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)