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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

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  • $\begingroup$ $\Bbb Z_n/\left<m\right>\cong\Bbb Z_{\gcd(m,n)}$. $\endgroup$ – Lord Shark the Unknown May 2 '18 at 5:31
  • $\begingroup$ Please explain a little bit more $\endgroup$ – Jandré Snyman May 2 '18 at 5:33
  • $\begingroup$ $\Bbb Z _n/\left<m\right>\cong \Bbb Z/\left<m,n\right>=\Bbb Z/\left<\gcd(m,n)\right>=\Bbb Z_{\gcd(m,n)}$. $\endgroup$ – Lord Shark the Unknown May 2 '18 at 6:02
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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 $p$.

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