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For matrices, $AB=I$ implies $BA=I$. However, in any ring $R$ with unity, does $ab=1$ ensures $ba=1$? I do not think so, but am not sure...

Keith
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    Try the endomorphism ring of an infinite-dimensional vector space. – Angina Seng Apr 10 '18 at 16:07
  • Do you mean there is a counterexample? – Keith Apr 10 '18 at 16:10
  • @LordSharktheUnknown: where you map the $n$-th component to the $n+1$th? – Raphael J.F. Berger Apr 10 '18 at 16:17
  • Yes. For example, the derivative and integral in a space of smooth functions – Veridian Dynamics Apr 10 '18 at 16:17
  • This question certainly had a duplicate at some point, but I can't seem to find it now. I thought I remember it being in lots of places. Oh well. We'll see if this one lasts. You should read this post https://math.stackexchange.com/questions/33491/why-are-dedekind-finite-rings-called-so which answers your question in its first sentence. – rschwieb Apr 10 '18 at 16:31
  • @Veridian Dynamics: A very nice example! Would be nice to appear as an answer. – orangeskid Apr 10 '18 at 16:35
  • Is the OP's question true in a finite ring? – Randall Apr 10 '18 at 16:54
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    @Randall Yes. The equation $ab=1$ implies that $b$ is not a zero divisor. The map $m_b:R\backslash{0}\to R\backslash{0}$, $m_b(x)=bx$ is injective (with left inverse $m_a$). If $R$ is finite, $m_b$ is surjective, so $bx=1$ for some $x$. Multiplying this equation by $a$ on the left gives $x=a$. – David Hill Apr 10 '18 at 18:59
  • @Randall yes, and THAT question is definitely duplicated numerous times on the site. – rschwieb Apr 11 '18 at 02:33

3 Answers3

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Let me explain the answer in the comments:

Let $V$ be an infinite dimensional vector with basis $\{v_1,v_2,v_3,\ldots\}$, and let $R=\mathrm{End}(V)$ be the set of linear transformations $T:V\to V$. Then $R$ is a ring under pointwise addition and composition of functions.

Now, let $T:V\to V$ be the map $T(v_i)=v_{i+1}$, and $S:V\to V$ be the map $$S(v_i)=\begin{cases}0&\mbox{if }i=1\\v_{i-1}&\mbox{otherwise.}\end{cases}$$ Then $ST=Id_V$, but $TS\neq Id_V$ since $TS(v_1)=0$.

David Hill
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    I don't understand your question. The ring axioms lead to examples of rings. One such example is the one I've explained above. – David Hill Apr 10 '18 at 16:22
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    +1 Another example in the same vein that usually comes up at the same time is using $\mathbb R[x]$ and letting $S$ be the differentiation operator, and $T$ map the function in the indefinite integral with constant $0$. – rschwieb Apr 10 '18 at 16:36
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What is true is that if $ab = 1$, and there exists some element $c$ with $ca=1$, then $c=b$. (This follows immediately from multiplying both sides of $ca=1$ on the right by $b$).

But there's no guarantee in a general ring that $a$ always has a left inverse, if it has a right inverse, as David Hill shows in his answer. (This is true in a commutative ring, of course, and more generally in a Noetherian ring.)

user7530
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The answer is, in general, no.

Let $\Bbb F$ be any field, and let $\mathcal V$ be the set of sequences of elements of $\Bbb F$; i.e.,

$\mathcal V = \{ (a_1, a_2, a_3, \ldots) \mid a_i \in \Bbb F, \forall i \in \Bbb N \}; \tag 1$

it is easy to see that $\mathcal V$ may be endowed with the structure of a vector space over $\Bbb F$ in the usual way, by defining addition component-wise and, for $\alpha \in \mathcal V$,

$\alpha = (\alpha_1, \alpha_2, \alpha_3, \ldots), \; \alpha_i \in \Bbb F, \; \forall i \in \Bbb N, \tag 2$

setting

$b\alpha = (b\alpha_1, b\alpha_2, b\alpha_3, \ldots), \tag 3$

where $b \in \Bbb F$.

Now the maps $R: \mathcal V \to \mathcal V$, $L:\mathcal V \to \mathcal V$ defined by

$R(\alpha_1, \alpha_2, \alpha_3, \ldots) = (0, \alpha_1, \alpha_2, \alpha_3, \ldots), \tag 4$

$L(\alpha_1, \alpha_2, \alpha_3, \ldots) = (\alpha_2,\alpha_4, \alpha_4, \ldots), \tag 5$

are easily seen to be in $\mathcal L(\mathcal V)$, the ring of linear endomorphisms of $\mathcal V$, and also it is easy to see that

$LR \alpha = \alpha, \tag 6$

for $\alpha \in \mathcal V$, but

$RL\alpha = RL(\alpha_1, \alpha_2, \alpha_3, \ldots) = R(\alpha_2, \alpha_4, \alpha_4, \ldots) = (0, \alpha_2, \alpha_3, \ldots) \ne \alpha \tag 7$

unless $\alpha_1 = 0$, so

$LR = I, \tag{8}$

the identity endomorphism, but

$RL \ne I, \tag{9}$

thus providing a counter-example.

Robert Lewis
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