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Fix a $\mathbb{Q}$-algebra $R$. Let's call an endomorphism $f : M \to M$ of an $R$-module $M$ locally nilpotent if for every $m \in M$ there is some $n \in \mathbb{N}$ such that $f^n(m)=0$. Equivalently, for every finitely generated submodule $N \subseteq M$ there is some $n \in \mathbb{N}$ with $f^n |_N = 0$. In particular, for a finitely generated $M$ we don't get anything new.

Now consider the following category $C$: Objects are pairs $(M,f)$, consisting of an $R$-module and a locally nilpotent endomorphism $f$ of $M$. A morphism is just a commutative diagram.

Question A. Is there a reference in the literature for the following observation: $C$ is isomorphic to the category of representations of the additive group scheme $\mathbb{G}_{a}$ over $R$.

This is not hard to prove. In fact, one can use the same method as in Example I.8.1 in Milne's script on algebraic groups; but there $R$ is a field and $M$ is supposed to be finite dimensional, so that locally nilpotent = nilpotent. But the same works in general, the $R[T]$-comodule structure on $M$ corresponding to $(M,f)$ is given by $M \to M[T]$, $m \mapsto \sum\limits_{n \geq 0} \frac{f^n(m) T^n}{n!}$.

Question B. How do infinite products look like in $C$?

Note that the description of $C$ above shows that $C$ is cocomplete as as well as complete. The forgetful functor $C \to \mathrm{Mod}(R)$ creates colimits, so they are easy to describe. The same is true for finite limits, so the only limits missing are infinite products. The forgetful functor doesn't create them: For a familiy of objects $(M_i,f_i)$, the endomorphism $\prod_i f_i$ of $\prod_i M_i$ does not have to be locally nilpotent. Thus they will be more complicated.

Question C. What are interesting examples of locally nilpotent endomorphisms which are not nilpotent?

Of course there are many examples: If $M$ is an $\mathbb{N}$-graded module and $f$ is of negative degree, then $f$ is locally nilpotent, but usually $f$ is not nilpotent. A specific example is the derivative $\partial : R[X] \to R[X]$. Are there more interesting examples which don't arise from gradings?

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I don't know a precise reference, but it's a very well known fact. I doubt that the category of representations of $\mathbb{G}_{\mathrm a}$ has infinite products: (algebraic) representations of algebraic groups are unions of finite-dimensional subrepresentations, and this property is not preserved under infinite products. – Angelo Feb 15 '12 at 16:51
This property is not preserved under the "naive product" which is no product at all (see details in Question B). The category has products by some abstract result (Todd, Theo or Mike probably can explain this?). – Martin Brandenburg Feb 15 '12 at 17:02
up vote 4 down vote accepted

Question A: Demazure-Gabriel, Groupes Algébriques, II, §2, 2.6

Question B: It is easily checked that the maximal submodule of $\prod M_i$ on which $\prod f_i$ acts locally nilpotent is the categorical product. This may also be described as $\bigcup_n \prod_i \ker f_i^n\subseteq\prod_i M_i$. Actually, since the set-valued functor $(M,f)\mapsto\ker f^n$ is represented by $(R[X]/(X^n),X)$, so commutes with products, and since the equality $M=\bigcup_n \ker f^n$ is equivalent to $f$ acting locally nilpotent, it is clear that the underlying set of the categorical product must be $\bigcup_n \prod_i \ker f_i^n$.

Question C: An example where the filtration $\ker f^n$ does not split is $R=k[[T]]$, $M=k((T))/k[[T]]$, $f=T$.

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Thanks for this answer! Another view on B: Actually $C$ is a coreflective subcategory of the category of $R$-modules equipped with some endomorphism, the coreflector is $(M,f) \mapsto (M,\cup_n \mathrm{ker}(f^n))$. Thus limits in $C$ may be computed by taking the coreflectors of the limits in the larger category. – Martin Brandenburg Feb 16 '12 at 8:00
Right. I should have thought of that. – Angelo Feb 17 '12 at 8:05

I address mainly Question C in the simplest special case where $R$ is $\mathbb Q$:

In this case you are looking at locally nilpotent endomorphisms of a vector space. Similarity classes of such endomoephisms correspond to isomorphism classes of torsion $\mathbb Q[[t]]$-modules. If you assume that the modules are of countable dimension and reduced, then a classification of isomorphism classes of such modules is given by Ulm's theorem: the isomorphism classes are in bijective correspondence with certain transfinite sequences of non-negative integers (the Ulm invariants). This should allow you to construct many interesting examples.

For the details, I can do no better than referring you to Kaplansky's book Infinite Abelian Groups where Ulm's theorem and its relation to the classification of locally nilpotent endomorphisms are beautifully explained. It is based on Kaplansky's paper with Mackey A generalization of Ulm's theorem, Sum. Bras. Math., 1951 (available in Kaplansky's selected papers).

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This is interesting, thank you. – Martin Brandenburg Feb 16 '12 at 8:02

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