Definition talk:Norm on Division Ring

After careful consideration, I think this definition may be generalised to a ring without zero divisors. This is good, as then it will incorporate the absolute value on $\Z$. --Lord_Farin 05:30, 9 December 2011 (CST)
 * I believe you can accomplish that by just dropping to ring. Positive definite + multiplicative will then prove it to have no proper zero divisors. That distinction will only come up if you decide to define a seminorm for rings. --Dfeuer (talk) 18:06, 17 January 2013 (UTC)

Wikipedia, at Berkovich space, says this:
 * A seminorm on a ring A is a non-constant function f→|f| from A to the non-negative reals such that |0| = 0, |1| = 1, |f + g| ≤ |f| + |g|, |fg| ≤ |f||g|. It is called multiplicative if |fg| = |f||g| and is called a norm if |f| = 0 implies f = 0.

I have no idea if that's right, of course. However, suppose $R$ is a division ring with a multiplicative seminorm.

Then since the seminorm is not constant, there is an $x$ such that $|x|≠0$. Let $y≠0_R$.

Then $|x|=|x \circ y^{-1}\circ y| \le |x \circ y^{-1}||y|$, so $|y|≠0$.

Thus a multiplicative seminorm on a division ring is a norm. --Dfeuer (talk) 20:53, 17 January 2013 (UTC)


 * Having discovered a definition of that site, it is now important that you find some corroborative evidence elsewhere to back it up. Not only is Wikipedia a tertiary source but it is well-known as being laughably unreliable. --prime mover (talk) 21:33, 17 January 2013 (UTC)

Change suggestion
Do you want this to look similar to the pages "Normed Vector Space" and "Norm (Vector Space)" or do you want to get rid of this page?


 * I moved the above comment from the redirect page onto here because it makes no sense there. --prime mover (talk) 04:27, 7 October 2018 (EDT)