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Why do we exclude $0$ and units from the definition of irreducible elements? More clearly how can it be shown that for the units and $0,$ $c=ab$ doesn't always imply either $a$ or $b$ is a unit (where $c = 0$ or $c$ is a unit)?

For $0$ my guess is that $0=0\cdot c$ for some non-unit $c$.

Stahl
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Sriti
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    Your question is a bit confusing to read: your title and first sentence ask two different questions, and it isn't clear what $c$ in the last sentence is supposed to be. I assume you mean "do" in the title and $c$ to be either $0$ or a unit. I've edited the question to read as I think you intended; if anything isn't as it should be, feel free to undo the edit. – Stahl May 19 '13 at 04:41
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    http://ncatlab.org/nlab/show/too+simple+to+be+simple – Zev Chonoles May 19 '13 at 04:41
  • @Stahl: You've right. – Sriti May 19 '13 at 04:58

1 Answers1

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Zero and units are excluded because it would generally make factorisation meaningless. For example, we say $6$ has two factors $2$ and $3$. These are the normal forms of the spaces of 2 by a unit, and 3 by a unit.

Were units to be included, the factorisation could be $2\cdot-3\cdot-1$, which suggests three factors. However $-3$ is a variety of $3$, in that each divides the other, and the product with $-1$ divides it without $-1$.

So factorisation is looking for numbers whose product divides the number and vice versa, but any lesser product is not divisible by the number. It is these that function as prime-like things.

The only time one includes units, if if the product of the primes in their normal forms do not give the expressed number, so eg $-6=-1\cdot2\cdot3$.

$0$ is also excluded, simply because it does not divide anything unless it's zero itself. So any non-zero thing is not a product of elements including $0$

wendy.krieger
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