 Automatic real bilinearity from a multiplicative norm

Suppose a ring has an underlying real normed vector space and its norm is multiplicative, without initially assuming real bilinearity of multiplication. For fixed $x$, distributivity makes $y\mapsto xy$ additive, while $\Vert xy-xz\Vert =\Vert x\Vert \Vert y-z\Vert$ makes it continuous. Additivity gives rational scalar compatibility, and continuity extends it to every real scalar. Apply the same argument in the other variable to obtain a bilinear map. The norm identity also excludes nonzero zero divisors.