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

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In summary: 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
  • #1
mathmari
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Hey! :eek:

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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  • #2
mathmari said:
Hey! :eek:

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)
 
  • #3
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)
 
  • #4
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)
 
  • #5
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)
 
  • #6
Is there a difference when the right ideal is the unique maximal right ideal? (Wondering)
 
  • #7
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 $\frac{5}{8}.$
 
  • #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.
 
  • #9
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 $\mathbb{Z}_{(3)}$.
 

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

Is there a definitive answer to the question "Is this statement true?"

No, there is not always a definitive answer to this question. It depends on the statement in question and the evidence or proof that is available to support it.

What is the difference between a true statement and a false statement?

A true statement is one that accurately reflects reality and can be verified through evidence or proof. A false statement, on the other hand, is one that does not align with reality and can be disproven through evidence or proof.

Can a statement be partially true and partially false?

Yes, it is possible for a statement to contain elements that are both true and false. This can happen when the statement is complex and contains multiple components that each have their own truth value.

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The truthfulness of a statement can be determined through critical thinking and evaluation of evidence or proof. It is important to consider the source of the statement, the context in which it was made, and any supporting evidence or proof.

Is it possible for a statement to be true at one point in time and false at another?

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