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What is an example of two injective homomorphisms $R \to A$, $R \to B$ of commutative rings such that $R \to A \otimes_R B$ is not injective?

Of course neither $R \to A$ nor $R \to B$ can be flat in this example (but this is not enough for an example).

A geometric reformulation is the following: What is an example of two morphisms of affine schemes $X \to S$, $Y \to S$ which are scheme-theoretical dense, but $X \times_S Y \to S$ is not?

Bonus question: What is an example where $R \neq 0$ but $A \otimes_R B=0$?

Martin Brandenburg
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Answer. Yes, there is an example. Let $k$ be a field, let $R = k [a, b] / (a^2, a b, b^2)$, $A = R [x] / (a - b x, x^2)$, $B = R [y] / (b - a y, y^2)$; then $$A \otimes_R B \cong R [x, y] / (a - b x, x^2, b - a y, y^2) \cong k [a, b, x, y] / (a^2, a b, b^2, a - b x, x^2, b - a y, y^2)$$ but \begin{align} a & = (a - b x) + (b - a y) x + (a - b x) x y + (x^2) b y \\ b & = (b - a y) + (a - b x) y + (b - a y) x y + (y^2) a x \end{align} so $(a, b) \subseteq \ker (R \to A \otimes_R B)$. (Actually, since $A \otimes_R B \ne \{ 0 \}$, the kernel is exactly $(a, b)$.) In particular, $R \to A \otimes_R B$ is not injective.

Notice that in the above example, the problematic elements of $R$ are nilpotent. In fact, this is the only possibility, at least when $R$ is noetherian and both $A$ and $B$ are finitely generated $R$-algebras. This is best understood geometrically.

Consider pullback diagram of noetherian affine schemes where all morphisms are of finite type, say: $$\require{AMScd} \begin{CD} X \times_S Y @>>> Y \\ @VVV @VVV \\ X @>>> S \end{CD}$$ Suppose $X \to S$ and $Y \to S$ are dominant (which is a weaker condition than having scheme-theoretic image equal to $S$). By Chevalley's theorem, the set-theoretic images of $X \to S$ and $Y \to S$ are constructible, so both contain a dense open subset of $S$. Since the intersection of dense open subsets is dense, it follows that $X \times_S Y \to S$ is dominant, as required. Thus, the ideal sheaf of the scheme-theoretic image of $X \times_S Y \to S$ be contained in the nilradical.

Bonus answer. A logical argument (namely, compactness) shows that if there is an example where $A \otimes_R B = \{ 0 \}$, then there must be an example where $R$, $A$ and $B$ are finitely presented as commutative rings. But then we are in the above situation, so $A \otimes_R B = \{ 0 \}$ implies that $1$ is nilpotent, i.e. $R = \{ 0 \}$.

Zhen Lin
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    First, it seems all you've shown that $R\to A\otimes_R B$ has kernel in the nilradical of $R$, since this is what is equivalence do dominance. – Alex Youcis Sep 02 '14 at 01:46
  • I agree with Alex Youcis, $\mathrm{Spec}(A) \to \mathrm{Spec}(R)$ is dominant (resp. scheme-theoretical dense) iff the kernel of $R \to A$ lies in the nilradical (resp. is zero). These only agree when $R$ is reduced. So in that case the proof works (which is already quite surprising!). – Martin Brandenburg Sep 02 '14 at 08:24
  • Hmmm, yes. So the only thing that can go wrong is that some nilpotent goes to zero in $A \otimes_R B$ but not in $A$ or $B$. Maybe there is an example after all. – Zhen Lin Sep 02 '14 at 08:37
  • @MartinBrandenburg As it turns out, there is an example! – Zhen Lin Sep 02 '14 at 22:29
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    Great! You can also see $a \in \ker(R \to A \otimes_R B)$ as follows: $a \otimes 1 = bx \otimes 1 = x \otimes b = x \otimes ay = ax \otimes y = bx^2 \otimes y = 0$. – Martin Brandenburg Sep 03 '14 at 10:06
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    By the way, this calculation doesn't use the relations between $a,b$ (as if this would also work for $R=k[a,b]$). But we really need these relations in order to ensure that $R \to A$ and $R \to B$ are injective. (This still has to be proven, but it's not hard.) – Martin Brandenburg Sep 03 '14 at 10:12
  • Yes, indeed. I used Gröbner bases in Mathematica for my calculations. – Zhen Lin Sep 03 '14 at 11:05