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In this wikipedia article it is said that set Z is a principal ideal domain, i.e. each one of its ideals can be generated by a single element. But if we consider set C of all composite integer numbers (Z without primes and 0), wouldn't it be an ideal? If we take arbitrary element from Z and take a product of a composite number and an arbitrary integer we will get a composite, thus an element of C? And, as far as I can understand, C can not be generated by a single element. So it is not a principle ideal.

I would appreciate pointing to my mistake very much.

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Not closed under addition. – user92843 Jun 21 '11 at 21:16
It is an ideal in the multiplicative semigroup. – Jonas Meyer Jun 21 '11 at 21:21
The set of all nonzero composite numbers is a semigroup ideal of the multiplicative semigroup of the integers, but is not an additive subgroup of the integers, so it's not an ideal in the sense of ring theory. – Arturo Magidin Jun 21 '11 at 21:22
Thanks to all of you guys! I got it now. – ikostia Jun 21 '11 at 21:35
up vote 9 down vote accepted

The problem is that the set of all composite integers does not form an ideal. For example if you add $21 - 4$ you get $17$ which is a prime and thus not composite. That's why there's no contradiction.

Remember that for a subset of a ring to be an ideal it must be closed under addition and under taking multiples by elements of the ring, and in this case the set of all composite integers is not closed under addition.

And the fact that $\mathbb{Z}$ is a principal ideal domain is because you have division with remainder in the set of integers.

It is a more general fact that any ring that has this sort of division, which is called an Euclidean Domain, is as a consequence a Principal Ideal Domain.

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Of course! I have missed the second condition - that (I,+) should form a subgroup of (R,+). Where is my attention?) Thanks a lot! – ikostia Jun 21 '11 at 21:21
@ikostia No problem, I'm glad it helped you. – Adrián Barquero Jun 21 '11 at 21:25

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