Ring with Multiplicative Norm has No Proper Zero Divisors

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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:

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

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:

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

a contradiction.

$\blacksquare$