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Thread: finite ring with an odd number of invertible elements

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    finite ring with an odd number of invertible elements

    Suppose $\displaystyle R$ is a finite ring with an odd number of invertible elements. Prove that $\displaystyle R$ has characteristic $\displaystyle 2$ .
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    Quote Originally Posted by petter View Post
    Suppose $\displaystyle R$ is a finite ring with an odd number of invertible elements. Prove that $\displaystyle R$ has characteristic $\displaystyle 2$ .
    I denote by $\displaystyle R^*$ the set of invertible elements of $\displaystyle R$.
    Assume that, for every $\displaystyle x\in R^*$, $\displaystyle x\neq -x$. Then, if we associate each element of $\displaystyle R^*$ with its opposite (which is in $\displaystyle R^*$ as well), we partition $\displaystyle R^*$ in pairs, contradicting the fact that the cardinality of $\displaystyle R^*$ is odd.
    As a consequence, there exists $\displaystyle x\in R^*$ such that $\displaystyle x=-x$. Multiplying by $\displaystyle x^{-1}$, this implies $\displaystyle 1=-1$ (where $\displaystyle 1$ is the unit element of $\displaystyle R$), or $\displaystyle 2=0$ (where $\displaystyle 2=1+1\in R$): the characteristic is $\displaystyle 2$.
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