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Suppose R is a commutative ring with identity, how do the elements of R relate to associates?

I know that for $\mathbb{Z}$, the units are $\pm$1 and thus the associates of any element n of $\mathbb{Z}$ are $\pm$n but I'm curious if every element is an associate of itself for any commutative rings with identity. Any feedback would be appreciated. Thanks in advance!

user26857
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asuhdude
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    Which definition of "associate" are you using? In general commutative rings it can mean $,a\sim b\iff a\mid b\mid a,,$ i.e. $(a) = (b)$ or it can mean $a = u b$ for some unit $u$. They are equivalent in domains but not in commutative rings It doesn't matter for your 2nd question but may for your first (depending on what it means - which is not clear). – Bill Dubuque Nov 19 '19 at 06:52
  • Suppose R is a commutative ring with identity, how do the elements of R relate to associates? That is kind of an ironic muddling to which I would say 'Being associates' is a relationship between elements. It sounds a bit like you're thinking of "associates" as almost being the same thing (or a special subset) of "elements" in the first line. – rschwieb Nov 19 '19 at 15:18

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Edited in response to comments:

To answer your question, you need to define "associate".

I will say elements $a$ and $b$ of a ring are associates if $a=ub$ for some unit $u$.

The multiplicative identity element (call it $1$) of a ring is always a unit, and $a=1a$,

so I would say that every element in a commutative ring with identity is an associate of itself.

This concurs with the reflexive property of associatedness as an equivalence relation.

The additive inverse of $1$ (denoted $-1$) is also always a unit,

so every element in a commutative ring with identity is also an associate of its additive inverse.

Some rings have additional units; for example, the Gaussian integers $\mathbb Z[i]$ also have $i$ and $-i$ as units.

J. W. Tanner
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