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Let $S$ be a set of all even integers. According to my text book, $(S,+,\cdot)$ is a ring which is not an integral domain. It is stated as a fact without an explanation and I fail to see the reason for this.

Why the ring from above is not an integral domain?

EDIT: It can't be because of the lack of $1$ element. In the next example, $(Z,+,\cdot)$ (where $Z$ is the whole set of integers) is stated to be an integral domain.

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Some (not all) definitions do not even allow this to be a ring since it lacks a multiplicative unit. – Baby Dragon Jun 14 '13 at 14:56
It is better to be more specific than "It is stated as a fact..." Who stated it? What definitions is that person/book/notes using? – Thomas Andrews Jun 14 '13 at 14:59
I'm not sure why your edit shows anything. $\mathbb Z$ has a multiplicative identity, so how does that contradict anything? – Thomas Andrews Jun 14 '13 at 15:00
Ohh, I can see it now. I thought that the multiplicative identity has to be denoted in the ring definition $(Z,+,\cdot,1,0)$. But now I get what you meant (that 1 is not even). – jakubka Jun 14 '13 at 15:03
up vote 5 down vote accepted

I suspect that your book's definition of integral domain requires a multiplicative identity element, which $(S,+,\cdot)$ does not have.

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and here's a thread explaining why:… – citedcorpse Jun 14 '13 at 14:56
@exitingcorpse: Thanks! I was just about to look for that. – Cameron Buie Jun 14 '13 at 14:58
Please see my edit.. It is possible that my book is simply wrong. – jakubka Jun 14 '13 at 14:59
@jakubka: $(\Bbb Z,+,\cdot)$ has a multiplicative identity element, but aside from that, there's no property that distinguishes it from $(S,+,\cdot).$ This suggests even more strongly to me that your book means that $(S,+,\cdot)$ isn't an integral domain precisely because it doesn't have a multiplicative identity. – Cameron Buie Jun 14 '13 at 15:02
Yes, I get it now. $\mathbb{Z}$ contains 1, but $S$ doesn't. Thanks a lot for the answer! – jakubka Jun 14 '13 at 15:06

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