This is a very elementary question. In Matsumura's book "Commutative ring theory", I've found the following isomorphism: $$\mathbb{Z}[\sqrt{-5}]\cong\mathbb{Z}[X]/(X^2+5).$$ As a quotient ring is involved, my idea is to use the first isomoprhism theorem, i.e. I want to define a homomorphism $\mathbb{Z}[X]\to\mathbb{Z}[\sqrt{-5}]$ whose kernel is exactly $(X^2+5)$, but here I get stuck.
Indeed, $$\sum_{i=0}^n a_nX^n\mapsto a_0+a_1\sqrt{-5}$$ is a group homomorphism w.r.t. $+$, but this is not a ring homomorphism as $$\Big(\sum_{i=0}^na_iX^i\Big)\Big(\sum_{i=0}^mb_iX^i\Big)=\sum_{i=0}^{n+m}(\sum_{k=0}^i a_kb_{i-k})X^i\mapsto a_0b_0+(a_1b_0+a_0b_1)\sqrt{-5}$$ which is different from $$(a_0+a_1\sqrt{-5})(b_0+b_1\sqrt{-5})=(a_0b_0-5a_1b_1)+(a_1b_0+a_0b_1)\sqrt{-5}.$$
What's the right idea to get this isomorphism using the first isomorphism theorem?
PS: Similarly, I've seen that $\mathbb{C}\cong \mathbb{R}[X]/(X^2+1)$ and the various proofs define explicitly an isomorphism, but how can I obtain this result from the first isomorphism theorem?
Try
$$\phi:\Bbb Z[x]\to\Bbb C\;,\;\;\phi p(x):=p(\sqrt{-5})$$
Observe that the image is $\;\Bbb Z[\sqrt{-5}]\;$ , and the kernel is...
For the second one, try
$$\psi:\Bbb R[x]\to\Bbb C\;,\;\;\psi p(x):=p(i)$$