Primes in $Z/12Z$

Question

I am trying to understand a solution given to me. So I have to find the prime elements in $Z/12Z$.

The following is given to me:

$3$ and $9 = −3$ are prime elements since $⟨3⟩ = ⟨−3⟩$ and $Z12/⟨3⟩$ is isomorphic to $(Z/12Z)/(3Z/12Z)$ which is isomorphic to $Z/3Z$ is an integral domain.

Similarly 2, 10 = −2 are primes since Z12/⟨2⟩ is isomorphic to $(Z/12Z)/(2Z/12Z)$ which is isomorphic to $Z/2Z$.

4, 8 and 6 are not primes. Why are these elements not prime. Can't we do the same thing with 4 and 8 as well? My question would also involve if there is a more simple way to figure this out.

Answer 1

Let $A$ be a ring.

A simple way of checking whether an element $p\in A,$ is a prime element or not is the following: $p\in A$ is prime iff for any $a,b\in A,$ if $p\mid ab,$ then, $p\mid a$ or $p\mid b.$ (Here, I define $\alpha\mid \beta,$ for an elements $\alpha, \beta\in A,$ as equivalent to there existing a $\gamma\in A$ such that $\beta=\alpha\gamma.$)

The condition above is easily proven to be equivalent to the definition you use, which is that $A/(p)$ is an integral domain.

For $a\in A,$ denote by $\bar a$ the image of $a$ in $A/(p)$. $A/(p)$ being an integral domain is equivalent to the following: for $\alpha,\beta\in A/(p),$ $\alpha\beta=0$ then $\alpha=0$ or $\beta=0.$ This is further equivalent to, for $a,b\in A,$ $\bar a\bar b=\bar{ab}=0$ then $\bar a=0$ or $\bar b=0.$
Now, this is equivalent to: for $a,b\in A,$ if $p\mid ab$ then $p\mid a$ or $p\mid b.$

Now, I will do one example: $4\in \mathbb Z/12\mathbb Z$ is not prime, since $4\mid2\cdot 2,$ but $4\not\mid2.$