I $A$ is a principal ideal domain and let N the set of non zero elements, that do not have an inverse and that cannot be decomposed in a product of irreductibles.

If $a \in N$

Let's create the set $I = aA$

I am trying to understand why a is not irreductible

(This is part of the proof of the fundamental theorem of arithmetics our professor gave us)

Is it just that if $a$ was irreductible then $a = a$ so it can be decomposed in a product of irreductibles?

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    $\begingroup$ I don't really understand your question. If $a$ cannot be decomposed into a product of irreducible elements, then $a$ can be both, irreducible and reducible. In a noetherian ring every element can be decomposed into irreducible elements. $\endgroup$ – Paul K May 31 '16 at 5:20
  • $\begingroup$ Let me change the question $\endgroup$ – aribaldi May 31 '16 at 5:22
  • $\begingroup$ Principal ideal domains are factorial $\endgroup$ – Paul K May 31 '16 at 5:35
  • $\begingroup$ @menag that's precisely what i'm trying to understand thanks though $\endgroup$ – aribaldi May 31 '16 at 5:36
  • $\begingroup$ @aribaldi What is the relevance of $I$ in your question? Is it part of your attempt at a solution. If so, how did you plan to use $I$? $\endgroup$ – M. Vinay May 31 '16 at 5:42

if A a factorial domain , for $a\in A$, a not 0 and not invertible, we have equivalence :

a irreductible

a prime

aA irreductible ideal

aA maximal ideal in the set of principal ideal and propre

aA prime ideal

  • $\begingroup$ But I'm asking about the element a that cannot be decomposed in a product of irreductibles, why is that very element not irreductible? $\endgroup$ – aribaldi May 31 '16 at 9:06
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    $\begingroup$ I tried to translate your question but I do not understand the problem. $\endgroup$ – m.idaya May 31 '16 at 9:27
  • $\begingroup$ is that N is the subset of A containing all the irreducible elements of A? $\endgroup$ – m.idaya May 31 '16 at 9:33
  • $\begingroup$ from the comments, i understant that you want to show a principal ring is factrial, is not it? $\endgroup$ – m.idaya May 31 '16 at 9:42
  • $\begingroup$ i want to know why an element that cannot be decomposed into irreductibles cannot be an irreductible $\endgroup$ – aribaldi May 31 '16 at 12:22

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