Let $A$ be a local ring, and suppose $B$ and $C$ are local $A$-algebras, by which I mean they are local rings that are $A$-algebras, and further, that the morphisms $A\rightarrow B$ and $A\rightarrow C$ that make them $A$-algebras are local ring maps.
Let $\kappa(A), \kappa(B), \kappa(C)$ be the residue fields of the local rings. Then there are canonical maps $A\rightarrow\kappa(A)$, etc. Taking the tensor product over $A$ of $B\rightarrow \kappa(B)$ with $C\rightarrow \kappa(C)$, and then invoking the fact that $A\rightarrow B,A\rightarrow C$ are both local maps, so that the maximal ideal of $A$ annihilates both $\kappa(B)$ and $\kappa(C)$, we get a ring homomorphism
$$\Phi: B\otimes_A C\rightarrow \kappa(B)\otimes_{\kappa(A)} \kappa(C).$$
Is $\ker \Phi$ contained in the Jacobson radical of $B\otimes_A C$? If so, what's the argument? If not, what's a counterexample? (Derogated, see Updates)
Update: More thought on the example below (see "Update 2") let me to the conclusion that it is actually a counterexample! In which case,
could natural conditions (essential integrality; flat-of-relative-dimension-zero-ness) be added to make the answer "yes"?
I hope that as nobody has attempted to answer the question yet, it is not poor form for me to change the goalpost in this way.
Example: Let $A=k$, an algebraically closed field. Let $B=k[x]_{(x)}$ and let $C = k[y]_{(y)}$. Then $\kappa(A)=\kappa(B)=\kappa(C)=k$, and thus $\kappa(B)\otimes_{\kappa(A)}\kappa(C) = k\otimes_kk = k$. Also, if I am thinking about this correctly, $B\otimes_A C$ is the localization of $k[x,y]$ in which all irreducible univariate polynomials with nonzero constant term are inverted. Since by Hilbert's Nullstellensatz, every maximal ideal of $k[x,y]$ has the form $(x-\alpha, y-\beta)$, the localization $B\otimes_A C$ has as maximals just these ideals that do not contain an inverted element. (Update: I no longer believe this. See below.) However, $(x-\alpha, y-\beta)$ contains an inverted element as soon as either $\alpha$ or $\beta$ is nonzero, and we conclude that in this case $B\otimes_A C$ is local, with unique maximal ideal $(x,y)$, so in this case, this is the Jacobson radical. The map $B\otimes_A C\rightarrow \kappa(B)\otimes_{\kappa(A)} \kappa(C)$ is exactly the quotient map by $(x,y)$ in this case, so its kernel is $(x,y)$, and in the end the kernel equals the Jacobson radical, so the answer to my question is "yes" in this case.
Update 2: Actually I think the reasoning above is wrong. It is true that in the localization $B\otimes_A C$, all the maximal ideals of $k[x,y]$ except $(x,y)$ are destroyed. However, I think many other maximal ideals are "uncovered": the prime ideals contained only in destroyed maximals are now maximal. In particular, in this ring, irreducible polynomials that are not univariate do not get inverted, thus they generate prime ideals. If the point $(0,0)$ does not lie on the curve defined by such a polynomial, then the prime ideal it generates is not contained in $(x,y)$. As all other maximals in $k[x,y]$ have been destroyed, this prime is now maximal. Thus $B\otimes_A C$ actually has many maximal ideals! In fact there is no ring element divisible by all such polynomials: so the intersection of these maximal ideals is trivial! In other words, I now believe $B\otimes_A C$ has trivial Jacobson radical. So $\ker \Phi$ is not contained in the Jacobson radical in this case.
Update 3: This is to record an interesting implication of the reasoning in Update 2. It is tangential to the question, so can be ignored.
As in Example and Update 2, we let $A=k$ (an algebraically closed field), $B=k[x]_{(x)}$, and $C=k[y]_{(y)}$. We interpret the ring $B\otimes_A C$ as the localization of $k[x,y]$ that inverts all univariate polynomials with nonzero constant term. Let $f$ be an irreducible polynomial in $k[x,y]$ that is not univariate and also has a nonzero constant term. The reasoning in Update 2 shows that the ideal generated by $f$ is maximal. This implies that the quotient by $f$ (call it $R$) is a field. I did not find this obvious, so I'll record a direct argument.
View $f$ as a polynomial in $x$ with coefficients in $k[y]$. Because $f$ has nonzero constant term as a bivariate polynomial, the constant term when it's viewed as a polynomial in $x$ is a univariate polynomial in $y$ with a nonzero constant term, which is a unit in the ambient ring. Thus, in the quotient $R$, where $f=0$ is an equation, we can do the standard trick of moving this unit to the other side of the equation and factoring out one factor of $x$ to obtain an inverse for $x$. (For example, if $f = x^2 + xy + y^2 - 1$, this looks like $x(x+y) = 1 - y^2$, obtaining $(x+y)(1-y^2)^{-1}$ as an inverse for $x$.) It follows that $k(x) = k[x]_{(x)}[x^{-1}]$ embeds in $R$. Call the image of this embedding $S$.
Furthermore, as $R$ is a quotient of a localization of $k[x,y]$, and as $S$ contains the image of $S$, $R$ is a localization of $S[y]\subset R$. (I have been playing fast and loose with using the symbols $x,y$ to represent their original meanings and also their images in $R$, I hope my meaning is clear.) But $S$ is a field, and, now viewing $f$ as an irreducible polynomial in $y$ with coefficients in $S$, $y\in R$ satisfies this polynomial, so $S[y]$ is a field! And $R$, a localization of it, is thus the same field!