> [!NOTE] Lemma (Irreducible Elements of Euclidean Domain are Prime) > Let $R$ be a [[Euclidean Domain|Euclidean domain]]. If $a\in R$ is [[Irreducible Elements of Integral Domain|irreducible]] then $a$ is [[Prime Elements of Integral Domain|prime]]. Remark: some authors call this Euclid's lemma for Euclidean domains. ###### Proof that Irreducible Elements of Euclidean Domain are Prime Let $R$ be a Euclidean domain and $a\in R$ be irreducible. Suppose $a\mid bc$ where $b,c\in R$. WTS that $a\mid b$ or $a\mid c$. Suppose $a\not\mid b$. If $a$ and $b$ are not coprime then there is some $c\in R$ dividing both $a,b$ that is not unit. As $a$ irreducible, we have $c\sim a$. Thus $a\mid c$ but $c\mid b$ therefore $c\mid b$ contradicting the fact that $a\not \mid b$. So $a$ and $b$ are coprime. Thus by [[Bézout's Lemma for Euclidean Domains|Bezout's lemma]], there are $x,y\in R$ such that $xa+yb=1$. Multiply by $c$ to obtain $xac+ybc=c$. Now $a\mid bc$ so we can write $bc=ka$ with $k\in R$. So $c=xac+ybc=xac+ka=a(xc+k)$, that is, $a\mid c$ as required.