What does primary decomposition of (sub) modules mean geometrically? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:38:09Z http://mathoverflow.net/feeds/question/7313 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7313/what-does-primary-decomposition-of-sub-modules-mean-geometrically What does primary decomposition of (sub) modules mean geometrically? Andrew Critch 2009-11-30T21:21:22Z 2010-11-17T05:42:49Z <p>I want to know how I should visualize <em>modules</em> in algebraic geometry. The way we visualize <em>rings</em>, via their spectra, automatically (or by the beauty of its design) depicts primary decomposition of <em>ideals</em>: the primary components of an ideal $I \triangleleft A$ cut out "primary subschemes" (irreducible and embedded components) whose union is $Z(I)=Spec(A/I)$. (See, for example, Eisenbud and Harris, <em>The Geometry of Schemes</em>, II.3.3, pp. 66-70). This aspect of scheme theory is essential to what makes it "geometric."</p> <p>By this standard, I think however we visualize <em>modules</em> should allow us to depict primary decomposition of <em>submodules</em>; otherwise I would say it's not a very good visualization. </p> <p>If we're happy taking quotients, WLOG we can just look at primary decompositions of $0$. So let $M$ be a finitely generated module over a Noetherian ring $A$, and $0=N_1\cap\cdots\cap N_n$ be a primary decomposition of $0$ in $M$, with primes $P_i$ co-associated to the primary modules $N_i$, i.e. associated to the coprimary modules $M/N_i$.</p> <blockquote> <p>How can one visualize the modules $M,N_1,\ldots,N_n$ in relation to $Spec(A)$ in a way that meaningfully depicts: <br><b>(1)</b> the primary decomposition of $0$ in $M$ (in particular that the $N_i$ are primary in $M$), and <br><b>(2)</b> the relationship of the modules $N_i$ to their co-associated primes, say <br>{ $P_i$ } $= Ass(M/N_i) \subseteq Spec(A)$?</p> </blockquote> <p>Some useful background results to make sense of the above (all rings and modules are Noetherian):</p> <ul> <li><p>The primes $P_i$ co-associated to $N_i$ are precisely the associated primes of $M$ (see <a href="http://www.math.uiuc.edu/~r-ash/ComAlg/ComAlg1.pdf" rel="nofollow">R. Ash, <em>Comutative Algebra</em></a>, Theorem 1.3.9)</p></li> <li><p>A module $Q$ is coprimary iff it has exactly one associated prime $P$, and then $P=\sqrt{ann Q}$. (see <a href="http://www.math.uiuc.edu/~r-ash/ComAlg/ComAlg1.pdf" rel="nofollow">R. Ash, <em>Comutative Algebra</em></a>, Corollary 1.3.11)</p></li> </ul> http://mathoverflow.net/questions/7313/what-does-primary-decomposition-of-sub-modules-mean-geometrically/8080#8080 Answer by Greg Kuperberg for What does primary decomposition of (sub) modules mean geometrically? Greg Kuperberg 2009-12-07T05:38:03Z 2009-12-07T05:43:17Z <p>I will assume that everything in sight is Noetherian and finitely generated. Then the primary decomposition amounts to a description of the geometric support of the module $M$ if you view it as a sheaf on $\text{Spec}(A)$. The scheme $\text{Spec}(A/\text{Ann}(M))$ is a subscheme of $\text{Spec}(A)$ that, by definition, supports $M$. If $N_i$ is minimal, then $\text{Spec}(A/P_i)$ is an irreducible component of $\text{Spec}(A/\text{Ann}(M))$. The module $M/N_i$ is also a quotient of $M \otimes (A/P_i^n)$ for $n$ large enough (and in a reduced decomposition I think they are equal), so it is part of $M$ on that irreducible component of its support. If $N_i$ is not minimal, then $\text{Spec}(A/P_i)$ is an irreducible scheme inside of a component of $\text{Spec}(A/\text{Ann}(M))$.</p> http://mathoverflow.net/questions/7313/what-does-primary-decomposition-of-sub-modules-mean-geometrically/46331#46331 Answer by roy smith for What does primary decomposition of (sub) modules mean geometrically? roy smith 2010-11-17T05:42:49Z 2010-11-17T05:42:49Z <p>Visualizing embedded primes: In P^2, a one dimensional scheme cannot have embedded points unless its ideal has more than one generator, by the unmixedness theorem of Macaulay. So imagine we have two polynomials that define a one dimensional scheme in P^2. We will imagine this scheme as a limit of zero dimensional schemes. First take two quadratic polynomials, one of which is a product of two linear factors, i.e. take one pair of lines meeting at p, and another irreducible conic. In general the irreducible conic C meets each of the lines twice, away from p. Thus the two qudratic polynomials define a zero dimensional scheme of 4 points. </p> <p>Now hold fixed the two intersections of C with one of the lines L, and let the two intersections of C with the other line M approach p, i.e. let C become tangent to M at p. When this occurs, the conic C now contains three distinct points of L, hence C has become reducible and contains L. Now the scheme defined by intersecting L+M with C has become one dimensional, reducible, and consists set theoretically only of the line L. I claim the point p is an embedded point of the component L of the scheme defined by L+M and C.</p> <p>This is easy algebraically, since the ideal of the given scheme is (xy,(x(x-y)) = (x^2, xy) which is the intersection of the primary ideals (x) and (x^2, xy, y^2), with associated primes (x) and (x,y). Hence (x,y) is an embedded prime. I.e. the origin is an embedded point on the y axis for this scheme. This also helps explain the apparent failure of Bezout's theorem for this intersection of two conics apparently not having degree 4.</p> <p>In general, in P^n, a scheme S with embedded subschemes must be defined by intersecting more hypersurfaces than the codimension of S. Thus such an S can always be viewed as a limit of lower dimensional schemes. It seems to me that embedded subschemes should arise when these lower dimensional schemes are reducible and some lower dimensional component comes to lie on a larger dimensional component of the limit. I do not know if this intuition is the only possibility, and since the world is wide, probably not.</p>