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Let $\mathfrak{a}$ be a monomial ideal in a polynomial algebra over some commutative ring $R$. If $R$ is reduced, then the radical $\sqrt{\mathfrak{a}}$ of $\mathfrak{a}$ is again a monomial ideal, and if $\mathfrak{a}$ is moreover finitely generated then so is $\sqrt{\mathfrak{a}}$. If $R$ is not reduced, then the nilradical of $R$ is contained in $\sqrt{\mathfrak{a}}$ and may make the latter somewhat less easy to handle. However, if $\mathfrak{a}$ and the nilradical of $R$ are finitely generated then so is $\sqrt{\mathfrak{a}}$.

This leads to the following question:

Is there a nice description of rings $R$ such that the nilradical of $R$ is finitely generated?

Or in more geometric terms:

Is there a nice description of schemes $X$ such that the associated reduced scheme $X_{{\rm red}}$ is locally of finite presentation over $X$?

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Fred: As Georges points out (and I'm sure you already knew), the good case includes all Noetherian rings. My impression is that in general in commutative algebra if you can show something for all Noetherian rings you are pretty happy...*unless* you have a specific class of non-Noetherian rings in mind. (As for me, the non-Noetherian rings I like are either domains or rings of functions, both of which are reduced.) Do you? –  Pete L. Clark Jul 15 '11 at 23:42
    
Dear Pete, this might be the case in "Noetherian commutative algebra". But there is so much more beyond Noetherianness (and not only abhorrent things, e.g. some tensor products of Noetherian rings...) that it seems unnatural to me to restrict oneself to a class of rings that is a priori totally unrelated to the notion in question. If Noetherianness turns out to be the "right" thing then I am also happy with it, but I am sure that this is not the case here. –  Fred Rohrer Jul 16 '11 at 1:58
    
Dear Fred, As someone who comes to commutative algebra via arithmetic geometry, I think I stand guilty as charged with a preference for Noetherian rings. But I wonder why you think there should be a nice description of rings for which the nilradical is finitely generated? Is there some analogous nice classification result that you can point to? –  Pete L. Clark Jul 16 '11 at 10:06
    
Dear Pete, I do neither think there is such a description, nor that there is not - I just do not know it! But I hoped (and still hope) that someone else might know whether there is one. –  Fred Rohrer Jul 16 '11 at 11:27

1 Answer 1

Good In a noetherian ring the nilradical, like any ideal, is finitely generated.

Bad Given a ring $A$ and an $A$-module $M$ you can construct an $A$-algebra $R=A\ast M $ whose underlying $A$-module is $A\oplus M$ and in which the multiplication is given by $(a,m).(a',m')=(aa',am'+a'm)$.
The first observation is that $M=0\ast M$ becomes an ideal of the ring $R$ satisfying $M^2=0$ and so definitely nilpotent.
In particular if $A$ is reduced the nilradical of $R$ is exactly $M$.
The second observation is that a subset $G\subset M$ generates $M$ as an ideal of the ring $R$ exactly when $G$ generates $M$ as an $A$-module.

And now you can construct rings with as badly non-finitely generated nilradicals as you like, for example by taking for $M$ a free $A$-module of dimension a huge cardinal.

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Dear Georges, thank you for your nice answer. The interesting task is, as so often, to find the line between the good and the bad. –  Fred Rohrer Jul 15 '11 at 8:52

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