# PID dosen't have to contain 1?

• pivoxa15

#### pivoxa15

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?

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?

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.

Most people *require* rings to have a unit. They use rng for a ring without I. How do you define <a>? That is entirely what matters. Even if you define it to be the set of things of the form xa for x in R, then it is prefectly possible that there is an x with xa=a, and for R to have no multiplicative identity. EG, take R any ring without unit, and a any idempotent (something that satisfies a^2=a). An important example of a ring without unit is the set of NxN matrices (here N means the cardinality of the natural numbers) over C the complex numbers, say, with finite trace. There are infinitely many idempotents in that ring.

So, the answer is, I suspect. <a> may or may not contain a. If R has an identity it certainly doesn't and R does not then you cannot say.