Let $R$ be ring, $I, J$ be ideals. I believe this exact sequence is valid. Can someone verify this for me?
$$ 0 \rightarrow IJ \xrightarrow{f} I \cap J \xrightarrow{g} I \oplus J \xrightarrow{h} (I + J) \rightarrow 0 \\ f: IJ \rightarrow I \cap J\quad f(r) \equiv r \\ g: I \cap J \rightarrow I \oplus J \quad g(r) \equiv (r \text{ mod } I , r \text{ mod } J) \\ h: I \oplus J \rightarrow I + J\quad h(i, j) \equiv i + j \\ $$
It is clear that $f: IJ \rightarrow I \cap J$ is injective, since $IJ \subseteq I \cap J$. (For a proof, see here)
It is clear that $h: I \oplus J \rightarrow I + J$ is surjective, since every element in $i + j \in I + J$ is the image of $(i, j)$: $(i, j) \overset{h}\mapsto (i + j)$.
The part I am unsure of is that of $(g,h)$. Let me attempt the proof: We need to show that $Im(g) = Ker(h)$. That is:
1. $Im(g) \subseteq Ker(h)$$$ g: I \cap J \rightarrow I \oplus J \quad g(r) = (r \text{ mod } I , r \text{ mod } J) \\ Im(g) = \{ (r \text{ mod } I, r \text{ mod } J) : r \in I \cap J \} \\ \text{(Since $r$ is in the ideals $I, J$, residue is $0$)} \\ Im(g) = \{ (0, 0) \} \quad \\ Im(g) \subseteq Ker(h) $$
2. $Ker(h) \subseteq Im(g)$$$ h: I \oplus J \rightarrow I + J\quad h(i, j) \equiv i + j \\ h((i, j)) = 0 \\ i + j = 0 \\ -i = j \\ \text{($j$ is a multiple of $i$; Ideals closed under multiplication)} \\ j \in I \\ (i, j) \in I \cap J \\ (i, j) \in Im(g) $$
Hence we conclude that $Im(g) = Ker(h)$.
Is this correct?
I was shown that this exact sequence is incorrect. Note that the image of $g: I \cap J \rightarrow I \oplus J$ is just $\{ 0 \}$, while the kernel of $h: I \oplus J \rightarrow I + J$ contains pairs $(i, -i)$. So, $Im(g) \neq Ker(h)$, and hence the sequence is not exact.