# $a$ is a prime and $R$ is an integral domain $\implies a$ is irreducible

I noticed the following statement in a proof of a theorem on Euclidean rings.

$a$ is a prime and $R$ is an integral domain $\implies a$ is irreducible

It seems like this part of the statement is superfluous - $R$ is an integral domain.

Aren't primes always irreducible?

No, if the ring is not an integral domain, primes need not be irreducible. Consider $R = \mathbb{Z}/(6)$.

The elements $\overline{2}$ and $\overline{3}$ are prime, but you have $\overline{2} = \overline{2}\cdot \overline{4}$ and $\overline{3} = \overline{3}^2$, so they are not irreducible.

The most common definition is that a nonzero element $p$ is prime if it not invertible and, for all $a,b\in R$, $p\mid ab$ implies $p\mid a$ or $p\mid b$.

Every prime in an integral domain is irreducible: if $p$ is prime and $p=ab$, then either $p\mid a$ or $p\mid b$. If $p\mid b$, we have $b=px$ and so $p=apx$, so $ax=1$ and $a$ is invertible.

Note that the hypothesis that $R$ is an integral domain has been used when canceling $p$ from $p=apx$. If you add in the definition that $p$ is not a zero divisor, then the reasoning can go on in the same fashion.

No, a prime element need not be irreducible.

Recall that an element $x$ in a ring $R$ is prime if $\langle x\rangle$ is a prime ideal, ie if $R/\langle x\rangle$ is an integral domain.

For an example of a non-reducible prime, take the element $x = (1,0)$ in $R = \mathbb{F}_2\times \mathbb{F}_2$, which is not irreducible, since $x = x^2$ is a non-trivial factorization. But the quotient $R/\langle x\rangle$ is isomorphic to $\mathbb{F}_2$ and hence an integral domain.