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I am trying to prove the following:

Let $R$ be a commutative ring with identity and $I,J\triangleleft R$. Show that $$R/I\otimes_R R/J\simeq R/(I+J).$$

I tried to prove the following hypothesis:

Let $M,N$ be $R$-modules and let $L\subset N$ be a submodule. Then the following holds: $$M\otimes_R (N/L)\simeq \frac{M\otimes_R N}{\langle m\otimes l\rangle_{l\in L, m\in M}}.$$

My approach is to construct $R$-bilinear mapping $$B:M\times N\to M\otimes_R (N/L)$$ defined as $$B(m,n):=m\otimes [n]_L.$$ Then it seems useful to use the tensor product universal property and the first isomorphism theorem (how to check surjectivity?)

But I am even not sure about the correctness of the hypotgesis.

UPD.

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  • $\begingroup$ The thing is, $\langle 0\otimes l\rangle_{l\in L}=0$. Perhaps you mean to quotient by $M\otimes_R L$? $\endgroup$ Aug 5 at 21:22
  • $\begingroup$ I think it will be easier to just construct a bilinear map $R/I\times R/J\to R/(I+J)$. $\endgroup$
    – Michael
    Aug 5 at 21:27
  • $\begingroup$ Yes, the span of elements $m\otimes l$, where $l\in L$. $\endgroup$ Aug 5 at 21:27
  • $\begingroup$ @Michael I wish to establish a general fact about quotients, if it is possible $\endgroup$ Aug 5 at 21:29

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Your hypothesis is correct, but unfortunately it is not a completely trivial thing. The technical term is to say that tensoring with $M$ is a "right exact" functor. Proving this directly is actually a little tricky, usually one would use the fact that it is a left adjoint. The accepted answer to this question also gives a more elementary way of showing that.

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