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Consider the free algebra $R=\mathbb{Z}\left<x,y\right>$ generated by noncommuting indeterminates $x,y$. Let $I$ denote the two-sided ideal generated by $xy-1$. Then the image of $x$ in $R/I$ has a right inverse but no left inverse.

This is an example mentioned in Lam's "A First Course in Noncommutative Rings" and is also mentioned in this thread: .

Here's a silly question that I can't seem to answer: How does one show this? If the image of $x$ had a left inverse, then it must be the image of $y$. Hence $yx-1$ must be contained in $I$. Thus we have $yx-1 = \sum a_i (xy-1) b_i$ for some $a_i,b_i \in R$. In examples similar to this (e.g., in showing the image of $x$ in $R/(xy)$ is a left zero-divisor but not a right zero-divisor), one easily derives a contradiction using the fact that every element of $R$ is represented uniquely as a polynomial in the noncommuting indeterminates. However, in the sum $\sum a_i (xy-1)b_i$, I may have some cancellations, which complicate things.

Lam says that this is "not hard to show (e.g. by specialization)," but I'm not sure what he means by "specialization." I have to be missing something simple...

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up vote 2 down vote accepted

To prove this claim "by specialization" means to find a ring $S$ with 2 elements $a,b$ such that $ab = 1$ but $a$ has no left inverse. Since $R = \mathbb{Z}\langle x,y \rangle$ is the free noncommutative ring with 2 generators, there exists a unique map $R \to S$ sending $x$ to $a$ and $y$ to $b$. Since $ab=1$, this map induces a map on the quotient $R/I \to S$. Now if $x$ had a left inverse in $R/I$, then $a$ would have a left inverse in $S$, contradiction.

One example of such a ring $S$ is the ring of linear transformations of the vector space of infinite sequences of real numbers $(r_1, r_2, r_3, \ldots)$. The shift map sending $(r_1, r_2, r_3, \ldots) \mapsto (r_2, r_3, \ldots)$ has an inverse on only one side.

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