TITLE: Rings with finitely generated nilradical
QUESTION [5 upvotes]: 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 $X$?

REPLY [6 votes]: 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.