What does primary mean geometrically? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:45:33Z http://mathoverflow.net/feeds/question/13248 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13248/what-does-primary-mean-geometrically What does primary mean geometrically? vandaele 2010-01-28T11:26:06Z 2010-01-29T04:48:05Z <p>Given a primary ideal I in a ring A, we can consider the subscheme V(I) of Spec(A). It is a nilpotentification (?) of the integral subscheme V(rad(I)) given by the radical rad(I) of I. My question is what kind of nilpotentifications you get that way. It is a vague question, but I'm asking because i am completely unable to understand the algebraic meaning of "primary ideal" (altho I learned it by heart) </p> http://mathoverflow.net/questions/13248/what-does-primary-mean-geometrically/13256#13256 Answer by Cam McLeman for What does primary mean geometrically? Cam McLeman 2010-01-28T13:02:24Z 2010-01-28T13:02:24Z <p>Geometrically, primary ideals (and most importantly, primary decomposition) gives you a way of visualizing, or at least identifying, embedded components in your variety/scheme. In fact, from the geometric point of view, it might be best to temporarily accept the definition of a primary ideal as slightly odd, make your way to the primary decomposition theorem, and reap the benefits of geometric intuition only after that point.</p> <p>To steal an example I found online, here's a primary decomposition of an ideal corresponding to a intersection of three varieties:</p> <p>$$I=\langle xy,x^3-x^2,x^2y-xy\rangle=\langle x\rangle\cap \langle x-1,y\rangle \cap \langle x^2,y\rangle$$</p> <p>so this intersection is seen to consist of the $y$-axis, the point $(0,0)$ ("embedded" on the $y$-axis), and the isolated point $(1,0)$ -- something not immediately discernible from the system of equations. Back to primary ideals, briefly: this nilpotentization you speak of is precisely the idea of giving extra fuzziness to this point $(0,0)$ as an embedded subscheme of the y-axis. So in some sense, it's just a slightly more nuanced version of the idea that, say, $(x,y)^2$ should correspond geometrically to a point of multiplicity 2, where you can now identify a point as a repeated point even though $(x,y)^2$ does not appear in the primary decomposition of this ideal.</p> http://mathoverflow.net/questions/13248/what-does-primary-mean-geometrically/13262#13262 Answer by Emerton for What does primary mean geometrically? Emerton 2010-01-28T17:00:35Z 2010-01-29T04:48:05Z <p>As Harry suggests in his answer, it is probably more intuitive to work with associated primes, rather than the slightly older language of primary decompositions. </p> <p>If $I$ is an ideal in $A$, an associated prime of $A/I$ is a prime ideal of $A$ which is the full annihilator in $A$ of some element of $A/I$. A key fact is that for any element $x$ of $A/I$, the annihilator of $x$ in $A$ is contained in an associated prime.</p> <p>The associated primes are precisely the primes that contribute to the primary decomposition of $I$. Geometrically, $\wp$ is an associated prime of $A/I$ if there is a section of the structure sheaf of Spec $A/I$ that is supported on the irreducible closed set $V(\wp)$. E.g. in the example given in Cam's answer, the function $x^2 - x$ is not identically zero on $X:=$ Spec ${\mathbb C}[x,y]/(x y, x^3-x^2, x^2 y - xy),$ but it is annihilated by $(x,y)$, and so is supported at the origin (if we restrict it to the complement of $(0,0)$ in $X$ then it becomes zero).</p> <p>The non-minimal primes of $I$ that play a role in the primary decomposition of $I$ (i.e. appear as associated primes of $A/I$) are the generic points of the so-called embedded components of Spec $A/I$: they are irreducible closed subset of Spec $A/I$ that are not irreducible components, but which are the support of certain sections of the structure sheaf.</p> <p>An important point is that if $I$ is radical, so that $A/I$ is reduced, then there are no embedded components: the only associated primes are the minimal primes (for the primary decomposition of $I$ is then very simple, as noted in the question: $I$ is just the intersection of its minimal primes).</p> <p>There is a nice criterion for a Noetherian ring to be reduced: Noetherian $A$ is reduced if and only if $A$ satisfies $R_0$ and $S_1$, i.e. is generically reduced, and has no non-minimal associated primes. Geometrically, and applied to $A/I$ rather than $A$, this says that if $A/I$ is generically reduced, then the embedded components are precisely the irreducible closed subsets of Spec $A/I$ over which the nilpotent sections of the structure sheaf are supported. This may help with your nilpotentification'' mental image.</p>