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So I'm trying to prove that in an integral domain every prime element is irreducible.

Let's suppose that $p$ is an element in my integral domain.

So now I can study $ p=ab $ (if is not in this form, $p=a*1 $ and this is an irreducible element).

  1. $ p $ divides itself so p|p so $ p=ab.$
  2. $p|a $ or p|b (because p is prime) and now, WLOG let's suppose that p|a.
  3. If $p|a $ need to exist an element $ q $ such that $a=pq.$
  4. I can multiply both sides with $ b $ so $ p=pqb.$
  5. $p(1-qb)=0 $ so $ qb=1$ (this is true because we are working in an integrity domain and if $ a*b=0 $ a needs to be $0 $ or $ b$ needs to be $0$) and now I can say that b is a unit and $ p=ab $ is an irreducible element.

My question is, why b needs to be a unit element of my domain? Cannot be an invertible element? Like $4*(1/4)$=1 in $\mathbb{Q}$. And if $ b$ can be an invertible element, why does that mean that $ p $ is irreducible?

Sorry for my english, any help would be appreciated! By the way, I'm using Artin's book and $ a $ in $ R $ where $ R$ is a ring is an irreducible element if is not a unit and if it has no proper divisors.

teo
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  • Is $\frac{1}{4} \in \mathbb{Z} $ true ? $p$ is irreducible in $R$ if $p$ can't be expressed as a product of two non zero non unit elements OF $R$. – Sourav Ghosh Dec 13 '21 at 18:57
  • No sorry, made a mistake, already edited. – teo Dec 13 '21 at 18:58
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    Try to read the definition of unit given in your book. $ a $ is unit if $a$ is Invertible i.e $a$ has multiplicative inverse. And $\frac{1}{4} $ is unit in $\Bbb{Q}$ – Sourav Ghosh Dec 13 '21 at 19:01
  • Yeah, you're right, Artin says that the element that has multiplicative inverse are units (page 412). Thank you so much! – teo Dec 13 '21 at 19:12
  • prime $,p = ab\Rightarrow p\mid a,$ or $,b.,$ If $,p\mid a,$ then cancelling $p,\Rightarrow, 1 = (a/p)b,$ so $,b,$ is a unit. By symmetry $,p\mid b,\Rightarrow, a,$ is a unit. Thus $p$ is irreducible by definition, i.e. if $,p = ab,$ then $a$ or $b$ is a unit. See the Theorem in the dupe for a proof that makes the inference clearer. – Bill Dubuque Dec 14 '21 at 09:53
  • A unit in $R$ is an element invertible in $R$ i.e. a divisor of $1$ in $R$, so $1 = qb,\ q\in R,$ means $b$ is a unit (with inverse $q).\ $ Maybe you are confusing "a unit" with "the unit" (neutral element for multiplication, usually denoted $1)\ \ $ – Bill Dubuque Dec 14 '21 at 10:00
  • Yes, I was confusing the additive or multiplicative unit of an element in a field (intuitively, I was thinking 1 in $\mathbb{Z}$) with the definition of a generically unit in Artin's book. Thanks! – teo Dec 17 '21 at 07:23

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