Ring with Multiplicative Norm has No Proper Zero Divisors

Theorem

Let $\struct {R, +, \circ}$ be a ring.

Let its zero be denoted by $0_R$.

Let $\norm {\,\cdot\,}$ be a multiplicative norm on $R$.

Then $R$ has no proper zero divisors.

That is:

$\forall x, y \in R^*: x \circ y \ne 0_R$

where $R^*$ is defined as $R \setminus \set {0_R}$.

Proof

Aiming for a contradiction, suppose:

$\exists x, y \in {R^*} : x \circ y = 0_R$

By positive definiteness:

$x, y \ne 0_R \iff \norm x, \norm y \ne 0$

Thus:

$\norm x \norm y \ne 0$

But we also have:

\(\ds \norm x \norm y\)

\(=\)

\(\ds \norm {x \circ y}\)

Multiplicativity

\(\ds \)

\(=\)

\(\ds \norm {0_R}\)

by hypothesis

\(\ds \)

\(=\)

\(\ds 0\)

Positive Definiteness

a contradiction.

$\blacksquare$