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Let \( \displaystyle R \) be an associative ring with unity \( \displaystyle 1_{R} \). Show that \( \displaystyle R \) is a division ring if and only if \( \displaystyle \forall a \in R \smallsetminus \{ 1_{R} \} \; \exists b \in R \; : \; ab = a + b \).

If \(R\) is a division ring, given \(a\neq 1\), consider the non-zero element \(a-1\) of \(R\). It has a multiplicative inverse \(x\). So \((a-1)x=1\). Taking \(b=x+1\), we get \(1 = (a-1)x = (a-1)(b-1) = ab-a-b+1\) giving \(ab=a+b\) as required.

Conversely, if \(R\) satisfies the property, given \(a\neq 0\) consider \(a+1 \neq 1\) and take \(b\) such that \((a+1)b = (a+1)+b\). Then \(ab=a+1\) so \(a(b-1)=1\). I.e. \(a\) has a right multiplicative inverse. To show that a left inverse exists, observe that if \(ax=1\) then \(xax=x\) and if \(y\) is the right inverse of \(x\) which we know that it exists then \(xaxy=xy\) giving \(xa=1\) i.e. that \(x\) is a left multiplicative inverse of \(a\) as well.

Associativity was used implicitly when writing things like \(xaxy\) without putting any brackets.

Thank you for the short and elegant solution.

The solution i found regarding the second part of the assertion is based on the same idea, but is rather longer than the one you suggested! So, thanks again!