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I'm trying to understand a little detail in this proof:

I didn't understand why in a ring we can always write an element as a products of non-units elements.

I need help.

Thanks in advance

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    $\begingroup$ That is what it means that $a$ is NOT irreducible. $\endgroup$ – N. S. Jul 16 '13 at 13:37
  • $\begingroup$ When $a$ is irreducible there is nothing to prove, so let's examine the non-irreducible case. Instead of "so" I'd probably use "otherwise". $\endgroup$ – egreg Jul 16 '13 at 13:43
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The question has been answered by the comments. In the sentence under question, the writer is dealing with the case where $a$ is reducible, so by definition of reducible, $a = bc$ for some nonunits $b$ and $c$.

It is untrue that all ring elements can be expressed as products of nonunits. The product $a = bc$ is a unit if and only if $b$ and $c$ are units.

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