I'm having some problems with tensor product.
I know that, in general, for $M$ an $R-$module and $I$ an ideal it holds $$M \otimes_R R/I = M/IM.$$ I also know that for $M_1$, $M_2$ and $N$ $R-$modules and $f: M_1 \longrightarrow M_2$ it holds $$Im(f \otimes Id_N) = Im(f) \otimes N$$
If I consider the short exact sequence: $$0 \longrightarrow I \longrightarrow R \longrightarrow R/I \longrightarrow 0$$ where the first map is the natural inclusion $\mathcal{i}$. Tensoring with $- \otimes_R R/J$, where J is an ideal of $R$, I obtain the exact sequence: $$ I \otimes_R R/J \longrightarrow R \otimes_R R/J \longrightarrow R/I\otimes_R R/J \longrightarrow 0$$ I have to prove that $$Ker(\mathcal{i} \otimes id_{A/J}) = (I \cap J)/IJ$$ Why is it wrong to say that $Im(\mathcal{i} \otimes id_{R/J}) =Im(\mathcal{i}) \otimes R/J =I \otimes R/J =I/IJ$ and conclude that $Ker(\mathcal{i} \otimes id_{A/J}) =0$?
Similarly, I have another problem looking at a counterexample that proves the fact that tensoring is not left exact. Put $R= K[X]$ and consider the exact sequence: $$0 \longrightarrow R \longrightarrow R \longrightarrow R/(X) \longrightarrow 0$$ where the first map is multiplication $\cdot X$ and the second is the natural projection. Tensoring we have an exact sequence $$ R \otimes_R R/(X)\longrightarrow R \otimes_R R/(X) \longrightarrow R/(X)\otimes_R R/(X) \longrightarrow 0$$ The first map is not injective since it is the zero map. This means that $Im(\cdot X \otimes id_{R/(X)})=0$. But using the two remarks above I have: $$Im(\cdot X \otimes id_{R/(X)}) = Im(\cdot X) \otimes R/(X) = (X) \otimes R/(X) = (X)/ (X)^2.$$
Where is my mistake? Can anyone help me?