invertible?

**clic-clac** · November 16th 2008, 06:09 AM

Hi
Let $A$ be a ring, $a,b \in A$ such that $ab=1$ . Suppose that $X=\{x \in A;\ ax=0\}$ is finite.
Prove that $ba=1$

We can see that $1-ba \in X$ , so if $X=\{0\}$ , it's over.
Any idea?

**NonCommAlg** · November 16th 2008, 10:52 AM

Originally Posted by clic-clac

Hi
Let $A$ be a ring, $a,b \in A$ such that $ab=1$ . Suppose that $X=\{x \in A;\ ax=0\}$ is finite.
Prove that $ba=1$

We can see that $1-ba \in X$ , so if $X=\{0\}$ , it's over.
Any idea?

we have: $a^{k-1} - ba^k \in X, \ \forall k \in \mathbb{N}.$ hence, since $X$ is finite, there exist $n, m \in \mathbb{N}$ such that $a^n-ba^{n+1}=a^m -ba^{m+1}, \ n > m,$ which gives us: $(1-ba)a^n=a^m-ba^{m+1}. \ \ \ \ (1)$

now multiply both sides of (1) from the right by $b^n$ and note that $\forall k \in \mathbb{N}: \ a^kb^k=1.$ then you'll get: $1-ba=(a^m-ba^{m+1})b^n=a^mb^n - ba^{m+1}b^n=b^{n-m}-b^{n-m}=0. \ \ \ \ \ \ \ \ \ \ \Box$

**clic-clac** · November 16th 2008, 11:16 AM

Thanks for your answer.

I don't see why there exists $n$ such that $a^n-ba^{n+1}=a^{n-1}-ba^n$ .
The conclusion is still the same with $a^n-ba^{n+1}=a^{k}-ba^{k+1}$ for a $k<n ,$ but there is something I'm missing...

**NonCommAlg** · November 16th 2008, 11:47 AM

Originally Posted by clic-clac

Thanks for your answer.

I don't see why there exists $n$ such that $a^n-ba^{n+1}=a^{n-1}-ba^n$ .
The conclusion is still the same with $a^n-ba^{n+1}=a^{k}-ba^{k+1}$ for a $k<n ,$ but there is something I'm missing...

you're right, thanks! it was just one of those really weird mistakes that i make sometimes! it's fixed now!

**clic-clac** · November 16th 2008, 12:13 PM

You're welcomed.
But for me the problem is that doing this, we assume that every element of $X$ is equal to a $a^{q}-ba^{q+1}$ , $q\in \mathbb{N}$ .
Why can we say that?

**NonCommAlg** · November 16th 2008, 12:23 PM

Originally Posted by clic-clac

we assume that every element of $X$ is equal to a $a^{q}-ba^{q+1}$ , $q\in \mathbb{N}$ .
Why can we say that?

i didn't say that! what we have is this: $A=\{a^{k-1} - ba^k: \ k \in \mathbb{N} \} \subseteq X,$ and we're given that $X$ is finite. so $A$ must also be finite and thus we can find two elements of $A$ which are equal.

**clic-clac** · November 16th 2008, 12:42 PM

Great NonCommAlg, I finally got it! Thank you.

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