# Nilpotent with infinite degree?

Consider the ring of polynomials over a field or ring: $R[x]$. We can define a formal differentiation operator: $D$ which maps $x^n$ to $nx^{n-1}$ etc.

For any particular element of $R[x]$, sufficient applications of $D$ will result in $0$. However, there is no $n$ such that $D^n$ maps all elements to $0$.

So, do we call $D$ nilpotent? If we do then do we say that it is nilpotent with infinite degree? If we don't then is there another standard term?

(Not homework, just an old man trying to exercise his ageing brain.)

• The operator is not nilpotent. It only gives zeroes for vectors with finite support, i.e. polynomial. It does not map power series to 0. May 4, 2017 at 8:38
• @KennyLau, that does not really make any sense: there are no series in $R[X]$. May 4, 2017 at 8:41
• And that is what does snot make sense. An operator has a ffixed domain. You cannot talk about what it does to things which are not in its domain, and series are not in the domain of the function $D$ that appears in the question. May 4, 2017 at 8:43
• @Mariano: Yes and no. I know what you're trying to say, but on the other hand the operators form a ring and rings can be made to act on lots of modules. And often, more general modules are more informative about the ring of operators than the original domain they were defined to act on.
– user14972
May 4, 2017 at 11:46
• @Hurkyl, that simply does not make sense, I am sorry. May 4, 2017 at 16:34

We say that $D$ is locally nilpotent.
• I suppose this is defined by requiring that for each point $p$ in the ring there exists an exponent $n$ (that is allowed to depend on $p$) such that $D^np$ is zero in the ring. Another approach that came to my mind is if we can define som kind of norm on the operators, then we could check if the sequence $\{ D^k \}_{k=1}^\infty$ converges to the zero operator $O$. Will this latter idea be "related" to the first notion (local nilpotence)? I know it is a vague question. With square matrices there is something called convergent matrix. May 4, 2017 at 13:39
• @badjohn When you said "infinite degree", I thought: ah, $A^\infty$ is the first power which is zero, but of course $A^\infty$ only makes sense if you can make the limit $\lim_{k\to\infty}A^k$ somehow, and that seems to require some norm (or other metric or topological structure) on the set of operators. May 4, 2017 at 15:38