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 \(\displaystyle x, y \in A\) ...

we have ... \(\displaystyle xy = 1 \Longrightarrow yx = 1\).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 \(\displaystyle x, y \in A\) ...

we have ... \(\displaystyle xy = 1 \Longrightarrow yx = 1\).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 ...

\(\displaystyle Aa \supseteq Aa^2 \supseteq Aa^3 \supseteq \ ... \ ... \ \supseteq Aa^r = Aa^{r + 1} = Aa^{r + 1} = Aa^{r + 2} = \ ... \ ... \)

Now ...

\(\displaystyle Aa^r = \{ ba^r \ | \ b \in A \}\)

and

\(\displaystyle Aa^{r + 1} = \{ ba^{r + 1} \ | \ b \in A \} \)

Now consider \(\displaystyle x \in A\) such that \(\displaystyle xy = 1\) ...

Given \(\displaystyle x \in A\) we have \(\displaystyle xa^r \in Aa^r\)

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

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

Thus we have

\(\displaystyle x = ha
\)

so

\(\displaystyle xy = hay\) so that \(\displaystyle 1 = hay\) ...

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 ...

\(\displaystyle Aa \supseteq Aa^2 \supseteq Aa^3 \supseteq \ ... \ ... \ \supseteq Aa^r = Aa^{r + 1} = Aa^{r + 1} = Aa^{r + 2} = \ ... \ ... \)

Now ...

\(\displaystyle Aa^r = \{ ba^r \ | \ b \in A \}\)

and

\(\displaystyle Aa^{r + 1} = \{ ba^{r + 1} \ | \ b \in A \} \)

Now consider \(\displaystyle x \in A\) such that \(\displaystyle xy = 1\) ...

Given \(\displaystyle x \in A\) we have \(\displaystyle xa^r \in Aa^r\)

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

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

Thus we have

\(\displaystyle x = ha
\)

so

\(\displaystyle xy = hay\) so that \(\displaystyle 1 = hay\) ...

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 $y = a$.

 

[blue_raver22]

, this is a very interesting problem! To get started, let's first recall some definitions. A ring is unital if it has a multiplicative identity element, denoted by 1. A left unital ring is a ring where the identity element 1 is only required to satisfy 1x = x for all x in the ring. This means that the identity element only acts as a left identity.

Next, we have the concept of an Artinian ring. An Artinian ring is a ring where every descending chain of left ideals (subsets of the ring that are closed under left multiplication) eventually becomes stationary. In other words, there is a finite number of left ideals in the chain before it repeats itself.

Now, let's consider the given statement: xy = 1 implies yx = 1 for any x and y in the left unital Artinian ring A. We can prove this by contradiction. Assume that there exist x, y \in A such that xy = 1 but yx \neq 1.

Since A is a left unital Artinian ring, we know that there exists a finite chain of left ideals I_1, I_2, ..., I_n in A such that I_1 \supseteq I_2 \supseteq ... \supseteq I_n = 0. This means that for any x in A, there exists some positive integer k such that xI_k = 0.

Now, let's consider the ideal I = yA. Since yx \neq 1, we know that y \notin I. This means that there must exist some element a \in A such that ya \notin I. Since yA is a left ideal, we know that yaA \subseteq yA = I. Therefore, yaI_k = 0 for some positive integer k.

Now, let's consider the element xy in A. Since xy = 1, we know that xyI_k = I_k. But we also know that xyI_k = x(yI_k) \subseteq xI_k = 0. This means that I_k = 0, which contradicts our assumption that the chain of left ideals is finite and eventually becomes stationary.

Therefore, our initial assumption that yx \neq 1 must be false, and we can conclude that yx = 1. This shows that in a left unital Artinian