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It will be shown now that every irreducible element in a factorial ring is also prime.

Let R be a factorial ring.

then let $a\in R$ be irreducible and $x,y \in R$ so that $a| xy$. It is now to show that $a|x$ or $a|y$. It is assumed that x and y are not units. Let $x=x_{1}…x_{r}, y=y_{1}…y_{s}$ be partitions into irreducible elements. Then it follows that $a|(x_{1}…x_{r}y_{1}…y_{s})$. Since this is unique it means that a as a irreducible element can be associated with a $x_{i}$ or a $y_{i}$. That's why $a|x$ or $a|y$ and a is a prime.

Tell me if this proof is correct. Please

V

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up vote 3 down vote accepted

Yes, this is correct.

Two minor comments: 1) You say "it is assumed that $x$ and $y$ are not units": maybe you should say why this assumption is warranted. 2) I and probably most other people would say "factorization" where you say "partition": in particular, "partition" makes it sound like the irreducible elements are distinct, which they need not be.

If for some reason you want more reassurance, you can find this standard argument written out in many places. For instance I do it on page 3 of this expository note in the paragraph beginning "Assume (FTA)". Technically speaking this argument takes place in the UFD $\mathbb{Z}$, but as I mention later it really works in the context of $R$ a UFD (so in particular works in $\mathbb{Z}$ once you know the fundamental theorem of arithmetic).

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if it is not assumed that it is not a unit then it must not be prime anymore in the end. ? –  VVV Nov 1 '11 at 22:13
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