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

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.

I would start with the homomorphism $R/\ker(\phi)\overset{\cong}{\longrightarrow}S\twoheadrightarrow S/\phi(J)$.
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 $\psi(r + \ker \phi) = \phi(r) + \phi(J). \ker \psi = \{ r + \ker \phi \in R/ \ker \phi | \phi(r) + \phi(J) = \phi(J) \}$.
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 $S/ \phi(J) \cong \frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi}$