0
$\begingroup$

I need some help with the following: I have already proved that $\mathbb Z[i]/(p) \cong \mathbb F_p[x]/(x^2+1)$ however, I cannot figure out how to prove that:

  1. (p) is prime in $\mathbb Z[i]$ if and only if there is no such $\mathbb a \in \mathbb F_p$ such that $\mathbb a^2+1=0 $
  2. It happens if and only if $\mathbb p\equiv 3\pmod 4$

    I managed to do one direction of the first of these. Since if there was such an $\mathbb a$ one could consider $\mathbb (x-a)(x+a)$ which gives $\mathbb 0$ as a result in $\mathbb F_p[x]/(x^2+1)$ while both of the factors are nonzero thus $\mathbb F_p[x]/(x^2+1)$ is not an integral domain, which means that $\mathbb Z[i]/(p)$ is not an integral domain either, which in turn gives us that $\mathbb (p)$ is not a prime ideal of $\mathbb Z[i]$.
    However, I am stuck with the other direction and the second part. Could you help me out? Thanks!
$\endgroup$
0
$\begingroup$
  1. You have already provided one direction. For the other, suppose there is no such $a$. $F_p[x]$ is a UFD so $x^2+1$ is prime if and only if it is irreducible, which since it is degree two, occurs if and only if it is has some linear factor, i.e. a root. But our assumption that there is no such $a$ exactly says that there is no root, so $x^2+1$ is prime, so $F_p[x]/(x^2+1)$ is an integral domain, so $\mathbb{Z}[i]/(p)$ is an integral domain.

  2. So we need to prove -1 is a non-square in $\mathbb{Z}/(p)$ if and only if $p$ is 3 mod 4. -1=1 is a square mod 2, so we can assume $p$ is odd. See any algebra book for a proof that the multiplicative group of $\mathbb{Z}/(p)$ is cyclic. It is necessarily of order $p-1$. Let $g$ be a generator. Then the (nonzero) squares are exactly the elements of the form $g^s$ for $s$ even. Write $-1 = g^r$. Then $r \neq 0$ mod $p-1$, but $2r = 0$ mod $p-1$, so we get $r = (p-1)/2$. This is odd if and only if $p = 3$ mod 4.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.