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Euclid's lemma: Let $A$ be a principal ring with $x$ and $y$ belonging to $A$, and $p$ an irreducible element of $A$. If $p$ divides the product $xy$, then $p$ divides $x$ or $p$ divides $y$.

Definition of irreducible element: A non zero element p is said to be irreducible in A (where A is an integral domain) if p is non invertible in A and all divisors of p in A are trivial. In other words a nonzero element p of A is irreducible iff p is non invertible and the only divisors of p in A are the invertible elements of A and the elements associated to p.

Question: Should $A$ be an integral domain too? And I need a justification why please. I'm not quite grasping why an integral domain can help with such lemmas.

Bill Dubuque
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Assil Chkeir
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  • You used the term irreducible element of $A$ without definition. Many authors will define that only in integral domains. See the Wikipedia article here for some background. You are asking if an irreducible element must be a prime element in a principal ring, not necessarily a domain. – hardmath Mar 13 '21 at 17:49
  • Fixed it....... – Assil Chkeir Mar 13 '21 at 17:56
  • The common proof of Euclid's Lemma still works here - see my answer in the linked dupe. – Bill Dubuque Mar 13 '21 at 19:34

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You don't need the assumption that $A$ is integral.

In fact, the following proof works for any commutative principal ring $A$ (understood as every ideal being generated by one element).

Proof: Suppose $p$ divides $xy$ with $p$ irreducible.

Since $A$ is principal, there exists $u\in A$ such that $(u) = (p, x)$.

It follows that there exists $t, r \in A$ such that $p = ut$ and $x = ur$. But $p$ is irreducible, hence either $u$ or $t$ is a unit.

If $t$ is a unit, then we have $x = ur = t^{-1}pr$ which means that $p$ divides $x$.

If $u$ is a unit, then we have $(p, x) = (u) = A$ and hence there exists $a, b \in A$ such that $pa + xb = 1$. It follows that $p$ divides $pay + xyb = y$.

WhatsUp
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  • I have a question regarding this, Shouldn't A be an integral domain for us to talk about an irreducible element in A. My grasp on what an integral domain helps us with isnt that great. So please if u could explain more im still a student – Assil Chkeir Mar 13 '21 at 18:09
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    Usually irreducible elements are defined only for integral domains. But you can define it for any commutative ring: $p$ is called irreducible if $p$ is not a unit and $p = ab$ implies either $a$ or $b$ is a unit. However, for non-integral rings, this notion is sometimes useless. Consider the ring $\Bbb Z / 6\Bbb Z$. There is no irreducible element in this sense! – WhatsUp Mar 13 '21 at 18:20