A Mind for Madness

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Infinitesimal Lifting Property

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I’ve basically recuperated from my test and I’m trying to get back into the AG frame of mind. I have about 5 posts half written, so I’m going to actually try to finish this one and start up a nice little series. I’m taking a class on deformation theory this quarter (which hasn’t actually started yet), so this series will review some of the very, very small amount of deformation theory scattered throughout the exercises of Hartshorne.

Let’s start with some basics on the infinitesimal lifting property. First assume {k} an algebraically closed field and {A} a finitely generated {k}-algebra with Spec {A} a nonsingular variety (over {k}). Suppose {0\rightarrow I\rightarrow B'\rightarrow B\rightarrow 0} is exact with {B'} a {k}-algebra and {I} an ideal with {I^2=0}. Then {A} satisfies the infinitesimal lifting property: whenever there is a {k}-algebra hom {f:A\rightarrow B}, there is a lift {g: A\rightarrow B'} making the obvious diagram commute.

First note that if {g, g':A\rightarrow B'} are two such lifts then, {\theta=g-g'} is a {k}-derivation of {A} into {I}. A quick subtlety is that a “{k}-derivation” is an {A}-module map that is a derivation and evaluates to zero on {k}. So we need to understand how {I} is an {A}-module. But {I^2=0}, so it is a {B}-module, which in turn is an {A}-module (via {g} and {g'} which will be used). The reason {im \theta\subset I} is that {\theta} is a lift of the zero map since {g} and {g'} both lift {f}. Since the sequence is exact and {\theta} lands in the kernel, it is in the image of the one before it, i.e. {I}.

Derivation:

\displaystyle \begin{array}{rcl} \theta(ab) & = & g(a)g(b)-g'(a)g'(b) \\ & = & g(a)g(b)-g(a)g'(b)+g(a)g'(b)-g'(a)g'(b)\\ & = & g(a)(g(b)-g'(b))+(g(a)-g'(a))g'(b)\\ & = & g(a)\theta(b) + \theta(a)g'(b)\\ & = & a\cdot\theta(b)+b\cdot\theta(a) \end{array}

Evaluates to 0 on {k}: Since {g(1)=g'(1)=1}, {\theta(1)=0}. Thus {\theta(k)=k\cdot 0 = 0}.

Since {\Omega_{A/k}} is a universal object, we can consider {\theta\in Hom_A(\Omega_{A/k}, I)}. Conversely, given any {\theta\in Hom_A(\Omega_{A/k}, I)}, we can compose with the universal map {d} to get {\theta'=\theta\circ d: A\rightarrow I} is a {k}-derivation. Compose this with the inclusion {I\rightarrow B'}, call this {\psi: A\rightarrow B'}. Since composing again with {B'\rightarrow B} gives {0}, {\psi} is a lift of {0} and hence {g'=\psi + g} is a lift of {f} (note we’ve only guaranteed {k}-linear so far, not algebra hom). Finally let’s check it preserves multiplication:

\displaystyle \begin{array}{rcl} \psi(ab)+g(ab) & = & \theta'(ab)+g(ab) \\ & = & \theta'(a)g(b)+g(a)\theta'(b)+g(ab)\\ & = & \theta'(a)\theta'(b) + \theta'(a)g(b)+g(a)\theta'(b)+g(a)g(b)\\ & = & (\theta'(a)+g(a))(\theta'(b)+g(b))\\ & = & g'(a)g'(b) \end{array}

Now let {P=k[x_1, \ldots , x_n]} for which {A=P/J} for some {J}. So we get another exact sequence {0\rightarrow J\rightarrow P\rightarrow A\rightarrow 0}. We now check that there is a map {h:P\rightarrow B'} such that the square {\begin{matrix} P & \stackrel{h}{\longrightarrow} & B' \\ \downarrow & & \downarrow \\ A & \stackrel{f}{\longrightarrow} & B \end{matrix}} commutes and this induces an {A}-linear map {\overline{h}: J/J^2\rightarrow I}.

Note a map out of {P} is completely determined by where the {x_i} go. Since {B'\rightarrow B} surjective, choose any {b_i\in B'} such that {b_i\mapsto f(\overline{x_i})}. Extend this to get {h}. By definition {h} makes the square commute. Chasing around exactness, we get that if {a\in J}, then considering {a\in P} gives {h(a)\in I}. Thus restricting gives {\overline{h}: J\rightarrow I}. Since {I^2=0} we have {h(a^2)=h(a)^2=0}, so this descends to a map {\overline{h}: J/J^2\rightarrow I}. It is clearly {A}-linear.

Let {X=} Spec {P} and {Y=} Spec {A}. The sheaf of ideals {\mathcal{J}=\tilde{J}} defines {Y} as a subscheme. Then by nonsingularity (Theorem 8.17 of Hartshorne) we have an exact sequence {0\rightarrow \mathcal{J}/\mathcal{J}^2\rightarrow \Omega_{X/k}\otimes \mathcal{O}_Y\rightarrow \Omega_{Y/k}\rightarrow 0}. Take global sections of this sequence to get the exact sequence {0\rightarrow J/J^2\rightarrow \Omega_{P/k}\otimes A\rightarrow \Omega_{A/k}\rightarrow 0} ({H^1} vanishes by Serre).

Now apply the functor {Hom_A(\cdot, I)} to get the exact sequence {0\rightarrow Hom_A(\Omega_{A/k}, I)\rightarrow Hom_P(\Omega_{P/k}, I)\rightarrow Hom_A(J/J^2, I)\rightarrow 0}. Exactness on the right is due to {\Omega_{A/k}} being locally free and hence projective, so {Ext^1} vanishes.

That surjectivity is exactly what we needed to say a lift exists. Take {\overline{h}\in Hom_A(J/J^2, I)} as constructed before. Then choose {\theta\in Hom_P(\Omega_{P/k}, I)} that maps to it. Compose with the universal map and inclusion to get a derivation {P\rightarrow B'} (we’ll just relabel this {\theta}). Set {h'=h-\theta}. Since if {j\in J} we have {h'(j)=h(j)-\theta(j)=\overline{h}(j)-\overline{h}(j)=0} it descends to a map {g:A\rightarrow B'} which is the desired lift.

Next time we’ll move on to infinitesimal extensions and rephrase what we just did in those terms.

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Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.

One thought on “Infinitesimal Lifting Property

  1. Pingback: √Čtale algebras « A Mind for Madness

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