How to show that
If $p$ is a prime and $p\equiv1\bmod4$ then there exists an ideal of the $R=\mathbb{Z}[i]$, the ring of Gaussian integers, such that $R/I$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$.
I don't know how to approach this problem. These are some things I noted while scribbling.
If we argue that the canonical homomorphism '$f$' from $R$ to $R/I$ takes $i$ to $f(i)$ such that $f(i)^2 = -1 \bmod p$ because $i^2 = - 1 \bmod p$.
The element $(p-1)/2 !$ in $\mathbb{Z}/p\mathbb{Z}$ satisfies this. Hence, $f(i) = (p-1)/2 !$, because using Wilson's Theorem that $(p-1)! = -1 \bmod p$ gives us $( (p-1)/2 ! )^2 = -1 \bmod p$.
I will really appreciate some help on this. Thank you so much.
You found an $a\in\mathbb Z$ such that $p\mid a^2+1$. If $p$ is prime in $\mathbb Z[i]$ then, from $p\mid a^2+1$ you get $p\mid a+i$ or $p\mid a-i$, and both cases lead to a contradiction. This shows that $p$ isn't prime in $\mathbb Z[i]$. Then $p=(m+ni)(m-ni)$, so $p=m^2+n^2$. Now just take the ideal generated by $m+ni$.