Suppose $A$ is a ring with proper ideals $I$ and $J$. I have showed the function $\theta:A\rightarrow A/I\times A/J$ given by $\theta(x) = (x+I,x+J)$ is a ring homomorphism with kernel $\ker \theta = I\cap J$. The First Isomorphism Theorem gives me that
$$A/\ker\theta = A/I\cap J \cong \text{im }\theta. $$
I am asked to define an injective ring homomorphism $A/I\cap J \rightarrow A/I\times A/J$. However, I have no clear idea of what $\text{im }\theta$ is supposed to be. Is seems like it is not necessarily equal to the codomain of $\theta$ and I am having a real mental block about what to do next.
Thanks!
The homomorphism $\theta$ need not be surjective always. For example, let $A = \mathbb{Z}, I = 2\mathbb{Z}, J = 4\mathbb{Z}$. Then $A/(I \cap J) = \mathbb{Z}_4,$ where as $A/I \times A/J = \mathbb{Z}_2 \times \mathbb{Z}_4$. If the ideals $I$ and $J$ are comaximal, i.e., if $I + J = A$ then $\theta$ is surjective.
But the induced map $\overline{\theta} : A/(I \cap J) \rightarrow A/I \times A/J$ is always injective. For that you don't need $\theta$ to be surjective.