Definition of a normed ring

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.

• Short maps seem like the most natural choice to me; for example the category of Banach spaces and short maps is the most well-behaved category of Banach spaces I know of. It's complete, cocomplete, closed monoidal wrt the projective tensor product... I imagine analogous statements hold for normed abelian groups. – Qiaochu Yuan Nov 26 '15 at 8:06
• @QiaochuYuan: Yes. The only problem is that $\mathsf{Ban}_1$ is not $\mathbb{Z}$-linear. – Martin Brandenburg Nov 26 '15 at 8:30
• I would like to point out that your definitions are not so different. Namely, starting with a bounded norm $\|.\|$, you can always find a short one that is equivalent to it. First, replace $\|.\|$ by $\|.\|' = \|.\|/\|1\|$ so as to ensure that $\|1\|'=1$. Next, for every $a\in R$, define $\|a\|'' = \sup(\{\|ab\|'/\|b\|', b\in R, b\ne 0\})$. This should do the job. – Jérôme Poineau Nov 26 '15 at 8:33
• Regarding "short maps" versus "bounded maps": I wonder if ncatlab.org/nlab/show/M-category is the best practical solution; certainly for Banach spaces and Banach algebras, it seems to reflect actual working practice – Yemon Choi Nov 26 '15 at 13:47
• @Martin: this isn't a big problem. This category is enriched over convex spaces, which is close enough to being enriched over abelian groups that you can still, for example, use it to replace (co)equalizers with (co)kernels. – Qiaochu Yuan Nov 26 '15 at 18:16

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 $$||y\cdot y^{-1}||'=||1||'=1> 1\cdot c^{-1}=||y||'\cdot ||y^{-1}||'.$$