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PID dosen't have to contain 1?

  • Thread starter pivoxa15
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  • #1
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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?
 

Answers and Replies

  • #2
Hurkyl
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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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  • #3
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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?
 
  • #4
Hurkyl
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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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  • #5
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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.
 
  • #6
matt grime
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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.
 

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