Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F\hspace{.02 in}$ be a field. $\:$ Let $E\hspace{.02 in}$ be a non-zero subring of $F$.
Let $\hspace{.03 in}\leq\hspace{.03 in}$ be a total order on $E\hspace{.02 in}$ that makes $E\hspace{.02 in}$ an ordered ring. $\;\;\;$ Let $\;\; |\cdot| \: : \: F\to E \;\;$ be a function.
I will require the following to hold for all members $x$ of $E$ and for all members $y$ and $z$ of $F\hspace{.03 in}$:

(a) $\qquad 0 \leq x \;\; \implies \;\; |x| = x$
(b) $\qquad 0 \leq |y|$
(c) $\qquad |y| = 0 \;\; \iff \;\; y = 0$
(d) $\qquad |y\cdot z| \;\; = \;\; |y| \cdot |z|$
(e) $\qquad |y+z| \;\; \leq \;\; |y| + |z|$

1. $\;$ Is there a name for what I've just set up?

Now, I'm wondering about the redudancy in those conditions. $\;$ I know that
neither (a) nor (b) follows from the other four, and that (c) follows from [(a) and (d)].

${}$2. $\;$ Does (d) follow from the other four?
3. $\;$ Does (e) follow from [(a) and (b) and (d)]?

share|cite|improve this question
I know that [(a) and (d)] forces $E\hspace{.02 in}$ to be a subfield of $F\hspace{.02 in}$. $\;$ – Ricky Demer Jan 26 '12 at 0:54
The property (d) does not follow from the other four, if I haven't made any mistakes. Take $E = \mathbb{R}$ and $F = \mathbb{C}$ with their usual structure. Define the function $|\cdot|_\mathrm{new}:\mathbb{C}\to\mathbb{R}$ to be $|z|_\mathrm{new}:=|\mathfrak{R}(z)|+|\mathfrak{I}(z)|$, that is: the sum of (usual) absolute values of the real and the imaginary part. This has properties (a), (b), (c), (e) but not (d). – Dejan Govc Jan 27 '12 at 0:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.