# PID dosen't have to contain 1?

Not all PIDs have to contain the multiplicative inverse?

If that is the case then the principle ideal <a> will not contain 'a' in it. I find that strange. But that is still possible isn't it?

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Hurkyl
Staff Emeritus
Gold Member
1 is the multiplicative identity.

It's principal, not principle.

In an algebra A without 1, <a> is, by definition, the smallest ideal of A containing a. So <a> has to contain a.

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Actually all PID lie in an integral domain which by definition must contain 1.

Are you suggesting that in non Integral domains, single ideals like <a> cannot exist?

Hurkyl
Staff Emeritus
Gold Member
Actually all PID lie in an integral domain which by definition must contain 1.
What do you mean by "lie in"?

Every PID is an integral domain. And there is a simpler reason why it contains 1: because it's a ring.

Are you suggesting that in non Integral domains, single ideals like <a> cannot exist?
No...

<a> exists because there is at least one ideal containing a, and the intersection of all such ideals is an ideal containing a. (And thus the smallest such ideal)

Hrm -- it's theorem that in a commutative ring (which does have 1), <a> is actually the set of all multiples of a. But that theorem doesn't hold for algebras without 1. (I don't remember if there are any complications for noncommutative rings)

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From all the axioms I have seen, a ring dosen't have to include a multiplicative identity. In a ring R without 1 then <a> will not include 'a' will it? With a in R.

matt grime