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How can I see that all irreducible elements in a principal ideal domain are prime?

$u$ is irreducible when $u_1 u_2 = u \implies u_1 $ or $u_2$ is a unit.
$u$ is prime when $u | ab \implies u|a$ or $u|b$.

Mark
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2 Answers2

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Suppose $u$ is irreducible in a principal ideal domain $R$. We will show that the principal ideal $(u)$ is maximal. Assume, to the contrary, that $(u)\subsetneq I\subsetneq R$ for some ideal $I$. Since $R$ is a PID, $I=(a)$ for some $a\in R$. Now, $u\in I$ gives $u=ar$ for some $r\in R$. Since $u$ is irreducible, and $a$ is not a unit (since $I=(a)\neq R$), it follows that $r$ is a unit. But then, $a=u r^{-1}$ and so $(u)=I$ which is a contradiction.

Conclusion: $(u)$ is a maximal ideal in $R$, and so in particular it is a prime ideal. Thus, $u$ is prime.

Prism
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  • Great! I needed a short proof of this fact for my answer to http://math.stackexchange.com/questions/1247762/a-proof-about-polynomial-division, and here it is! Cheers! – Robert Lewis Apr 23 '15 at 05:27
  • Is it true that the converse is false ? i.e is it false that "If all irreducibles of an integral domain are prime then the ring is a principal ideal domain" ? Because I read that in $\mathbb C[x,y]$ all irreducibles generate a prime ideal, hence they are prime right ? Yet $\mathbb C[x,y]$ isn't a principal ideal domain since $(x,y)$ isn't principal. – James Well May 23 '17 at 16:22
  • Yeah you are correct. The statement "If all irreducible of an integral domain are prime then the ring is a principal ideal domain" is false. Indeed, any example of UFD which is not a PID will be a counterexample (for example, $\mathbb{C}[x, y]$ is such a ring, as you indicated). In a UFD (unique factorization domain), it is always true that all irreducible elements are prime. It would be interesting to find an example of a ring $R$ which is not a UFD, but yet all the irreducible elements of $R$ are prime. – Prism May 23 '17 at 16:58
  • @Prism I don't think that can happen. If we take an element and express it in terms of irreducible elements, in two different ways say $a = i_1i_2\cdots i_n = j_1j_2\cdots j_m$ as each of the irreducible elements are prime one notes that $j_1j_2\cdots j_m$ is divisible by $i_1$ so $i_1$ must divide one $j_k$ for some k, but that implies that $j_k=i_1$. One continues this process for each of the i's and finds that if $n\neq m$ then some of the $j_k$'s must have been 1 for some $k$ and so $n=m$ by irreducibility, so the factorisation was unique after all. – Kelechi Nze Apr 02 '18 at 08:55
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Lemma: In a PID, maximal ideals are prime.

Proof: Let $R$ be a PID, let $M$ be maximal, then $R/M$ is a field and thus an integral domain which implies $M$ is prime. $\square$.

Theorem: If $p$ is irreducible in a PID, then $p$ is prime.

Proof: Let $R$ be a PID. Suppose $p$ is irreducible. It is enough to show $(p)$ is maximal. Suppose there exists an ideal $I \subset R$ with

$$(p) \subsetneq I \subsetneq R.$$

As $R$ is a PID, there exists an $m \in R$ such that $I = (m)$. As $(p) \subset (m)$, there exists an $r \in R$ such that $p = rm$. As $p$ is irreducible, either $r$ a unit or $m$ a unit, if $m$ a unit then $I=R$ and $(p)$ is maximal. If $r$ is a unit, then $I=(p)$ and we are done. $\square$

homosapien
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