I am struggling slightly with proving the following;
$A = \langle p \rangle$ (principal ideal generated by $p$) is a prime ideal of $\mathbb{Z}_n$ if and only if $p$ is a prime divisor of $n$.
I am working on proving $A = \langle p \rangle$ is a prime ideal of $\mathbb{Z}_n$ if $p$ is a prime divisor of $n$.
I am using the fact that
$\mathbb{Z}_n/\langle p \rangle$ is an integral domain $\iff$ $\langle p \rangle$ is a prime ideal of $\mathbb{Z}_n$.
Now I already know the following;
- $\mathbb{Z}_n$ is a commutative ring with unity.
- $\langle p\rangle$ is an ideal of $\mathbb{Z}_n$.
- The above points imply $\mathbb{Z}_n/\langle p \rangle$ is a commutative ring with unity.
So all that we need to show is that $\mathbb{Z}_n/\langle p \rangle$ has no divisors of zero, that is if $(a +\langle p \rangle),(b +\langle p \rangle) \in \mathbb{Z}_n/\langle p \rangle$ such that
$$ (a +\langle p \rangle)(b +\langle p \rangle) = 0 + \langle p\rangle \implies (a +\langle p \rangle)= 0 + \langle p\rangle \text{ or } (b +\langle p \rangle) = 0 +\langle p \rangle $$
This is as far as I can get at the moment on this direction.
I have not yet start the only if direction
This boils down to $$ p\mid ab\implies p\mid a\lor p\mid b$$
Suppose $p$ divides neither $a$ nor $b$, so the gcd with either is $1$. Then by the extended Euclidean algorithm, we can write $1=xp+ya=zp+wb$ with integers $x,y,z,w$. By multiplication, $$ 1=(xp+ya)(zp+wb)=(xtp+xwb+ayz)\cdot p+yw\cdot ab$$ so that $ab$ cannot be a multiple of $p$.