# invertible?

• Nov 16th 2008, 06:09 AM
clic-clac
invertible?
Hi
Let $\displaystyle A$ be a ring, $\displaystyle a,b \in A$ such that $\displaystyle ab=1$. Suppose that $\displaystyle X=\{x \in A;\ ax=0\}$ is finite.
Prove that $\displaystyle ba=1$

We can see that $\displaystyle 1-ba \in X$, so if $\displaystyle X=\{0\}$, it's over.
Any idea? :)
• Nov 16th 2008, 10:52 AM
NonCommAlg
Quote:

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

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

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

now multiply both sides of (1) from the right by $\displaystyle b^n$ and note that $\displaystyle \forall k \in \mathbb{N}: \ a^kb^k=1.$ then you'll get: $\displaystyle 1-ba=(a^m-ba^{m+1})b^n=a^mb^n - ba^{m+1}b^n=b^{n-m}-b^{n-m}=0. \ \ \ \ \ \ \ \ \ \ \Box$
• Nov 16th 2008, 11:16 AM
clic-clac

I don't see why there exists $\displaystyle n$ such that $\displaystyle a^n-ba^{n+1}=a^{n-1}-ba^n$.
The conclusion is still the same with $\displaystyle a^n-ba^{n+1}=a^{k}-ba^{k+1}$ for a $\displaystyle k<n ,$ but there is something I'm missing...
• Nov 16th 2008, 11:47 AM
NonCommAlg
Quote:

Originally Posted by clic-clac

I don't see why there exists $\displaystyle n$ such that $\displaystyle a^n-ba^{n+1}=a^{n-1}-ba^n$.
The conclusion is still the same with $\displaystyle a^n-ba^{n+1}=a^{k}-ba^{k+1}$ for a $\displaystyle k<n ,$ 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!
• Nov 16th 2008, 12:13 PM
clic-clac
You're welcomed.
But for me the problem is that doing this, we assume that every element of $\displaystyle X$ is equal to a $\displaystyle a^{q}-ba^{q+1}$, $\displaystyle q\in \mathbb{N}$.
Why can we say that?
• Nov 16th 2008, 12:23 PM
NonCommAlg
Quote:

Originally Posted by clic-clac

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

i didn't say that! what we have is this: $\displaystyle A=\{a^{k-1} - ba^k: \ k \in \mathbb{N} \} \subseteq X,$ and we're given that $\displaystyle X$ is finite. so $\displaystyle A$ must also be finite and thus we can find two elements of $\displaystyle A$ which are equal.
• Nov 16th 2008, 12:42 PM
clic-clac
Great NonCommAlg, I finally got it! Thank you.