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