When is the product of two ideals equal to their intersection? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:38:11Z http://mathoverflow.net/feeds/question/49259 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection When is the product of two ideals equal to their intersection? evgeniamerkulova 2010-12-13T13:50:27Z 2013-03-06T15:54:43Z <p>Consider a ring $A$ and an affine scheme $X=SpecA$ . Given two ideals $I$ and $J$ and their associated subschemes $V(I)$ and $V(J)$, we know that the intersection $I\cap J$ corresponds to the union $V(I\cap J)=V(I)\cup V(J)$. But a product $I.J$ gives a new subscheme $V(I.J)$ which has same support as the union but can be bigger in an infinitesimal sense. For example if $I=J$ you get a scheme $V(I^2)$ which is equal to "double" $V(I)$.</p> <p>Vague Question : What is geometric interpretation of $V(I.J)$ in general?</p> <p>Precise question : When is $I\cap J=I.J$? Everybody knows the case $I+J=A$ but this is absolutely not necessary. For example if $A$ is UFD and $f,g$ are relatively prime then $(f).(g)=(f)\cap(g)$ but in general $(f)+(g)\neq A$ (e.g. $f=X, g=Y \in k[X, Y]$)</p> <p>Thank you very much. </p> http://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection/49261#49261 Answer by David Speyer for When is the product of two ideals equal to their intersection? David Speyer 2010-12-13T14:14:45Z 2010-12-13T14:14:45Z <p>Answer to the precise question: When $\mathrm{Tor}^1(A/I, A/J)=0$. </p> <p><strong>Proof:</strong> We have the exact sequence <code>$$0 \to I \to A \to A/I \to 0$$</code> Tensoring with $A/J$, we get <code>$$0 \to \mathrm{Tor}^1(A/I, A/J) \to I/(I \cdot J) \to A/J \to A/(I+J) \to 0.$$</code> The left hand term is $0$ because $A$ is flat as an $A$-module.</p> <p>Now, what is the kernel of $I \mapsto A/J$? Clearly, it is $I \cap J$. So the kernel of $I/(I \cdot J) \to A/J$ is $(I \cap J)/(I \cdot J)$. We see that $I \cap J = I \cdot J$ if and only if $\mathrm{Tor}^1(A/I, A/J)=0$.</p> http://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection/49299#49299 Answer by Hailong Dao for When is the product of two ideals equal to their intersection? Hailong Dao 2010-12-13T19:23:21Z 2010-12-13T19:34:20Z <p>To add to David Speyer's answer, since this story continues with a rather interesting and illustrious history:</p> <p>When $A$ is regular, the Tor functor satisfies the following property:</p> <blockquote> <p>(1) $\text{Tor}_1^A(M,N) = 0$ implies $\text{Tor}_i^A(M,N) = 0$ for $i>0$ for any two finitely generated modules. </p> </blockquote> <p>(this is a theorem by Auslander in the geometric and unramified case and Lichtenbaum in the ramified case. (1) is called the <em>rigidity</em> of Tor). </p> <p>It turns out that when $A$ is regular and local (so one can talk about depth), (1) implies </p> <blockquote> <p>(2) $\text{depth} (M) + \text{depth}(N) = \dim A + \text{depth} {M\otimes_AN}$</p> </blockquote> <p>This stunning formula looks exactly the same as the property of "proper intersection" in intersection theory, except that one uses depth instead of dimension. Note that if $M=A/I, N=A/J$ then $M\otimes N = A/(I+J)$, which represents the intersection of $V(I)$ and $V(J)$, so this is very geometric. </p> <blockquote> <p>(3) Talking about intersection theory, by <a href="http://en.wikipedia.org/wiki/Serre_multiplicity_conjectures" rel="nofollow">Serre formula for intersection multiplicity</a>, as all the Tors vanish, one can compute the intersection multiplicity of $V(I), V(J)$ by counting the length at the minimal components (i.e. the naive way). So you will have a generalization of Bezout theorem.</p> </blockquote> <p>Finally, if $V(I)$ and $V(J)$ only intersect at isolated closed points, (2) implies (1) locally on the support of the intersection, so </p> <blockquote> <p>(4) If <code>$V(I) \cap V(J)= \{m_1, \cdots, m_n \}$</code> then $I\cap J = IJ$ if and only if $A/I, A/J$ are locally Cohen-Macaulay at the points $m_i$s. </p> </blockquote> <p>You can find the last statement in Serre's Local Algebra book, V.6, Theorem 4, p 110 of the English version.</p> <p>PS: Also, David did not mention his own interesting contribution, <a href="http://front.math.ucdavis.edu/0601.5202" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection/123777#123777 Answer by Guillermo Peñafort for When is the product of two ideals equal to their intersection? Guillermo Peñafort 2013-03-06T15:54:43Z 2013-03-06T15:54:43Z <p>A vague answer to the vague question:</p> <p>When you want the union of $V(I)$ and $V(J)$ to behave well under deformations and to `count with multiplicity', then you may prefeer to use the ideal $IJ$ rather than $I\cap J$. Let me give an example:</p> <p>Take $V=V(x)$, $W=V(x-t)$ and $T=V(t)$ denote $V_0:=V\cap T=V(x,t)=W\cap T=:W_0$. If you use intersection of ideals for the union of varieties you will get:</p> <p>$(V\cup W)\cap T=V(x^2)$, and</p> <p>$(V_0\cup W_0)\cap T=V(x)$.</p> <p>While using product you will get:</p> <p>$(V\cup W)\cap T=V(x^2)=(V_0\cup W_0)\cap T$.</p>