# Ring with Multiplicative Norm has No Proper Zero Divisors

## Theorem

Let $\left({R, +, \circ}\right)$ 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

Assume otherwise:

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

Thus:

$\norm{x} \norm{y} \ne 0$

But we also have:

 $\displaystyle \norm{x} \norm{y}$ $=$ $\displaystyle \norm{x \circ y}$ Multiplicativity $\displaystyle$ $=$ $\displaystyle \norm{0_R}$ by assumption $\displaystyle$ $=$ $\displaystyle 0$ Positive definiteness

$\blacksquare$