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I'm trying to learn the theory of algebraic groups using "modern" algebraic geometry, going past the traditional, very standard books by Borel, Springer, and Humphreys, all of which try to avoid the use of group schemes. One of the motivations for using schemes is that in the classical setting, "nilpotents are not allowed," a sentence I don't understand. For example, on page 18 in Milne's Algebraic Groups (which is still in draft form), he writes, "Since we allow nilpotents in the structure sheaf, the points of an algebraic group with coordinates in a field, even algebraically closed, do not convey much information about the group. Thus, it is natural to consider its points in a $k$-algebra. Once we do that, the points capture all information about the algebraic group."

This is clearly not meant to be totally precise. But I still wonder what he means. I'm thinking of the curves $x^2=0$ and $x=0$ as one example which is sometimes used to motivate the definition of scheme (the behavior at 0 is different in those two examples).

Here are my questions, presumably related: By "nilpotents in the structure sheaf," does he mean nilpotents in the coordinate ring? What does "allowing nilpotents" mean, and what is wrong with not allowing nilpotents? Why doesn't allowing nilpotents force the points to "convey much information" about the group? What information does thinking of algebraic groups as functors capture?

I should add that I'm mostly used to thinking of algebraic groups as functors. For example, $SL_n$ is thought of as a functor that takes a ring $R$ to the matrix group $SL_n(R)$.

(I can post this as more than one question if it would be better, but I don't know enough about the answer I'm looking for to know whether or not that is a good idea.)

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  • $\begingroup$ When you view algebraic groups as schemes, there is a lot going on at once. Please do ask more questions as you get further. $\endgroup$ – D_S Jun 6 '17 at 20:29
  • $\begingroup$ I'm also trying to understand the scheme approach. I learned algebraic groups by reading Borel/Springer/Humphreys, and now I'm going back and doing it again using schemes. $\endgroup$ – D_S Jun 6 '17 at 20:30
  • $\begingroup$ I agree with @D_S These are good questions so keep them coming. $\endgroup$ – Tobias Kildetoft Jun 7 '17 at 8:41
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Let $X = \textrm{Spec } A$ (or $\textrm{Spm } A$ if you want) for a finitely generated algebra $A$ over a given field $k$. Not allowing nilpotents means only considering the radical ideals $I$ of $A$, i.e. only considering those ideals $I$ for which $A/\sqrt{I}$ is reduced. In particular, we want the ideal corresponding to $X$ to be radical, i.e. we want $A$ to be a reduced ring. This is how things are done in the classical theory. Here are a couple of reasons why:

1 . A function on $X$ is completely determined by its points.

Let $f \in A$. For each point $\mathfrak p \in X$, you can think talk about evaluating $f$ at the point $\mathfrak p$: that is, $f(\mathfrak p)$ is the element $f + \mathfrak p A_{\mathfrak p}$ in the field $A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}$. If $f(\mathfrak p) = 0$ for all $\mathfrak p$, then $f$ lies in

$$\textrm{Nil } A = \bigcap\limits_{\mathfrak p \in \textrm{Spec } A} \mathfrak p = \bigcap\limits_{\mathfrak m \in \textrm{Spm } A} \mathfrak m$$

and so you can conclude that $f(\mathfrak p) = 0$ if $A$ is reduced. Otherwise, you cannot conclude this.

2 . Closed subvarieties are easy to define.

If $E$ is a closed subset of $X$, then you automatically know how to think of $E$ as a subvariety of $X$: $E$ is the variety corresponding to the reduced ring $A/\mathcal I(E)$. More generally when $X$ is an arbitrary variety, you can use an affine open cover of $E$ to give a canonical subvariety structure on $E$. If you do allow nilpotents, then there are many variety structures you can define on $E$ (corresponding to the ideals of $\mathcal O_X(X)$ whose radical is $\mathcal I(E)$ in the affine case), and the definition of closed subvariety is more delicate.

Now, here is a reason why you would want to allow nilpotents. Let $f:G \rightarrow H$ be a morphism of group schemes over a field $k$. From the functor of points view, the kernel of $f$ should be a closed subvariety $K$ of $G$, with the property that for all $k$-algebras $R$, $K(R)$ is the kernel of the homomorphism $G(R) \rightarrow H(R)$.

Let $e_H: \textrm{Spec } k \rightarrow H$ be the identity section of $H$. Let

$$K = G \times_H \textrm{Spec } k$$

the fiber product of $G$ and $\textrm{Spec } k$ over $H$ with respect to the morphisms $f, e_H$. You can argue that the projection $K \rightarrow G$ is a closed immersion (this follows since closed immersions are stable under base change, and $e_H$ is a closed immersion). I don't want to type this all out, but you can see that for all $k$-algebras $R$, under the identification of $K(R)$ as a subset of $G(R)$ via $K \rightarrow G$, $K(R)$ is exactly the kernel of the homomorphism $G(R) \rightarrow H(R)$.

This shows that $K$ is the "correct" object to look at as the kernel of the homomorphism $G \rightarrow H$.

On the other hand, the coordinate ring of $K$ need not be reduced! For example, if $k$ has characteristic $p$, and if $G = H = \mathbb{G}_m = \textrm{Spec } k[T,T^{-1}]$, and $f$ is the morphism $G \rightarrow H$ given on points by $x \mapsto x^{p^n}$, then $K$ is the closed subscheme of $G$ corresponding to the ideal generated by $T^{p^n} - 1 = (T-1)^{p^n}$. Its coordinate ring is not reduced.

If you try to replace $K$ by the closed subscheme $K'$ corresponding to the radical ideal $T - 1$, then $K'(R)$ is no longer going to be the kernel of the homomorphism $G(R) \rightarrow H(R)$, in general.

So basically, the reason we allow nilpotents is because in the categories we are considering, basic objects such as kernels do not exist without them.

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  • $\begingroup$ Very good answer. I added one with some examples of why these kernels are nice to have. $\endgroup$ – Tobias Kildetoft Jun 6 '17 at 20:24
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This will be more of a supplement to the answer by D_S than a stand-alone answer, but I felt like commenting a bit on some reasons we are very happy to be able to work with precisely those kernels he mentions, namely those of the iterates of the Frobenius endomorphism, and which are called Frobenius kernels.

First let me mention that the special case of just the kernel of the Frobenius morphism is indeed somewhat special. The representation theory of this group scheme is equivalent to that of the Lie algebra of the group, seen as a restricted Lie algebra.

But in the general case, there is no such connection to the Lie algebra, and yet these kernels play an immense role in so many parts of the representation theory of reductive algebraic groups. One example I like is that developing the representation theory of them leads to a fairly straightforward proof of one of the most beautiful theorems of the subject, known as Steinberg's Tensor Product Theorem. It states that if we write the dominant weight $\lambda$ as $\lambda_0 + p\lambda_1$ with $\lambda_0$ $p$-restricted then the simple module $L(\lambda)$ is isomorphic to the tensor product $L(\lambda_0)\otimes L(\lambda_1)^{(1)}$ where the ${}^{(1)}$ denotes the modules twisted by the Frobenius morphism.

So this theorem allows us to reduce a lot of problems to the study of simples with restricted highest weights, which for semisimple groups is a finite set of modules.

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