Let $R $ be an integral domain with identity. Prove that if $p $ is irreducible and $u$ is a unit, then $pu $ is irreducible.

Question

Let $R $ be an integral domain with identity. Prove that if $p $ is irreducible and $u$ is a unit, then $pu $ is irreducible.

My proof:

Let $ Pu=nm$

$P=u^{-1} nm$

Then $u^{-1} n$ or m is unit since $P $ is irreducible .

Thus $Pu $ is irreducible.

Is my proof correct ( I did not use that $R$ is $ID$).

Answer 1

You are assuming commutativity in your proof, otherwise you might be multiplying $u^{-1}$ on the wrong side since we do not know which of $m$ and $n$ is the unit, so commutativity is needed for this proof. I think the proof is correct,you have shown every factorization of $Pu$ in two factors must have a factor that is a unit, this is one of the definitions of irreducibility.

When I look up "Irreducible element" in wikipedia the definition is given for Integral Domains, so this may be the reason it is an integral domain.

Answer 2

The proof is correct, but it should be supplemented by some remark.

Suppose $pu=nm$; we have to show that either $n$ or $m$ is a unit.

From $pu=nm$ we have $p=(u^{-1}n)m$. By irreducibility of $p$, either $u^{-1}n$ or $m$ is a unit. If $m$ is a unit, we are done. If $u^{-1}n=v$ is a unit, then $n=uv$ is a unit too.

So we have proved that either $m$ or $n$ is a unit. Hence $pu$ is irreducible.