2

Let $R$ be a ring with ideals $I$ and $J$. I want to show that

$$R/I \otimes_R R/J = R/(I+J)$$

I'm not sure how to do it and I find working with tensors and quotients together quite difficult so some guidance would be appreciated.

I have constructed a map $\phi : R/I \times R/J \to R/(I + J)$ as $\phi(a + I,b +J) = ab + I + J$ and showed that this is well defined and bilinear, this gave me a map $\bar \phi : R/I \otimes_R R/J \to R/(I+J)$ from the universal property of tensor products.

In order to get a map back the way I have considered the kernel of the map $\psi : R \to R/I \times R/J$ defined by $\psi(m) = (m + I)\otimes (m + J)$. I think the kernel will be $I + J$ and I wanted to show that all elements of $I + J$ get mapped to zero by this and then secondly show that anything gets mapped to zero is within $I + J$.

For the first part I reason that any element of $I + J$ is of the form $a i + b j$ and so $\psi(a i + b j) = (a i + b j + I) \otimes (a i + b j + J) = (b j + I) \otimes (a i + J)$. I don't know how to continue to show that this $= 0$.

For the second part I don't know how to start.

And then once I have the two maps how would I show they form an isomorphism?

  • 2
    Rather try to prove that $M/(I+J)$, with your map $\phi$, satisfies the universal property of tensor products. – Berci Apr 17 '20 at 16:41
  • Thanks @Berci, To do that I think I would need to take an arbitrary bilinear map from $M/I \times M/J \to P$ and show that it factors uniquely through $M/(I + J)$ but I can't figure out how to do that. Some hints would be great! –  Apr 17 '20 at 16:44
  • 2
    Wait. First, is $R$ commutative? Second, $I,J$ are not submodules of $M$ but ideals of $R$. Maybe you wanted to write $R/I\otimes R/J\cong R/(I+J)$? It is more probable to be true.. – Berci Apr 17 '20 at 17:19
  • Thanks, I see from your comment that I have tried to prove something too strong by bringing in M. I'll change it to R. –  Apr 17 '20 at 17:53
  • https://math.stackexchange.com/questions/742473/tensor-product-of-quotient-rings – Viktor Vaughn Apr 17 '20 at 23:47

2 Answers2

3

I assume $R$ is a commutative ring with unity $1$.

The key observation is that $R/I$ is a cyclic $R$-module, generated by $[1]_I=1+I$, for any ideal $I$.

Let $\beta:R/I\times R/J\to M$ be a bilinear map, and define $b_0:R\to M, \ r\mapsto r\cdot \beta([1]_I,\,[1]_J)$.

Show that $I+J\subseteq \ker b_0$, so that $b_0$ induces a map $b:R/(I+J) \to M$.
Prove that $b$ is the unique homomorphism making $\beta=b\circ\phi$.

Berci
  • 92,013
1

Let‘s show that $R/(I+J)$ satisfies the universal property of the tensor product; then we‘re done by the usual universal property argument. First, we need a bilinear map $\phi: R/I \times R/J\rightarrow R/(I+J)$, which is the map you defined. Now let S be any ring together with a bilinear map $\varphi: R/I \times R/J\rightarrow S$; we need to see that this factors uniquely through $R/(I+J)$. Consider the map $\alpha: R\rightarrow R/I\times R/J\rightarrow S$ with $r\mapsto (r+I,r+J)\mapsto\varphi(r+I,1)$. Since $I$ and $J$ are in the kernel of $\alpha$, so is $I+J$ and we get an induced map $\beta:R/(I+J)\rightarrow S$ defined by $r+(I+J)\mapsto \varphi(r+I,1)$ which is the one we were looking for.

Thanks.
  • 696