Rings $R/I \cong h(R)/h(I)\,$ for injective ring hom $h$ [duplicate]

Solution 1:

It follows from the first isomorphism theorem for rings, like this:

Suppose $\phi: R\to S$ is any injective ring homomorphism. Then $R/I\cong \phi(R)/\phi(I)$ as rings.

Proof: Let $\psi: R\to \phi(R)/\phi(I)$ be given by $\psi(r)=\phi(r) + \phi(I)$. It can be checked that this is a ring homomorphism.

Moreover, its kernel is precisely $I$. Just chase it down: $\psi(r)=0+\phi(I)$ iff $r-i\in \ker \phi$, and $\phi$ was injective so...

Now, it's fairly obvious that there is an isomorphism between the rings in the "numerators" of the latter half, and the denominator of the second quotient is the image of the denominator in the first quotient, so you can apply this lemma above.