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.
 A: I know of some rings with nonzero Jacobson radical for which I see nothing "wrong" whatsoever: for instance, a discrete valuation ring (DVR) is such a ring and is pretty much the best possible ring after a field.
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 $X$.
I think you should try to ask a more precise question.  
