Showing that the map $R/I \otimes_R R/J \to R$, $(r+I) \otimes (s+I) \mapsto rs$ is well defined 
Let $R$ be a commutative ring, and $I,J\subseteq R$ ideals. I'm trying to show that the map $$\phi: R/I\otimes_R R/J\rightarrow R$$ given by $$(r+I)\otimes (s+J)=rs(1+I)\otimes (1+J)\mapsto rs$$ is well defined. 

This is what I've thought so far:
Let $r+I=r'+I$, and $s+J=s'+J$. Then I need to show that
$$
  (r+I)\otimes (s+J) = (r'+I)\otimes (s'+J)
  \implies
  rs = r's'.
$$
First I thought I could write $$(rs-r's')(1+I)\otimes (1+J)=(r+I)\otimes (s+J)-(r'+I)\otimes (s'+J)=0$$ and say that the LHS maps to $rs-r's'$ and the RHS maps to $0$, and that this implies $rs-r's'=0$. But this doesn't seem right, since I'm then assuming $\phi$ is well defined.
I'm also not sure if I can say that $(1+I)\otimes (1+J)$ generates $R/I\otimes_R R/J$, so if $(rs-r's')(1+I)\otimes (1+J)$ then $rs-r's'=0$.
Am I on the right track? Any hint would be appreciated.
 A: Your proposed map
$$
         R/I \otimes_R R/J
  \to    R,
  \quad  \overline{x} \otimes \overline{y}
  \mapsto xy
$$
is well-defined only if $I = J = 0$, in which case it is the usual isomorphism of $R$-modules $R \otimes_R R \to R$.
This follows from the universal property of the tensor product:
The above map is well-defined if and only if
$$
          R/I \times R/J
  \to     R,
  \quad   (\overline{x}, \overline{y})
  \mapsto xy
$$
is a well-defined $R$-bilinear map.
By fixing the argument $\overline{1} \in R/I$ we see that the map
$$
          R/J
  \to     R,
  \quad   \overline{y}
  \mapsto y
$$
must be well-defined, which it is only for $J = 0$ (because it follows for every $y \in J$ from the well-definedness and $\overline{y} = \overline{0}$ that $y = 0$).
We find in the same way that we need $I = 0$.

What does hold is that $R/I \otimes_R R/J \cong R/(I+J)$:
It holds for every $R$-module $M$ that
$$
        R/I \otimes_R M
  \cong M/IM,
$$
and it therefore follows that
$$
        R/I \otimes_R R/J
  \cong (R/J) / (I \cdot R/J)
  =     (R/J) / ((I+J)/J)
  \cong R/(I+J).
$$
This isomorphism is on elements given by
$$
          R/I \otimes_R R/J
  \to     R/(I+J),
  \quad   \overline{x} \otimes \overline{y}
  \mapsto \overline{xy}.
$$
Note that the well-definedness of your proposed map is therefore equivalent to the well-definedness of the map
$$
          R/(I+J)
  \to     R,
  \quad   \overline{z}
  \mapsto z,
$$
which holds only for $I + J = 0$, i.e. for $I = J = 0$.
This confirms the above result.
