Question about ideals of quotient ring

Question

I found in a book that $\frac{Z[i]}{(3)} \approx \frac{\frac{Z[X]}{(X^2+1)}}{(\overline{3})}$, where $(\overline{3})$ is ideal generated by $\overline{3}$ in $\frac{Z[X]}{(X^2+1)}$. And after that, it says that $(\overline{3}) = \frac{(3,X^2+1)}{(X^2+1)}$. My question is: Why $(\overline{3}) = \frac{(3,X^2+1)}{(X^2+1)}$ ? What Isomorphism theorem is this?

Answer 1

$\langle\overline{3}\rangle = \frac{\langle 3, X^2 + 1\rangle}{\langle X^2 + 1\rangle}$ is not merely an isomorphism; the two objects are literally equal as sets.

So consider any $\overline{a}(X) \in \langle\overline{3}\rangle$. Then for some $b(X) \in \mathbb Z[X]$ $$ \overline{a}(X) = \overline{3} \ \overline{b}(X) = [3 + \langle X^2 + 1\rangle][b(X) + \langle X^2 + 1\rangle] = 3b(X) + \langle X^2 + 1\rangle $$ which is in $\frac{\langle 3, X^2 + 1 \rangle}{\langle X^2 + 1\rangle}$ because $3b(X) \in \langle 3, X^2 + 1 \rangle$.

Conversely consider any $\overline{a}(X) \in \frac{\langle 3, X^2 + 1 \rangle}{\langle X^2 + 1\rangle}$. Then for some $b_1(X), b_2(X) \in \mathbb Z[X]$ $$ \overline{a}(X) = [3b_1(X) + (X^2 + 1)b_2(X)] + \langle X^2 + 1\rangle = 3b_1(X) + \langle X^2 + 1\rangle $$ which is in $(\overline{3})$.