Let $p\in \mathbb Z$ be a prime congruent to $1\mod 4$. Let $p=\pi\overline \pi$. Show that $\mathbb Z[i]/(p)\simeq \mathbb Z[i]/(\pi)\times\mathbb Z[i]/(\overline\pi )$. Show that the quotient ring $\mathbb Z[i]/(p)$ has order $p^2$ and conclude that the above factors are both fields of order $p$.
To apply the Chinese remainder theorem, I need to prove that $(\pi)+(\overline \pi)=1$. Let $\pi=a+bi$. Then I need to show that $1\in (a+bi)+(a-bi)$. But I don't see why this must be true. The RHS contains $2a$, but $a,b\in \mathbb Z$, why should it contain $1$?
Further, $\pi,\overline \pi$ are irreducible (=prime), so the factors are fields. But how do I show that they contain $p $ elements? According to the question, I'm supposed to deduce it from the fact that $\mathbb Z[i]/(p)$ has order $p^2$. Can I show the latter by saying that $\mathbb Z[i]/(p)\simeq \mathbb Z_p[x]/(x^2+1)=\{x+iy:x,y\in \mathbb Z_p\}$? And then, am I supposed to conclude that since neither of the factors $\mathbb Z[i]/(a\pm bi)$ has order $1$, each of them must have order $p$?