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I would be really happy if someone got an idea because I have no clue at all.

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    This has been asked here several times.... – 2017-01-13

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Let $R$ be the ring of endomorphisms of $Z=\{(z_n):\ z_n\in\mathbb Z,\ \forall n\}$. Define $$ a(z_1,z_2,\ldots)=(z_2,z_3,\ldots) $$ and $$ b(z_1,z_2,\ldots)=(0,z_1,z_2,\ldots). $$ Then $ab=1$, but $$ ba(z_1,z_2,\ldots)=(0,z_2,z_3,\ldots) $$ so $ba\ne 1$.

Of course you don't need $\mathbb Z$ to build the underlying ring; sequences over any ring would do.

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    hey, thanks alot already.. but could you maybe explain a bit more whats happening? why is ab=1? I'm sorry but I can't really understand the calculations. – 2017-01-13
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    The ring $R$ are functions over $Z$. The operations are pointwise addition and composition. The unity is the identity function. So $$ab(z_1,z_2,\ldots)=a(0,z_1,z_2,\ldots)=(z_1,z_2,\ldots)$$ and $ab $ is the identity function. – 2017-01-13
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    But wouldn't then be ba: ba(z1,z2,…)=b(z2,z3,…)=(0,z2,z3,…) – 2017-01-13
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    Exactly. That's why $ba\ne1$. – 2017-01-13
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    Alright thank you very much but then you got a little mistake in your calculation thats why I got confused. :) – 2017-01-13
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    Oh, I just noticed the typo. My bad. – 2017-01-13
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Here is a classic example you can use in many situations:

Let $S$ be the space of sequences of real numbers. Let $R$ be the space of linear functions from $S$ to $S$. This is a ring. Then we let $a$ and $b$ be two shift operators:

$a$ takes a sequence and appends a zero to the front. $b$ removes the first element of the sequence, reindexing so that everything comes one index earlier.

Then we have $ab=1$ but $ba\ne 1$.

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If this is possible, then the following is an example.

  • Let $S$ be the free ring $\mathbb{Z}\langle x, y \rangle$ generated by two elements.
  • Let $I$ be the two sided ideal generated by $xy - 1$.
  • Let $R = S/I$
  • Let $a = x + I$
  • Let $b = y + I$

This is the universal example in the following sense: given any other triple $(R', a', b')$ that also has this property, there is a unique ring homomorphism $\varphi : R \to R'$ with the property that $a' = \varphi(a)$ and $b' = \varphi(b)$.

We can check directly that $ba \neq 1$; in fact, I assert that the additive group of $R$ is a free abelian group generated by the monomials $b^m a^n$ where $m,n$ range over all natural numbers.

There are a number of ways one might go about proving this fact. I think the way I would do so is to prove that $R$ is (isomorphic to) the monoid ring $\mathbb{Z}[M]$, where $M$ is the monoid presented by $M = \langle a,b \mid ab=1 \rangle$, and checking that the elements of $M$ are precisely the words $b^m a^n$.