MHB Problem Regarding Left Unital Artinian Ring (set by Euge)

[Math Amateur]

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

 

[Euge]

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 $a \in A$.

 

[Math Amateur]

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

 

[Euge]

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 $y = a$.