$(2)$ is a prime ideal in $\mathbb{Z}_6$

Question

How does one show that $(2)$ is a prime ideal in $\mathbb{Z}_6$?

With it being modular arithmetic I’m not sure how this is done. I know I have to show that if $ab \in (2)$ then either $a \in (2)$ or $b\in (2)$.

Answer 1

$ab \in (2)$ implies that $ab \in \{0,2,4\}$. Now suppose $a,b \not\in \{0,2,4\}$, same as saying that $a,b \in \{1,3,5\}$.

Now you can check that $\{1,3,5\}$ is closed under multiplication . This shows that $ab \in \{1,3,5\}$, a contradiction.

Another way:

Consider $\mathbb{Z}_6/\langle 2 \rangle=\{0+\langle 2 \rangle,\, 1+\langle 2 \rangle,\, 3+\langle 2 \rangle,\, 5+\langle 2 \rangle\}$. Now show that this ring is an integral domain.