$R/I \otimes_A R/I \cong (R \otimes_A R)/(I \otimes_A I)$?

Question

Let $f: A \to R$ be a homomorphism of commutative rings, and let $I$ be an ideal of $R$.

Is it true that $R/I \otimes_A R/I \cong (R \otimes_A R)/(I \otimes_A I)$ ?

After obtaining the surjection $g: R \otimes_A R \to R/I \otimes_A R/I$, I have a problem of showing that the kernel of $g$ equals $I \otimes_A I$; I can only show directly that $I \otimes_A I \subseteq Kerg$.

Any help will be appreciated.

Edit: Actually, my question is a special case of an already asked and answered question: Why does $(A/I)\otimes_R (B/J)\cong(A\otimes_R B)/(I\otimes_R 1+1\otimes_R J)$? (this reference appears in a comment of @user26857).

So this is not true... If I am not wrong, on the left we will have $R/I$, while on the right we will have $R/(I \otimes_R I)$ ? — user237522, Jun 08 '15 at 23:03

score 1 · Answer 1 · answered Jun 08 '15 at 23:01

1

First of all, let's be careful: $I\otimes_A I\to R\otimes_A R$ may not be injective, so presumably $(R\otimes_A R) / (I\otimes_A I)$ is supposed to be the quotient by the image of this map.

Anyway, this is false even when $f$ is the identity: then the left-hand side is isomorphic to $R/I$, while the right-hand side is isomorphic to $R/I^2$.

answered Jun 08 '15 at 23:01

Andrew Dudzik

30,074

Thank you both for your answers! @user26857: Thanks for the correct isomorphism. – user237522 Jun 08 '15 at 23:16
@user237522 Count me as very skeptical that the other answer gives the correct isomorphism. – Andrew Dudzik Jun 08 '15 at 23:26
:-) Though I still wish to know why this is exactly the kernel of $g$. – user237522 Jun 08 '15 at 23:33

score 1 · Accepted Answer · answered Jun 08 '15 at 23:02

1

In fact, the kernel of $g$ is $I\otimes_AR+R\otimes_AI$, so the correct isomorphism is $$(R\otimes_AR)/(I\otimes_AR+R\otimes_AI)\simeq R/I\otimes_AR/I.$$

answered Jun 08 '15 at 23:02

user26857

52,094

How can we show that this is the entire kernel without some kind of flatness assumption? – Andrew Dudzik Jun 08 '15 at 23:06
@Slade This is a notation for the ideal of $R\otimes_AR$ generated by the monomials $a\otimes b$ with $a\in I$ or $b\in I$. – user26857 Jun 08 '15 at 23:08
@Slade If you prefer, $I\otimes_AR$ stands for the image of the map $I\otimes_AR\to R\otimes_AR$. – user26857 Jun 08 '15 at 23:16
@user26857: There is not need to assume that $R$ is flat over $A$? – user237522 Jun 08 '15 at 23:21
@user26857 I understand the notation. How do we show that this is the whole kernel? I don't think this is obvious—there may be many ways to write a given tensor. – Andrew Dudzik Jun 08 '15 at 23:26
@Slade For a proof see here. – user26857 Jun 08 '15 at 23:33
@user26857: Great! thanks. – user237522 Jun 08 '15 at 23:35

$R/I \otimes_A R/I \cong (R \otimes_A R)/(I \otimes_A I)$?

2 Answers2