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

Question

I still struggle with elementary knowledge about quotient rings.

For example, why can I just write down the second following isomorphism:

$$\frac{\mathbb{C}[x, y, z]}{(x^2 + y^2 - 1, y \pm iz)} \simeq \frac{\frac{\mathbb{C}[x, y, z]}{(y \pm iz)}}{\frac{(x^2 + y^2 - 1,y \pm iz)}{(y \pm iz)}} \simeq \frac{\mathbb{C}[x, y]}{(x^2 + y^2 - 1)} $$

I know the first isomorphism is right by the third isomorphism theorem. But why can I write down the second one?

Answer 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.

Answer 2

Well, let $R$ be a commutative ring with ideals $J\subseteq I$. Then $R/I$ is isomorphic to $(R/J)/(I/J)$.

In the first isomorphism, you use $I=\langle x^2+y^2-1, y\pm iz\rangle$ and $J=\langle y\pm iz\rangle$. In the second isomorphism, you use the same result.