I want to show that if the ring homomorphism $f:R \rightarrow S$ is onto, $I,J$ two ideals of $R$ and $\ker f\subseteq I$ then $f(I\cap J)=f(I)\cap f(J)$.
To prove $f(I)\cap f(J)\subseteq f(I\cap J)$, I started with $f(x)\in f(I)\cap f(J)$, thus $f(x)\in f(I)$ and $f(x)\in f(J)$, so $x\in I\cap J$ and $f(x)\in f(I\cap J)$.
This is clearly wrong because I have not used $\ker f\subseteq I$ and without this condition this statement is not true. Can you point out where I did wrong and hint at the right path?
There are multiple theorems with the condition $\ker f\subseteq I$. How do I use this condition? I wanted to find $f(x)-f(y)=f(x-y)=0$ to have $x-y\in\ker f$ and go from there.
Note that I have not yet studied the homomorphism theorems and quotient rings; I know the correspondence theorem.
Let $y\in f(I)\cap f(J).$ Then $y\in f(I)$ and $y\in f(J),$ that is, there exist $a\in I$ and $b\in J$ such that $y=f(a)=f(b),$ so
$$f(a-b)=0\implies a-b\in Ker(f)\implies a-b\in I\implies b\in I.$$
Therefore, $b\in I\cap J\implies y=f(b)\in f(I\cap J).$
The other inclusion is easy. I think that the onto condition is not necessary.