# Ring with Multiplicative Norm has No Proper Zero Divisors

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## 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

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 |

a contradiction.

$\blacksquare$