Right invertible element

Question

My question is as follows: $a,b \in L$ where $L$ is a ring. We are given that $ab=1$.

I need to prove or disprove that $a$ is invertible (meaning, there is an element $x \in L : ax=xa=1$).

But how can we say that for sure? just because $ab=1$ doesn't mean that $ba=1$, ba could be something else altogether, we werent given that this is a commutative ring.

Please give advice on how to solve, thank you.

Word by word translation of the question to make it clearer: Prove or disprove that if $a$ and $b$ are elements of $L$ and $ab=1$, then $a$ is invertible.

Yes, indeed, @nik ! I wanted to delete my answer but since Oria started a dialogue about it I'll leave my post (also because I wrote a comment on some students' reaction to certain similar examples). — Georges Elencwajg, Nov 02 '13 at 22:51
You didn't quantify whether this was for all $a$ there exists a $b$ such that $ab = 1$. Is that what you mean? This makes it evident it is invertible (it is the def'n). — Don Larynx, Nov 02 '13 at 22:17
I'll quote the question word by word: prove or disprove that if a and b are elements of L and ab=1 then a is invertible. — Oria Gruber, Nov 02 '13 at 22:19
Admitting an inverse. An object that is invertible is referred to as an invertible element in a monoid or a unit ring, or to a map, which admits an inverse map iff it is bijective. This means it has a left and right inverse in the inverse part, if that is what you are asking @OriaGruber — Don Larynx, Nov 02 '13 at 22:25

score 1 · Accepted Answer · answered Nov 02 '13 at 22:31

1

Suppose $L$ is the ring of $\mathbb R$-linear maps $l:\mathbb R[X]\to \mathbb R[X]$ (with multiplication being composition).
Consider $D,I\in L$ defined by $D(P(X))=P'(X)$ and $I(P(X))=\int _0^XP(t)dt$.
Then $D\circ I=Id$ but nevertheless $D$ is not invertible because...

answered Nov 02 '13 at 22:31

Georges Elencwajg

150,790

Because I (composition) D is not the identity transformation? – Oria Gruber Nov 02 '13 at 22:33
Yes. Or because $D$ is not injective: $D(r)=0$ for all constant polynomials $r\in \mathbb R\subset \mathbb R[X]$ – Georges Elencwajg Nov 02 '13 at 22:37
By the way, could you complete the equality $I(D(P(X)))=?$ , in other words compute $I\circ D$ ? – Georges Elencwajg Nov 02 '13 at 22:40
My students are often puzzled by this kind of example because it mixes linear algebra and calculus. Have teachers on this site had the same reaction from some of their students? – Georges Elencwajg Nov 02 '13 at 22:45
I'm not sure I (composition) D is a function, since it can have many images for the same source. for example the polynomial 2x can go to 2x or 2x+1 or 2x-1...I(D(2x)) = I(2) = 2x+c, c is an element of R – Oria Gruber Nov 02 '13 at 22:50
Dear Oria, no: $I$ is not any old antiderivative but the antiderivative which vanishes when $X=0$. So $I(2)=\int _0^X 2dt=2X$ and not $I(2)=2X+1$, which is false. – Georges Elencwajg Nov 02 '13 at 22:56
in that case why isn't I*D identity? can you give an example where I(D(P(X))) is not P(X)? – Oria Gruber Nov 02 '13 at 22:57
Here is an example: $(I\circ D)(2X+1)=I(2)=2X$, so $I\circ D\neq Id$ – Georges Elencwajg Nov 02 '13 at 22:59
That's right! I should have thought of that. Thank you Georges you have helped me a lot and I understood what you said and the examples you gave. Thank you. – Oria Gruber Nov 02 '13 at 23:00
You are welcome, dear Oria, and I'm very happy that I could help you understand these rather subtle concepts. – Georges Elencwajg Nov 02 '13 at 23:04

Right invertible element

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