ring $R$ is a cancellation ideal if and only if $a$ is not a zero divisor.

Also by a result in the below paper $K$ is a cancellation ideal if $K$ is locally a regular principal ideal.

D. D. Anderson and Moshe Roitman; A Characterization of Cancellation Ideals. Proc. Amer. Math. Soc. 125 (1997), 2853-2854

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Thanks @Yazdegerd for the reference! Still, it is not necessary to my purpose that $K$ is a cancellation ideal, because I just need (1) to hold for fixed $I$ and $J$, not for arbitrary ideals. Thus, what are the conditions (involving both $I,J$ and $K$) for (1) to hold? – Guillermo Peñafort Mar 6 '13 at 22:14

There's a good chance you already know this, but $IK\subset JK$ at least implies $I\subset\sqrt{J}$.

Proof: Let $k_1,\ldots,k_n$ generate $K$. Then for any $x\in I$, we have $xk_\alpha=\sum j_{\alpha\beta}k_\beta$ for some $j_{\alpha\beta}\in J$. Putting these together gives a matrix equation $$(x\cdot 1-M)k=0$$

where $1$ is the identity matrix, $M$ has all its entries in $J$, and $k$ is the column vector consisting of the $k_\alpha$.

This implies that $(x\cdot 1-M)$ has determinant zero, but clearly this determinant is of the form $x^n-j$ with $j\in J$. So $x\in \sqrt{J}$.

I realize this is unlikely to be all you need, since you went out of your way to say that $J$ might be nonreduced.

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Thanks Steven for the interest, but I really need the equality of ideals, not of its zeros, that is, of its radical ideals. – Guillermo Peñafort Mar 7 '13 at 11:01

My point with question (2) was $IK\subseteq JK$ if and only if $I\subseteq(JK):K$ and, if we assume $(JK):K=J$, then $I\subseteq J$. Thus, under what circumstances a pair of ideals $J,K$ is such that $(JK):K=J$?

As a consequence of what @Yazdegerd III pointed out, in the local case the answer depends on both $J$ and $K$ unless $K$ is a regular principal ideal. It turns out that the ideal $K$ of the question is regular but not principal.

At least we know that the answer to (3) is negative.

Could someone provide a counterexample to $(JK):K=J$ with $K$ regular?

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Hi Guillermo,

If we mean an ideal generated by a regular sequence by "regular", then there is an example.

Let $R = k[x,y]$ be a polynomial ring over a field and $m = (x,y)R$. Let $J = (x^2, y^2)$ and $K = m^2$. Then one can check that $J K = m^4$. Notice that $JK : K = m^4 : m^2 = m^2$, but $xy$ is not in $J =(x^2, y^2)$. Hence $JK : K \neq J$.

I believe the questions you have might have a close connection to the integral dependence of ideals (or reduction of ideals). For instance, the condition (1) implies that the ideal $J$ is integral over $I$ if $K$ contains a regular element. The technique is exactly the same as in Steven's proof. Also, check out the m-full property which has to do cancellations in some cases. I recommend an excellent book by Huneke, Swanson on this subject for reference. Chapter 1 contains a good overview of the theory.

I would like to suggest you to move your answer to the question.

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Assume we're in an integral domain. (I realize your example is actually a polynomial ring over ${\mathbb C}$, but let's work in a more general domain for now.) Let's also suppose our domain to be an algebra over a field of characteristic $\neq 2$.

Let's look for ideals $J$ and $K$, and an element $x$ such that $xK\subset JK$ but $x\notin J$. (This would give counterexamples to both (1) and (2).)

This is surely impossible if $K$ is principal, so let's investigate the case where $K$ is generated by 2 elements.

Then I claim the following are equivalent:

1) The ring $S$ contains a counterexample to your (1) and/or (2) with $K$ two-generated.

2) The ring $S$ contains elements $A,B,C,D,F$ with $(A-D-F)(A-D+F)=4BC$ and $F\notin (A,B,C,D)$.

${\bf Proof:}$ Let $\alpha, \beta$ generate $K$. Then given a counterexample, we can write $$x\pmatrix{\alpha\cr\beta\cr}=\pmatrix{A&B\cr C&D\cr}\pmatrix{\alpha\cr\beta\cr}$$ for some $A,B,C,D\in J$, which we might as well assume generate $J$. Thus $x$ is an eigenvalue of the displayed two-by-two matrix and so satisfies its characteristic equation, whence there exists $F$ with $x=A-D-F$ and $(A-D-F)(A-D+F)=4BC$. Also, $(\alpha,\beta)$ must be the transpose of an eignvector, which we can take to be $(A+D-F,-C)$.

This will be a counterexample iff $x\notin J$, hence iff $F\notin J$. QED.

Thus, for algebras over a field $k$ of characteristic $\neq 2$, the universal counterexample is given by $$R=k[A,B,C,D,F]/((A-D-F)(A-D+F)-4BC)$$ $$x=A-D-F$$ $$J=(A,B,C,D)$$ $$K=(A+D-F,-C)$$

Your ring $S$ will contain a counterexample (with $K$ two-generated) iff it contains a homomorphic image of $R$ in which $F\notin (A,B,C,D)$. When $S$ is a polynomial ring, I'm not sure whether this is the case but it might not be too hard to settle.

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