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)

Recent change to link to Norm
I'm not sure that the recent edit to redirect "norm" to "Complex modulus" is correct, and "Trace" has not been disambiguated, so I've put that template back. --prime mover (talk) 21:59, 9 November 2013 (UTC)