Why nilpotent elements must be allowed in modern algebraic geometry? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:27:17Z http://mathoverflow.net/feeds/question/55244 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry Why nilpotent elements must be allowed in modern algebraic geometry? minimax 2011-02-12T22:16:37Z 2011-02-17T13:03:38Z <p>On the Wikipedia page about algebraic varieties <a href="http://en.wikipedia.org/wiki/Algebraic_variety" rel="nofollow">http://en.wikipedia.org/wiki/Algebraic_variety</a>, a sentence reads as follows:</p> <p>[[A more significant modification is to allow nilpotents in the sheaf of rings. A nilpotent in a field must be 0: these if allowed in coordinate rings aren't seen as coordinate functions.</p> <p>From the categorical point of view, nilpotents must be allowed, in order to have finite limits of varieties (to get fiber products).]]</p> <p>So I am wondering if there is an intuitive example to get non-reduced 'schemes' from reduced 'algebraic varieties' (probably by taking fiber product or alike)? </p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55246#55246 Answer by Donu Arapura for Why nilpotent elements must be allowed in modern algebraic geometry? Donu Arapura 2011-02-12T22:31:31Z 2011-02-12T22:37:03Z <p>The easiest example I can think of is to consider the map $\mathbb{A}^1\to \mathbb{A}^1$ by sending $x\to x^2$, where $x$ is the coordinate. Then take the fibre over the origin, which would be $Spec k[x]/(x^2)$. The idea is that $x=0$ in the fibre should be counted "twice".</p> <p>There is lot more that one can say about why this is a useful, and perhaps someone else will. But let me just refer you to the book by Eisenbud and Harris for further discussion.</p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55249#55249 Answer by Ramsey for Why nilpotent elements must be allowed in modern algebraic geometry? Ramsey 2011-02-12T22:44:38Z 2011-02-16T15:07:15Z <p>Here's an example that highlights why nilpotents are at least informative: Consider the intersection of the curves in $\mathbb{A}^2$ defined by $y=x^2$ and $y=0$. Set-theoretically, there is only the point $(0,0)$, but this doesn't account for the fact that the multiplicity of the intersection is $2$. </p> <p>However, one can phrase this simple case in terms of fiber products (as you request) as follows: Let $C$ denote the curve in $\mathbb{A}^2$ defined by $y=x^2$ and let $\pi:C\to \mathbb{A}^1$ denote the projection to the $y$ coordinate. Then the fiber product of $\pi$ along the inclusion of $0\hookrightarrow \mathbb{A}^1$ is precisely (the spectrum of) $K[x]/(x^2)$, and this exponent $2$ (which is the <em>ramification index</em> of $\pi$ at $(0,0)\in C$) "sees" this higher multiplicity via. nilpotence!</p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55250#55250 Answer by Harry Gindi for Why nilpotent elements must be allowed in modern algebraic geometry? Harry Gindi 2011-02-12T22:46:49Z 2011-02-12T23:22:47Z <p>Non-reduced schemes have the very interesting geometric property that they effectively equip points with "infinitesimal orientations" similar to directional derivatives in differential geometry. It's worth actually drawing some non-reduced affine curves and seeing how they behave by looking at the algebra. The motivation for the definition of formal smoothness, for instance, is based on this observation, namely that given a map $X_{red}\to Y$ of $S$-schemes, with $X$ an affine scheme, we want to be able to lift this map to a map $X\to Y$. That is, formal smoothness of $Y$ says that $Y$ is locally nice enough to accomodate infinitesimal deformations of maps into it from affines.</p> <p>I'm not sure that this was the original motivation for allowing nilpotents, but it's clear that nilpotents give us a much richer picture of the geometry.</p> <p>Edit: To avoid any confusion, please note that the thickenings of the form $X_{red}\to X$ will not always work if $X=Spec(A)$ does not satisfy good enough finiteness properties (in particular, the nilradical should be a nilpotent ideal, which can fail spectacularly away from Noetherian rings). In general, the requirement is that we have, for any square-zero nilpotent thickening of affine schemes over $S$, $Spec(T/J)\to Spec(T)$ (where $J^2=0$ is a nilpotent ideal of $T$) and any map of $S$-schemes $f:Spec(T/J)\to Y$, there exists a map of $S$-schemes $\tilde{f}:Spec(T)\to Y$ extending the map $f$. </p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55255#55255 Answer by Ramsey for Why nilpotent elements must be allowed in modern algebraic geometry? Ramsey 2011-02-12T23:08:05Z 2011-02-12T23:17:24Z <p>Elaborating on the first sentence of Harry's answer - here's another motivation for considering nilpotents. The scheme $D=Spec(K[t]/(t^2))$ set-theoretically consists of a single point, but this point has a non-trivial (one-dimensional) tangent space with a distinguished non-zero tangent vector. For a scheme $X/K$, a map $D\to X$ amounts to a point $x\in X$ and an element of the tangent space $T_x(X)$. </p> <p>This observation is used to characterize tangent spaces as fibers of the maps $X(D)\to X(Spec(K))$. This is useful in the study of Lie algebras of algebraic groups, for example.</p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55257#55257 Answer by Charles for Why nilpotent elements must be allowed in modern algebraic geometry? Charles 2011-02-12T23:54:03Z 2011-02-12T23:54:03Z <p>Another example of this necessity.</p> <p>If $G$ be an algebraic group, you would like it center $\mathcal{Z}(G)$ to be an algebraic group. </p> <p>But for example if you take $SL_n$ defined over a field $k$ (which is absolutely reduced), its center is $\mu_n=Spec\left(\frac{k[X]}{(X^n-1)}\right)$, which is not reduced if $char(k)$ divides $n$.</p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55258#55258 Answer by Sándor Kovács for Why nilpotent elements must be allowed in modern algebraic geometry? Sándor Kovács 2011-02-12T23:57:10Z 2011-02-13T00:46:56Z <p>Suppose you want to do moduli theory or to put it simpler, you are interested in deformations and degenerations. Often the degenerate objects have a natural non-reduced structure. In fact it is possible that taking the corresponding reduced scheme screws things up. </p> <p>Here are two simple examples:</p> <p><strong>Example #1</strong>: Consider the morphism $\mathbb A^2\to \mathbb A^1$ defined by $(x,y)\mapsto x^2$. The fibers are the curves defined by $x^2=\lambda$. For $\lambda\neq 0$ this is a parabola and for $\lambda=0$ a (double) line. If we only consider reduced schemes, then this is just a line, but otherwise we would expect that the members of a family of plane curves have the same intersection numbers (counted properly and also counting intersections at infinity) with other curves. Taking another line in general position one can see easily that the parabola intersects it in $2$ points while the line in only $1$. Considering the scheme theoretic fiber $x^2=0$ which is a double line resolves this problem.</p> <p><strong>Example #2</strong>: Let <code>$X=\{(1,\lambda t, t^2,t^3)\vert (t,\lambda)\in \mathbb A^2\}\subset \mathbb A^3$</code>. This is a surface defined "classically". Consider its projection to $\mathbb A^1$ by mapping the point $(1,\lambda t, t^2,t^3)$ to $\lambda$. Denote this by $f:X\to\mathbb A^1$. Still pretty classical. Now notice that the (classical=reduced) fiber of $f$ over $\lambda=0$ is a nodal cubic curve while for $\lambda\neq 0$ it is a twisted cubic. Also notice that this family can easily be compactified to be a projective family, so we get a family of $\mathbb P^1$'s degenerating to a projective nodal curve. However, without nilpotents this leads to severe headache.</p> <p>Since $X$ is irreducible and $\mathbb A^1$ is non-singular, $f$ should be <em>flat</em>. But fibers of a flat morphism have constant Hilbert polynomials, in particular their arithmetic genus is constant. The arithmetic genus of a twisted cubic (i.e., $\mathbb P^1$) is $0$ while that of a nodal cubic is $1$. If you want a completely classical argument, then one could say that the nodal cubic is also an obvious degeneration of non-singular plane cubic curves. Working over the complex numbers a plane cubic is a torus while a twisted cubic is a sphere. So this would suggest that it is possible to deform a sphere to a torus....</p> <p>The resolution of this dilemma is that if you compute the scheme theoretic fiber (using fibre products) then you'll see that the <em>correct</em> fibre over $\lambda=0$ is actually the nodal cubic, with a nilpotent sitting at the singularity. So the fibre is a non-reduced scheme and its arithmetic genus is $0$ so we can all sleep peacefully.</p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55259#55259 Answer by Andrea Ferretti for Why nilpotent elements must be allowed in modern algebraic geometry? Andrea Ferretti 2011-02-12T23:58:05Z 2011-02-13T00:04:28Z <p>As others have mentioned, nilpotent elements show up (at least) in the structure rings of varieties counted with multiplicities. Why should we want to have such objects? I can think of at least three reasons:</p> <ul> <li>The concept of family is easier to deal with. For instance, in the context of schemes, it is easy to speak of a family of conics degenerating to a double line. If we replace the double line with the same line counted once, the family behaves more badly (it is not flat)</li> <li>As rings have fibered coproducts (tensor products), schemes have fibered products. This is a general construction with good categorical properties, and it generalizes a variety of contexts (fibers, intersections, pullbacks of vector bundles...). If you want to stick with varieties, this construction will not be available, as the tensor products of reduced rings can be non-reduced</li> <li>A particular non-reduced scheme $k[x]/(x^2)$ is very useful in deformation theory. In deformation theory you want to study a given map up to the first order (or maybe higher orders, so rings like $k[x]/(x^n)$ appear). The existence of non-reduced schemes allows you indentify such an object (a first order approximation to some map) with an actual map from $k[x]/(x^2)$. This is quite handy and simplifies many arguments.</li> </ul> <p>One more reason to be happy in keeping schemes the way they are is the existence of the Quot scheme. This is a general construction due to Grothendieck which allows you to have schemes which parametrizes a manifold of objects: subschemes, morphisms and so on. Moreover, most other moduli space in use in algebraic geometry are constructed starting from a Quot scheme, typically as a GIT quotient.</p> <p>There is no corresponding general existence theorem in the context of varieties, so moduli theorists would have a pretty hard time abandoning schemes. Of course often it happens that the relevant Quot schemes are actually varieties, but we do not know how to construct them directly. It is easier to produce something (a scheme) and <em>then</em> show that it is nice (a variety), than producing a nice object in one step.</p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55279#55279 Answer by Emerton for Why nilpotent elements must be allowed in modern algebraic geometry? Emerton 2011-02-13T03:32:53Z 2011-02-17T13:03:38Z <p>I'm a bit confused by the quoted wikipedia entry, because the category of reduced rings also has coproducts (take the tensor product, and then pass to the quotient by the nilradical), and hence the category of reduced schemes, the category of varieties over a field, and so on, all admit fibre products. [Added: See Jim's Borgers series of comments below for a discussion of why, nevertheless, there may be a purely categorical description of the sense in which constructions in the category of reduced rings can be "wrong", while constructions in the category of all rings are the right ones.] </p> <p>So the answer to the question of why we need nilpotents is not that it is necessary for the existence of fibre products. </p> <p>Grothendieck introduced nilpotents for many reasons, a number of which are discussed in the other answers: to get correct counting in degenerate situations, it is typically necessary to allow nilpotents; they are also the bedrock of deformation theory and other applications of analytic ideas in algebraic geometry.</p> <p>It might be helpful to recall another motivation, which forms a significant part of Grothendieck's overall strategy for studying algebraic geometry: Suppose that we want to prove a property about a morphism $f: X \to S$. A typical approach is to first show that it is a local property, in some sense, so that we can reduce to the case when Spec $\mathcal O_{S,s}$ for some point $s \in S$, and hence assume that $S$ is local; and then to use a flat descent argument to pass from $\mathcal O_{S,s}$ to its completion, and thus assume that $S$ is the Spec of a complete local ring. We then write this complete local ring as the projective limit of the quotients by its maximal ideals, and so reduce to the case when $S$ is the Spec of an Artinian local ring. Since such a Spec has a single point, we can then hope to reduce to checking our property on the fibre over this one point, which reduces us to the case when $S$ is the Spec of a field.</p> <p>This is a powerful method, which absolutely requires us to be able to do geometry over an Artinian ring (and hence requires us to allow nilpotents). It comes up in lots of places, e.g. in establishing basic properties of abelian schemes, by reducing to the abelian variety case. See the anwers to <a href="http://mathoverflow.net/questions/16292/the-central-role-of-varieties-a-comment-from-mumfords-red-book/" rel="nofollow">this question</a> for some examples.</p>