$a$ is unit if and only if $a+I$ is unit.

Asked Apr 16 '24 at 10:23

Active Apr 16 '24 at 10:23

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Let R be a ring and I is ideal of it. We want to prove or disprove the statement.

Let $a$ be a unit. Then there is $a'$ such that $aa'=1$. The quotient ring R/I implies, $(a+I)(a'+I)=aa'+I=1+I$ which makes $a+I$ is unit.

Let $a+I$ be unit. Then, $(a+I)(a'+I)=1+I$ and $aa'-1\in I$.

I'm stuck right here.I know I can not say $r+I=1+I$ if and only if $r=1$ . But I can't find any other idea.

asked Apr 16 '24 at 10:23

Do you know what is $(a+I)(b+I)$? Do you know how it is defined? – Arctic Char Apr 16 '24 at 10:27
2

You are so close to a disproof. – David Lui Apr 16 '24 at 10:29
@ArcticChar it's the casual quotient ring multiplication. – Apr 16 '24 at 10:40
Take $R=Z$ and $I=2Z$. Observe that $3+2Z=1+2Z$ but $3\neq 1$. Does this answer your query? – Dots_and_Arrows Apr 16 '24 at 10:42
@MathCosmo So it was simple as that. – Apr 16 '24 at 10:53
@ArdaYonet Also see this question https://math.stackexchange.com/questions/4015907/prove-that-a-langle-b-rangle-is-a-unit-in-the-quotient-ring-r-langle-b?rq=1 – Dots_and_Arrows Apr 16 '24 at 11:00
A true variant of this question is:if $x\in R$ and $J(R)$ is the Jacobson radical of $R$, then $x$ is a unit of $R$ iff $x+J(R)$ is a unit in $R/J(R)$. – rschwieb Apr 16 '24 at 15:35

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