Problem Regarding Left Unital Artinian Ring (set by Euge)

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Discussion Overview

The discussion revolves around a problem related to left unital Artinian rings, specifically addressing the implications of the equation $$xy = 1$$ on the relationship between $$x$$ and $$y$$ in the context of the ring's structure. The focus is on exploring the properties of such rings and the application of the descending chain condition.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Peter seeks assistance in proving that for a left unital Artinian ring $$A$$, if $$xy = 1$$, then $$yx = 1$$.
  • Euge suggests applying the descending chain condition to the chain $$Ax \supseteq Ax^2 \supseteq \cdots$$ and using the equation $$xy = 1$$ to show that $$1 = ax$$ for some $$a \in A$$.
  • Peter elaborates on the implications of the descending chain condition, noting that there exists a natural number $$r$$ such that the chains stabilize, leading to the conclusion that $$Aa^r = Aa^{r + 1}$$.
  • Peter expresses uncertainty about the next steps after deriving $$1 = hay$$ from his analysis.
  • A later reply from Euge corrects a previous typo and reiterates the use of the chain $$Ax \supseteq Ax^2 \supseteq \cdots$$, emphasizing that since this chain stabilizes, there exists an $$r$$ such that $$Ax^r = Ax^{r+1}$$, leading to $$x^r = ax^{r+1}$$ for some $$a \in A$$.

Areas of Agreement / Disagreement

Participants are engaged in a collaborative problem-solving effort, with no explicit consensus reached on the final steps of the proof. There is a shared understanding of the application of the descending chain condition, but uncertainty remains regarding the implications of the derived equations.

Contextual Notes

Limitations include the need for further clarification on the implications of the derived equations and the specific conditions under which the properties of the ring apply. The discussion does not resolve the mathematical steps leading to the conclusion.

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Can someone please help me get started on the following problem:

Show that if A is a left unital Artinian ring, then:

... whenever $$x, y \in A$$ ...

we have ... $$xy = 1 \Longrightarrow yx = 1$$.Peter
 
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Peter said:
Can someone please help me get started on the following problem:

Show that if A is a left unital Artinian ring, then:

... whenever $$x, y \in A$$ ...

we have ... $$xy = 1 \Longrightarrow yx = 1$$.Peter

Hi Peter,

Apply the descending chain condition to the chain $Ax \supseteq Ax^2 \supseteq \cdots$ and use the equation $xy = 1$ to show that $1 = ax$ for some $a \in A$.
 
Last edited:
Euge said:
Hi Peter,

Apply the descending chain condition to the chain $Aa \supseteq Aa^2 \supseteq \cdots$ and use the equation $xy = 1$ to show that $1 = ax$ for some $a \in A$.
Thanks for the help so far, Euge ... but I am going to need some more help ...

Analysis so far ...Given that A is Artinian, the descending chain condition implies that all descending chains become stationary at some point ... so there is some natural number r such that we have ...

$$ Aa \supseteq Aa^2 \supseteq Aa^3 \supseteq \ ... \ ... \ \supseteq Aa^r = Aa^{r + 1} = Aa^{r + 1} = Aa^{r + 2} = \ ... \ ... $$

Now ...

$$Aa^r = \{ ba^r \ | \ b \in A \}$$

and

$$Aa^{r + 1} = \{ ba^{r + 1} \ | \ b \in A \} $$

Now consider $$x \in A$$ such that $$xy = 1$$ ...

Given $$x \in A$$ we have $$xa^r \in Aa^r$$

But $$Aa^r = Aa^{r + 1}$$ so there exists a $$g \in Aa^{r + 1}$$ such that

$$xa^r = g = ha^{r + 1}$$ for some $$h \in A$$

Thus we have

$$x = ha
$$

so

$$xy = hay$$ so that $$1 = hay$$ ...

BUT ... where to from here?

Can you help?

Peter
 
Peter said:
Thanks for the help so far, Euge ... but I am going to need some more help ...

Analysis so far ...Given that A is Artinian, the descending chain condition implies that all descending chains become stationary at some point ... so there is some natural number r such that we have ...

$$ Aa \supseteq Aa^2 \supseteq Aa^3 \supseteq \ ... \ ... \ \supseteq Aa^r = Aa^{r + 1} = Aa^{r + 1} = Aa^{r + 2} = \ ... \ ... $$

Now ...

$$Aa^r = \{ ba^r \ | \ b \in A \}$$

and

$$Aa^{r + 1} = \{ ba^{r + 1} \ | \ b \in A \} $$

Now consider $$x \in A$$ such that $$xy = 1$$ ...

Given $$x \in A$$ we have $$xa^r \in Aa^r$$

But $$Aa^r = Aa^{r + 1}$$ so there exists a $$g \in Aa^{r + 1}$$ such that

$$xa^r = g = ha^{r + 1}$$ for some $$h \in A$$

Thus we have

$$x = ha
$$

so

$$xy = hay$$ so that $$1 = hay$$ ...

BUT ... where to from here?

Can you help?

Peter

Sorry Peter, I had a typo. Use the chain $Ax \supseteq Ax^2 \supseteq \cdots$. I made the correction in my earlier post. Since this chain stabilizes, there is an $r$ such that $Ax^r = Ax^{r+1}$. Since $x^r\in Ax^r$, $x^r = ax^{r+1}$ for some $a\in A$. Deduce from this that $1 = ax$ and hence $y = a$.
 

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