Definition of a normed ring

Question

A normed ring "should" be a monoid object in the monoidal category of normed abelian groups. There are (at least) two choices of morphisms of normed groups, namely bounded or short homomorphisms, resulting in two definitions of a normed ring: Concretely, a normed ring is a ring $R$ (not necessarily commutative) together with a map $R_{\mathsf{Set}} \to \mathbb{R}_{\geq 0},~ x \mapsto |x|$ such that for all $x,y \in R_{\mathsf{Set}}$

• $|x+y| \leq |x|+|y|$
• $|{-}x|=|x|$
• $|x|=0 \Leftrightarrow x=0$,

and now either (choosing short homomorphisms)

• $|1| \leq 1$ and $|x y| \leq |x| |y|$ for all $x,y$,

(which easily implies $R = 0$ or $|1|=1$), equivalently

• $|x_1 \cdot \dotsc \cdot x_n| \leq |x_1| \cdot \dotsc \cdot |x_n|$ for every finite sequence of elements $x_1,\dotsc,x_n$ (including the empty one!),

or (choosing bounded homomorphisms)

• there is a constant $K \geq 0$ with $|xy| \leq K|x||y|$ for all $x,y$

(but no condition for the unit). In all the references that I have found so far, these two definitions are somewhat mixed. For example, $|xy| \leq |x||y|$ is assumed, but nothing about the unit. Isn't this a mistake? Also, I have almost never found the condition $|{-}x|=|x|$, but this doesn't seem to follow from the rest, right? What do you think about the stronger multiplicativity condition $|xy|=|x||y|$? This prevents zero divisors, which is certainly useful, but should one really put this into the definition? Also, which of the two definitions (short/bounded) is preferred in which contexts, and why? I would like to ask similar questions for normed modules over a normed ring, but maybe this will be a separated post.

Answer 1

Throughout, I'll work with your "short" axioms. I'll begin by addressing your questions about independence of the axioms.

First, the axiom $||1||\leq 1$ does not follow from the other axioms. To see this take $R=\mathbb{Z}$, let $c>1$ be a real number, and let your norm be given by $||x||=|x|\cdot c$ (where by $|x|$, I mean the usual absolute value of $x\in \mathbb{Z}$).

Second, the axiom $|-x|=|x|$ does not follow from the others. To see this again take $R=\mathbb{Z}$, let $c_1\neq c_2>1$ be real numbers with $c_2^2\geq c_1$, and let your norm by given by $$||x||=\begin{cases}|x|\cdot c_1 & \text{ if }x\geq 0\\ |x|\cdot c_2 & \text{ if }x<0\end{cases}.$$ A quick check should show this satisfies all of the other axioms, if I didn't make any mistake.

Third, I believe that Jérôme Poineau's comment is incorrect, in that by dividing by $||1||$ you can lose the norm property. To see this, take $R=\mathbb{Z}[y,y^{-1}]$, let $c>1$ be a real number, and let your norm be given by $$||\sum_{\rm finite}a_m y^m|| = \sum_{m<0}|a_m|+\sum_{m\geq 0}|a_m|\cdot c.$$ Again, if I didn't make any mistakes, this should satisfy all the norm axioms except that $||1||=c>1$. However, if you try to create a new norm $||f||'=||f||/||1||$, this loses the multiplicative property since $$||y\cdot y^{-1}||'=||1||'=1> 1\cdot c^{-1}=||y||'\cdot ||y^{-1}||'.$$

As for your question about which definition is preferred, I'm not an expert and will leave that to others.