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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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- Thread starter pivoxa15
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- #1

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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?

- #2

Hurkyl

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1 is the multiplicative **identity**.

It's princip**al**, not princip**le**.

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

It's princip

In an algebra A without 1, <a> is,

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- #3

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

- #4

Hurkyl

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

Every PID

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

<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

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- #6

matt grime

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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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