Prove that $\frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi} \cong \frac{S}{\phi(J)}$

Question

Let $\phi: R \to S$ be a surjective homomorphism. Prove that $\frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi} \cong \frac{S}{\phi(J)}$ for an ideal $J$ of $R.$

Obviously, $S \cong R/ \ker \phi$ (first isomorphism theorem) and $\phi(J)$ is an ideal of $S$, since $\phi$ is surjective ($\forall s \in \phi(J), s = \phi(j)$ for some $j \in J, \forall s' \in S, s' = \phi(r)$ for some $r \in R$, so $ss' = \phi(j)\phi(r) = \phi(jr) \in \phi(J)$, and $ss' = \phi(r)\phi(j) = \phi(rj) \in \phi(J)$).

However, I'm having trouble with going on. If it's important, I mean a unitary ring by a ring.

Answer 1

I would start with the homomorphism $R/\ker(\phi)\overset{\cong}{\longrightarrow}S\twoheadrightarrow S/\phi(J)$.

Answer 2

Thanks are due to Batominovski's answer. I'm writing this to expand on that, because there are some(at least one) nontrivial steps in my opinion. Still, I'm accepting Batominovski's answer.

So we have an isomorphism $\phi': R/ \ker \phi \to S$ defined by $\phi' (r + \ker \phi) = \phi(r)$. Composing it with the projection mapping we get $\psi: R/ \ker \phi \to S/ \phi(J)$ defined by $\psi(r + \ker \phi) = \phi(r) + \phi(J)$. $$\ker \psi = \{ r + \ker \phi \in R/ \ker \phi \mid \phi(r) + \phi(J) = \phi(J) \}.$$

Now, here goes logic: $\phi(r) + \phi(J) = \phi(J) \Rightarrow \phi(r) \in \phi(J) \Rightarrow \ \ \exists j \in J: \ \phi(r) = \phi(j) \Rightarrow \ \exists j \in J: r-j \in \ker \phi \Rightarrow \ \exists j \in J, k \in \ker \phi: r = j + k \Rightarrow r \in J + \ker \phi$.

So, $\ker \psi = \ker \phi + J$. And by first isomorphism theorem we get $S/ \phi(J) \cong \frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi}$.