Square of tensor product

Question

Let $A$ be an integral domain and $B$ be an $A$ algebra. Let $I$ be and ideal of $B$.

Something has been bugging me : Is it true that

$$(B\otimes I)/(B\otimes I)^2 \cong B \otimes (I/I^2)$$

We obviously have a surjective map $B\otimes I \to B \otimes (I/I^2)$ given by $$ x \otimes y \mapsto x \otimes \bar{y}$$

But I fail to see why it should be injective (maybe it isn't).

what is the definition of the square of B tensor I? – user119882 Mar 29 '14 at 15:47 — user119882, Mar 29 '14 at 15:47

score 1 · Answer 1 · answered Mar 28 '14 at 15:19

1

Clearly $(B\otimes I)^2=B\otimes I^2$. Now we have a short exact sequence $$0\to I^2\to I\to I/I^2\to 0$$ If we tensor this with $B$ and $B$ is flat, we obtain a short exact sequence $$0\to B\otimes I^2\to B\otimes I\to B\otimes I/I^2\to 0$$ The same is true if $Tor_1^A(B,I/I^2)=0$. But if this torsion group is not zero, then the leftmost map may not be injective.

answered Mar 28 '14 at 15:19

Mariano Suárez-Álvarez

135,076

It seems that we only need right-exactness. – user119882 Mar 29 '14 at 15:44

Square of tensor product

1 Answers1