Let $B$ be a commutative ring (with 1). The Jacobson radical can be defined as $$ J(B) = \{b \in B \mid \forall a \in B \colon \quad 1 + a\cdot b \text{ is a unit in } B \} $$ or $$ J(B) =\{ b \in B \mid \forall a \in B, u \in B^\times \colon \quad u + a\cdot b \text{ is a unit in } B \}\text{.}$$ If $A \subset B$ is a ring extension (or more generally, $B$ an $A$-algebra), we can define two relative variants of the Jacobson radical: $$ \bar J_A(B) =\{b \in B \mid \forall a \in A \colon \quad 1 + a\cdot b \text{ is a unit in } B \} $$ and $$ \hat J_A(B) =\{ b \in B \mid \forall a \in A, u \in B^\times \colon \quad u + a\cdot b \text{ is a unit in } B \}\text{.}$$

The first variant is rather nice, the functor $\bar J_A(\bullet) \colon CRing/A \rightarrow Set$ is representable by $S^{-1}A[X]$ where $S$ is the multiplicative set generated by linear polynomials of the form $1 + a\cdot X$. So, if e.g. $A$ is an algebraically closed field, then $S^{-1}A[X] = A[X]_{(X)}$.

The second variant is stranger; for starters, it doesn't form a functor at all. Obviously $\bar J_B(B) = \hat J_B(B) = J(B)$ and $\bar J_A(B) \supseteq \hat J_A(B) \supseteq J(B)$. I think I can show that $\hat J_A(B) = J(B)$ if B is integral over A.

I can generate examples where $\hat J_A(B) \not= J(B)$: let $A$ be any ring with $J(A) \not= Rad(A)$ and take $b \in J(A)$ not nilpotent and $B = A[X]$. It is well known that units in $B$ are the polynomials where the constant term is a unit and all other coefficients are nilpotent. So $b \in \hat J_A(B)$ but $b \notin J(B)$ since $1 + b \cdot X$ is not a unit.

These examples are all using non-Jacobson rings, so my question is:

Question Is there a ring extension $A \subset B$ with $\hat J_A(B) \not= J(B)$ and $B$ a Jacobson ring? Furthermore, how nice can $A$ and $B$ be? E.g., is it possible that $B$ is finitely generated over $A$ and $A$ is a localization of a finitely generated algebra over $\mathbb{C}$?

Edit: I previously asked whether $A$ and $B$ can be finitely generated over $\mathbb{C}$. This can't be the case: Suppose $b \in \hat J_A(B)$ and let $\mathfrak{m}$ be a maximal ideal of $B$. By the Nullstellensatz, $B/\mathfrak{m} \cong \mathbb{C}$. If $b \notin \mathfrak{m}$, we easily obtain $z \in \mathbb{C} \subset A$ with $1 + z \cdot b \in \mathfrak{m}$, which would be a contradiction. With a bit of work I think I can extent it to exclude $A$ Jacobson and $B$ finitely generated over $A$. In fact $A$ is Jacobson if and only if $\hat J_A(B) = J(B)$ for all finitely generated $A$-algebras.

Edit2 (in response to darij's comment):

$\bar J_A(B)$ is in general not even closed under addition: If $k \subset l$ is any field extension, then $J_k(l) = l \setminus k^\times$.

$\hat J_A(B)$ is obviously a $A$-submodule of B. It is an ideal of B if and only if $\hat J_A(B) = J(B)$ holds. With some calculation you can show that $\hat J_A(B)$ is closed under multiplication in the sense that if $b_1, b_2 \in \hat J_A(B)$, then $b_1 \cdot b_2 \in \hat J_A(B)$. In particular, for $b \in \hat J_A(B)$, all positive powers $b^n$ are again in $\hat J_A(B)$.

This actually motivates a third variant: $$ \tilde J_A(B) =\{ b \in B \mid \forall f \in A[X], f(0) = 1 \colon \quad f(b) \text{ is a unit in } B \}\text{.}$$ This fits in as follows: $$ \bar J_A(B) \supseteq \tilde J_A(B) \supseteq \hat J_A(B) \supseteq J(B)$$ It is again representable but not closed under addition in general (consider $\mathbb C \subset \mathbb C(X)$), and I can show the following:

  1. If $B$ is an integral $A$-algebra, then $\tilde J_A(B) = \hat J_A(B) = J(B)$.
  2. $A$ is Jacobson if and only if $\tilde J_A(B) = \hat J_A(B) = J(B)$ for all finitely generated A-algebras B.
  • $\begingroup$ The ordinary Jacobson radical is also the intersection of all maximal ideals. Do you have an analog of this for your relative version? $\endgroup$ – მამუკა ჯიბლაძე Apr 3 '16 at 21:11
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    $\begingroup$ Are your two generalizations ideals to begin with? $\endgroup$ – darij grinberg Apr 4 '16 at 0:26

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