# Show that a certain right ideal in a non-commutative ring is infinite

Let $R$ be a unitary (associative and non-commutative) ring with elements $a,b,c$ such that $a\cdot b=1\not=0=a\cdot c$ and $c\not=0$ where $1$ is the multiplicative identity. Then the set $I=R\cap\{x:a\cdot x=0\}$ is a nontrivial right ideal, and one has to show that $I$ is infinite. How is this done? (To my knowledge, this is in a slightly different form as an exercise in Jacobson's Basic Algebra I.)

Added as response to comments. Actually, it is trivial that $I$ is a right ideal, and I did not ask this. The only problem is to show that $I$ is necessarily infinite under the assumptions given. I do not know what the exact exercise in Jacobson's book is, but I think it is somewhere in the chapter introducing rings, i.e., somewhere around page 100. (I don't have the book but I have seen the contents page.) The exercise is to show that if an element has two distinct right inverses, then it has infinitely many.

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The Jacobson problem and your problem are indeed connected. If $\{b_i\}$ is an indexed set of distinct elements inverting $a$ on the right, with $b_1=b$, then $\{b_i-b\mid i>1\}$ is a set of distinct zero divisors for $a$ on the right.

Do you need help proving the Jacobson problem? If so I'll add that in.

Actually, as I half expected, there is already a question for that here. Check out this question for the solution to the infinitely many right inverses question.

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Thanks for the Kaplansky link. It solves the problem. –  The-unKnowN Jan 20 '13 at 17:48

The set $I=R\cap\{x:a\cdot x=0\} = \{x \in R: a \cdot x = 0 \}$ is known as the right annihilator of $a$ in $R$. Showing that it is an ideal is simply an application of the definitions: if $x, y \in I$, then $a \cdot (x-y) = ax - ay = 0$ so $x - y \in I$. Also, if $r \in R$, then $a \cdot (xr) = (ax)r = 0r = 0$. So $I$ is a right ideal. That $I$ is infinite cannot be deduced unless we assume $R$ is infinite, and even then I'm not convinced it should be true...

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If $R$ is finite, then $ab=1$ implies $ba=1$. In fact, this is true for any Noetherian ring. –  rschwieb Jan 21 '13 at 19:43

I'd put this as a comment, but I can't. The formulation of your exercise is a little off. It should say if $a \cdot b=1=a \cdot c$, $1 \neq 0$, and $b \neq c$, then $I=\{r \in R \ \vert \ a \cdot r=1\}$ is infinite, in accordance with your own comment. To do this, first prove that neither $b$ nor $c$ can be a left inverse and notice that $1-ba+b$ is another right inverse of $a$. Try using this to construct an infinite number of non-equal right inverses.

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