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Recently, I have encountered a problem:

Given a ring $R$. Prove that:

1) If $x \in R$ has two distinct right inverses, it has infinitely many right inverses

2) If $x \in R$ has a unique right inverse, it is invertible.

3) If $R$ doesn't have any zero divisors, every left or right unit is a unit.

For the first task, I can only see an example: Given ring $R$ with the multiplication $x . y = x$, then it is easily seen that $1$ has infinitely many right inverses. However, it doesn't provide any insights for me to reach the solution.

For the second task, I guess that if $x$ has the unique right inverse $y$, then $y$ must also be the inverse of $x$. However, I don't have a way to get it.

For the third task, I think that it is not necessary that the left/right inverse of $x$ must be the inverse of $x$. However, for this case, I can't go any further.

Please consider the problem and give me some hints. Any help is appreciated. Thank you for reading.

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1 Answer 1

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Since Callus' comment provides a proof for (1), I only consider (2) and (3).

(2). Suppose that $xa = 1$. Then $xax = x$, whence $x(ax-1) = 0$ and $x(a+(ax-1)) = 1$. Since $x$ has a unique right inverse, $a = a + (ax-1)$ and thus $ax = 1$. Thus $a$ is an inverse of $x$.

(3) Same proof, but use the fact that there is no zero divisor to conclude directly that $ax - 1 = 0$.

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  • $\begingroup$ Thank you, sir. However, there is still one thing I'm wondering about the proof for (1). I can't prove the fact that if $x \in R$ has a right inverse and a left inverse, then $x$ has a unique inverse. Please show me. $\endgroup$
    – ElementX
    Commented Dec 31, 2017 at 6:28
  • $\begingroup$ Also, I'm still working on how the author found out the expression $a + (1 - ax)x^n$. I can "temporally" explain that because $x(1 - ax) = x - xax = x - x.1 = 0$, so $x(1 - ax)x^n = 0$, thus $xa + x(1 - ax)x^n = 1$. But starting from scratch, is there a way to find this out without relying on experience? $\endgroup$
    – ElementX
    Commented Dec 31, 2017 at 6:31
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    $\begingroup$ For your first comment, see this question. $\endgroup$
    – J.-E. Pin
    Commented Dec 31, 2017 at 7:01
  • $\begingroup$ That way, it's possible to prove that the left inverse is equal to the right inverse. But how about proving the uniqueness of the inverse? It can be easily done on a group, but on a ring, I have difficulties. Please enlighten me, sir. $\endgroup$
    – ElementX
    Commented Dec 31, 2017 at 7:37
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    $\begingroup$ The proof shows that if $a$ is any left inverse and $b$ is any right inverse, then $a=b$. If $c$ is any other right inverse, then $c=a$ so $c=b$. The proof really shows that if there is a a right inverse and a left inverse, then it is unique. $\endgroup$ Commented Jan 1, 2018 at 3:56

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