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!
 A: 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.  
A: 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 
A: 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?
A: 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. 
A: 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.  
