Is $1-y$ always lacking a right inverse when $y$ is a non-unit in a right ideal?

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Discussion Overview

The discussion revolves around the question of whether the element $1-y$ lacks a right inverse when $y$ is a non-unit in a right ideal. Participants explore this concept within the context of various rings and ideals, including specific examples and counterexamples.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that if $I$ is a right ideal and $y \in I$ with $y \neq 1$, then $1-y$ has no right inverse.
  • One participant provides an example using $R=\mathbb{Z}$ and $I=2\mathbb{Z}$, questioning whether $1-y$ has a right inverse in this case.
  • Another participant suggests that $(1-y) \cdot -1 = 1$, indicating that $1-y$ can have a right inverse in certain instances.
  • A participant introduces the unique maximal right ideal in the localization of $\mathbb{Z}$ at the ideal $(3)$, arguing that $1-y$ can have a right inverse in this context.
  • One participant attempts to establish a general claim about the lack of right inverses for $1-ir$ when $I$ is a unique maximal right ideal, but this is challenged by another participant.
  • Another participant points out that the assumption made in the previous argument is incorrect, referencing the unique maximal ideal being the set of non-units in $\mathbb{Z}_{(3)}$.

Areas of Agreement / Disagreement

Participants express differing views on whether $1-y$ can have a right inverse, with some examples supporting the idea that it can, while others argue against it. The discussion remains unresolved, with multiple competing perspectives presented.

Contextual Notes

Participants highlight the importance of specific ring structures and the nature of ideals in determining the properties of $1-y$. There are unresolved assumptions regarding the conditions under which $1-y$ may or may not have a right inverse.

mathmari
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Hey! :o

When $I$ is a right ideal and $y\in I$ with $y\neq 1$, does it holds that $1-y$ has no right inverse? (Wondering)
 
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mathmari said:
Hey! :o

When $I$ is a right ideal and $y\in I$ with $y\neq 1$, does it holds that $1-y$ has no right inverse? (Wondering)

Hey mathmari! (Smile)

Suppose we pick $R=\mathbb Z,\ I=2\mathbb Z,\ y=2$... (Thinking)
 
I like Serena said:
Suppose we pick $R=\mathbb Z,\ I=2\mathbb Z,\ y=2$... (Thinking)

All of these elements, the 1-y, has ho right inverse, right? (Wondering)
 
mathmari said:
All of these elements, the 1-y, has ho right inverse, right? (Wondering)

Isn't $(1-y)\cdot -1 = (1-2)\cdot -1 = 1$? (Wondering)
 
I like Serena said:
Isn't $(1-y)\cdot -1 = (1-2)\cdot -1 = 1$? (Wondering)

Oh yes... So, we cannot say anything in general, can we? (Wondering)
 
Is there a difference when the right ideal is the unique maximal right ideal? (Wondering)
 
Hi,

Is still false. Think on $\mathbb{Z}_{(3)}$, which is the localization of $\mathbb{Z}$ at the ideal $(3)$.

Then $\mathbb{Z}_{(3)}=\{\dfrac{a}{b} \in \mathbb{Q} \ : \ 3\nmid b\}$.

This ring has a unique maximal ideal $\mathfrak{m}=\{\dfrac{a}{b}\in \mathbb{Q} \ : \ 3\nmid b, \ 3 \mid a\}$.

And now, for example, $\dfrac{3}{8}\in \mathfrak{m}$ and $\dfrac{8}{5}\in \mathbb{Z}_{(3)}$ is the inverse of $\frac{5}{8}.$
 
Ah ok... (Thinking)

But is the following correct? (Wondering)

Let $I$ be the unique maximal right ideal of the ring $R$.
We have that $ri=1$, for some $i\in I, r\in R$ and $ir\in I$.
Then $1-ir$ has no right inverse.
Suppsoe that $1-ir$ has a right inverse, then there is a $x\in R$ such that $(1-ir)x=1$.
$$(1-ir)x=1 \Rightarrow r(1-ir)x=r \Rightarrow (r-rir)x=r \Rightarrow (r-(ri)r)x=r \Rightarrow (r-r)x=r \Rightarrow 0x=r \Rightarrow 0=r \Rightarrow 0i=ri \Rightarrow 0=1$$ A contradiction.
 
No, it isn't corect because you are assuming that

mathmari said:
Ah ok... (Thinking)
We have that $ri=1$, for some $i\in I, r\in R$ and $ir\in I$.

And this isn't true. In my example above you can see that the unique maximal ideal is exactly the set of non-units of the ring $\mathbb{Z}_{(3)}$.
 

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