Talk:Frobenius's Theorem

As for the recent edit, it seems to presume that $A$ is embedded in $\C$ or so. This needn't be the case. If I misunderstand, please correct me, but I'd approach this formally and simply reason contrapositively that two $t(x),n(x)$ and $t'(x),n'(x)$ satisfying the equation are equal iff $t(x)=t'(x)$, and then write $x = \frac{n-n'}{t-t'}$ if it isn't; the latter is in $\R$. --Lord_Farin (talk) 22:12, 29 November 2012 (UTC)