If $I$ and $J$ are ideals of an $R$-algebra $A=I+J$ then $I\oplus J\simeq A\oplus (I\cap J)$?

Question

Let $I$ and $J$ be two left ideals of an $R$-algebra $A$ such that $A=I+J$. Here $R$ is commutative ring with identity $1_R$. How can I show $$I\oplus J\simeq A\oplus (I\cap J)?$$

I've tried several mappings but none worked out.

Thanks.

Answer 1

The point is exactly that $A$ itself is a projective $A$-module, but you can understand the proof as long as you know that $A$ is a free $A$-module with basis $\{1\}$. First the inclusion maps $I\subset A$ and $J\subset A$ give a homomorphism $\pi:I\oplus J\to A$. Since $I+J=A$, $\pi$ is a epimorphism and there exist $i\in I$ and $j\in J$ such that $\pi(i,j)=i+j=1$. Define $\iota: A\to I\oplus J, 1\mapsto(i,j)$. Then we have $\pi\iota=1_A$ and $\ker\pi=\{(i,-i)\mid i\in I\cap J\}\cong I\cap J$, which force $I\oplus J\cong\text{im }\pi\oplus\ker\pi\cong A\oplus (I\cap J)$.

Answer 2

Consider the $A$-module (or $R$-module) epimorphism $$I\oplus J \rightarrow I+J=A, \;\; (i, j)\mapsto i+j \in A,$$

compute its kernel and use the first isomorphism theorem.

Now, if you know that $A$ is a projective $A$-module and know what properties projective modules have, this should give you the isomorphism you are looking for. If not, the isomorphism is still not too hard to guess.