(Proof explanation) $\mathbb{Z} /\left(a^{2}+b^{2}\right) \mathbb{Z} \cong \mathbb{Z}[i] /\langle a+b i\rangle$ doesn't hold

Question

I was trying to solve the following exercise. I found the official solution to be too concise for me.

If $a$ and $b$ are not relatively prime, then the $\operatorname{map} \mathbb{Z} /\left(a^{2}+b^{2}\right) \mathbb{Z} \rightarrow \mathbb{Z}[i] /\langle a+b i\rangle$ is neither injective nor surjective.

Proof: Suppose $d = gcd(a,b)$ divides $a$ and $b$. Then the ideal $\left\langle X^{2}+1, d\right\rangle$ contains the ideal $\left\langle X^{2}+1, a+b X\right\rangle .$ We thus get a surjection of $\mathbb{Z}[X] /\left\langle X^{2}+1, a+b X\right\rangle$ onto $\mathbb{Z}[X]\left\langle X^{2}+1, d\right\rangle \cong (\mathbb{Z} / d \mathbb{Z})[X] /\left\langle X^{2}+1\right\rangle .$ But it is clear that the map from $\mathbb{Z}$ to $(\mathbb{Z} / d \mathbb{Z})[X] /\left\langle X^{2}+1\right\rangle$ is not surjective, so the map from $\mathbb{Z}$ to $\mathbb{Z}[X] /\left\langle X^{2}+1, a+b X\right\rangle$ cannot be surjective either.

Here are the questions that I had:

1. Is the reason that the map from $\mathbb{Z}$ to $(\mathbb{Z} / d \mathbb{Z})[X] /\left\langle X^{2}+1\right\rangle$ is not surjective due to counterexamples like $x + \left\langle X^{2}+1\right\rangle$? (Nothing seems to get mapped to this, although I can't prove it for sure)

2. Then why is the above map not injective? (I don't think this was addressed in the proof?)

Answer 1

1. Yes. Since the map factors through $\mathbb{Z}/d\mathbb{Z}\to(\mathbb{Z}/d\mathbb{Z})[X]$ (i.e., $\mathbb{Z}\to\mathbb{Z}/d\mathbb{Z}\to(\mathbb{Z}/d\mathbb{Z})[X]\to(\mathbb{Z}/d\mathbb{Z})[X]/\langle X^2+1\rangle$) and there are no nonzero $n_0+n_1X\in(\mathbb{Z}/d\mathbb{Z})[X](X^2+1)$.

2. Both (the underlying sets of) $\mathbb{Z}/(a^2+b^2)$ and $\mathbb{Z}[i]/(a+bi)$ are finite sets of $a^2+b^2$ elements, and we know $\mathbb{Z}/(a^2+b^2)\to\mathbb{Z}[i]/(a+bi)$ is not surjective, hence it is also non-injective.