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Let $R \to A$ and $R \to B$ be two homomorphisms of commutative rings whose kernels are nil (i.e. consist only of nilpotent elements). Then the kernel of $R \to A \otimes_R B$ is also nil.

See SE/916173 for Zhen Lin's proof of this fact, which uses algebraic geometry, notably Chevalley's Theorem. This is used to show that if $X \to S$ and $Y \to S$ have dense image, then $X \times_S Y \to S$ too.

I wonder, is there a direct and elementary algebraic proof of this fact? After looking at $R_{red}$ etc., we may assume that $R,A,B$ are reduced and that $R \to A$, $R \to B$ are injective, so that we have to prove that $R \to A \otimes_R B$ is injective. Is the special case $R=\mathbb{Z}$ simpler? In order to avoid nasty tensor calculations, we may rephrase the question as follows: How to construct an injective ring homomorphism $R \to C$ which factors through $R \to A$ and $R \to B$?

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If $A$ is an $R$-algebra, and $P\subset R$ a prime ideal, then $A_P \neq 0$ exactly when the kernel of $R\to A$ is contained in $P$. So $R\to A$ has nil kernel exactly when $A_P \neq 0$ for all minimal primes $P$.

But, when $P$ is a minimal prime, $A_P\neq 0$ exactly when there is a map $f: A\to L$ to a field, with $R\cap \ker f = P$.* As in the previous argument, if we have such a map for $A$ and $B$, we can construct one for $A\otimes B$.


* If $A_P \neq 0$, then $A_P$ has a maximal ideal, so there is a map $f:A\to A_P \to L$ with $R\cap \ker f$ a prime ideal contained in $P$. Since $P$ is minimal, $R\cap \ker f = P$. Conversely, if $f:A\to L$ is a map with $R\cap \ker f = P$, it factors $A\to A_P \to L$, since it inverts everything not in $P$. So $A_P\neq 0$.


Previous argument:

If $R$ is an integral domain, $K$ its field of fractions, I think we can argue as follows.

$R\to A$ is injective if and only if $A_K = A\otimes K \neq 0$. But then there is a maximal ideal of $A_K$, with quotient field $L$ extending $K$. This gives us a map $A\to L$ whose kernel meets $R$ trivially.

Similarly, there is a map $B\to L'$ whose kernel meets $R$ trivially. By passing to a common field extension, we may assume $L=L'$. The universal property of the tensor product then gives us a map $A\otimes B \to L$. Since the kernel meets $R$ trivially, it follows that $R\to A\otimes B$ is injective.

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  • $\begingroup$ Alright. If $R$ is just reduced, wlog noetherian, can we use the finitely many minimal prime ideals somehow in order to reduce to the case of an integral domain? $\endgroup$ Sep 5, 2014 at 10:07
  • $\begingroup$ @MartinBrandenburg I've updated the answer with a more general approach. If I haven't overlooked something obvious, this may resolve the question. Chevalley seems to have been overkill, geometrically speaking. $\endgroup$ Sep 5, 2014 at 21:08
  • $\begingroup$ @MartinBrandenburg Added to the answer. $\endgroup$ Sep 5, 2014 at 21:36
  • $\begingroup$ If $A_P \neq 0$, then $A_P$ has a maximal ideal, yes, but why one containing $P A_P$? It could happen that $P A_P = A_P$. (This is the main difference to the case $A=R$) $\endgroup$ Sep 5, 2014 at 22:04
  • $\begingroup$ @MartinBrandenburg Can this really happen when $A_P\neq 0$? In any case, I think I only require $R\cap \ker f \subset P$. $\endgroup$ Sep 5, 2014 at 22:11

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