# What happens geometrically when the Jacobson radical is non-zero?

I'm interested in intuition about the affine schemes of rings with a non-vanishing Jacobson radical.

In the ring of real-valued continuous functions on a topological space the Jacobson radical is reduced to zero. Thinking of an arbitrary ring as the global sections on an affine scheme the situation is quite different, but there is also the addition of generic points (more stuff) and a very weak topology (fewer continuous functions).

In short: How does a non-vanishing Jacobson radical manifest itself geometrically? What happens, and how should I think about this? If $A$ is a ring with a non-vanishing Jacobson radical $J$, how does for instance the space $Spec(A/J)$ look compared to $Spec(A)$?

Thanks.

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There is no need to sign your posts, since your user name always appears at the bottom as a "signature" of sorts. "Thanks" is certainly fine, and welcomed, but you don't have to put your name down as well. –  Arturo Magidin Jun 18 '11 at 19:59
Old habit from writing mails, but I might as well stop if anyone feels it worth commenting on. –  Eivind Dahl Jun 18 '11 at 20:08

If $R$ is finitely generated over a field, then it is a Hilbert-Jacobson ring and its Jacobson radical is equal to its nilradical. A ring with nonzero nilradical is indeed not so nice: it has nonzero nilpotent elements so cannot be a "ring of functions" in the literal sense -- i.e., it is not a subring of the ring of all functions from a set $S$ to a field $k$ under pointwise addition and multiplication.
With regard to rings of functions: let $X$ be any topological space whatsoever, and let $C(X)$ be the ring of continuous functions $f: X \rightarrow \mathbb{R}$. For any $x \in X$, let $\mathfrak{m}_x = \{f \in C(X) \ | \ f(x) = 0\}$. Then evaluation at $x$ induces an isomorphism $C(X)/\mathfrak{m}_x \cong \mathbb{R}$, so $\mathfrak{m}_x$ is a maximal ideal. Clearly $\bigcap_{x \in X} \mathfrak{m}_x = 0$, hence the Jacobson radical of $C(X)$ is equal to zero. It does not have anything to do with the separation properties of $X$.
Naturally I do not think there is actually anything wrong with it. I removed the comment on separation; oops. I also tweaked the question, since we're not actually looking at the ring of real-valued functions on an affine scheme. I'm looking for intuition, and phenomena which can be directly attributed to a non-vanishing Jacobson radical $J$. How does for instance the space $Spec(A/J)$ look compared to $Spec(A)$. Hopefully the question is in better shape now. Thanks for the feedback! –  Eivind Dahl Jun 18 '11 at 23:11
@Eivind: I don't really know what to say about your revised question. For instance, if $R$ is a local ring, then passing from $\operatorname{Spec} R$ to $\operatorname{Spec} (R/J)$ just gives you the spectrum of the residue field, a one-point space. This is a pretty drastic change. At the other extreme, for finitely generated algebras over a field (say) modding out by the radical induces a homeomorphism on Spec's. Between these two extremes I would imagine that just about anything could happen... –  Pete L. Clark Jun 21 '11 at 23:47