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I'm trying to prove this property of inverses in Artin's algebra text.

An element $a$ may have a left inverse or a right inverse, though it is not invertible.

I don't know if there is a general proof of this fact. What I'm ultimately trying to establish is that there exists a $b$ such that $ab = e$ but $ba \neq e$. Does it suffice to give an example of such an element?

For example, since $a$ can be any element, I can write a $1 \times 2$ matrix, \begin{align*} a = \begin{bmatrix} 2 & 3 \end{bmatrix}. \end{align*} Then, I have a right inversre, \begin{align*} b = \begin{bmatrix} x \\ y \end{bmatrix}, \end{align*} where $2x + 3y = 1$. Picking $y = 0$ and $x = \frac{1}{2}$ easily gives a right inverse, $\begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix}$, so $ab = 1$, though it is not the case that $ba = 1$, since $ba$ isn't even defined. Similarly, $a$ is a left inverse of $b$, but not a right inverse, so neither $a$ nor $b$ are invertible.

Does this suffice as a proof, or should I aim for something more general?

John P.
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  • Asked and answered in this post. But the attempted solution has enough to critique that maybe it's not a duplicate. – rschwieb Mar 11 '20 at 19:42
  • By the way, you are not really tasked with "proving a property," you are supposed to be giving an example. – rschwieb Mar 11 '20 at 19:44
  • I don't think it's a duplicate because this isn't necessarily a ring. The problem in Artin asked to prove these properties for some law of composition. I would be interested in knowing what there is to critique in my answers, though. (Not suggesting it is perfect by any means, but I'd like to improve it.) – John P. Mar 11 '20 at 20:03
  • It doesn't matter that a ring was given as an example. The multiplicative group of that ring is still a monoid, the minimal object for which this question makes any sense. And Jose Carlos Santos has already explained the problem with what you've said: you have not explained the monoid your two elements live in. – rschwieb Mar 11 '20 at 20:05

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My guess is that the context of that sentence is the study of rings. In that case, your example will not work, since you did not say which ring contains those two matrices (and I don't imagine which ring it could be).

You can take the ring of the linear maps from the space $L$ of sequences of real numbers into itself and define $f\colon L\longrightarrow L$ as$$f(a_1,a_2,a_3,a_4,\ldots)=(0,a_1,a_2,a_3,\ldots).$$It has a left inverse:$$\begin{array}{rccc}g\colon&L&\longrightarrow&L\\&(a_1,a_2,a_3,\ldots)&\mapsto&(a_2,a_3,a_4,\ldots),\end{array}$$but no right inverse.

J. W. Tanner
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José Carlos Santos
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  • The problem is in the second chapter on groups, so I don't think Artin was referring to rings. My understanding is that he's referring to laws of composition generally, even those which don't admit a group structure. – John P. Mar 11 '20 at 18:25
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    Then my guess is that Artin is talking about monoids (that is, semigrouos with an identity element), and my example still applies in that context. – José Carlos Santos Mar 11 '20 at 18:47
  • This makes more sense, and your example makes sense to me. Could you tell me whether the example I gave also holds? – John P. Mar 11 '20 at 18:56
  • Yes, if you can get a semigroup which contains those matrices. Otherwise, no. – José Carlos Santos Mar 11 '20 at 18:59