Prove that $x = uy$ for some unit $u \in R^{\times}$ iff $(x)=(y)$ as ideals.

Asked Nov 16 '20 at 15:28

Active Nov 16 '20 at 15:28

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Prove that:In a commutative ring $R,$ $x = uy$ for some unit $u \in R^{\times}$ iff $(x)=(y)$ as ideals.

My thoughts:

I feel like the idea is similar to the proof of the following proposition:

Proposition. Let $I$ be a left or right ideal of a ring $R$ with unity. Then $I = R$ if and only if $I$ contains a unit.

Proof: Let $I$ be a left ideal of $R$. If $I = R$ then $1 \in I$, so $I$ contains a unit. Conversely, if $u \in I$ is any unit, let $u^{-1}$ be its inverse. For any $r \in R$, we have $r = r (u^{-1} u) = (r u^{-1}) u \in I$, because $I$ is a left ideal. Thus $I = R$. This logic is easily adapted for right ideals to show the same result. Certainly it then is true for two-sided ideals and ideals of unital commutative rings.

But still I do not know how to write a rigor proof, could anyone help me in doing so please?

asked Nov 16 '20 at 15:28

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You will find from the duplicate that "$(x)=(y)$ implies associate" (in your sense, as unit multiples of each other) is false. – rschwieb Nov 16 '20 at 16:02
but it is correct if we are in an integral domain ..... am I correct? @rschwieb – Nov 16 '20 at 16:11
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Yes, for an integral domain it is correct. – rschwieb Nov 16 '20 at 16:13
@rschwieb that is all what I need ..... thank u! – Nov 16 '20 at 16:14
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The answer for domains would also be available on the site in a search. – rschwieb Nov 16 '20 at 16:14

Prove that $x = uy$ for some unit $u \in R^{\times}$ iff $(x)=(y)$ as ideals.

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