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

[mathmari]

Hey!

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)

 

[I like Serena]

Hey!

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

Isn't $(1-y)\cdot -1 = (1-2)\cdot -1 = 1$? (Wondering)

Oh yes... So, we cannot say anything in general, can we? (Wondering)

 

[mathmari]

Is there a difference when the right ideal is the unique maximal right ideal? (Wondering)

 

[Fallen Angel]

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 $\frac{5}{8}.$

 

[mathmari]

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.

 

[Fallen Angel]

No, it isn't corect because you are assuming that

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 $\mathbb{Z}_{(3)}$.