Thread: Integral domain

1. Integral domain

Originally Posted by The problem
Suppose D is an integral domain with the property that there is a map $v\setminus \{0\} \to \mathbb{N}" alt="v\setminus \{0\} \to \mathbb{N}" /> satisfying the properties (i) For all a,b in D\0, v(a)v(b) = v(ab); and (ii) v(a) = 1 if and only if a is a unit of D. Show that D[x] also has a map with this property.
I've tried all sorts of things, such as (call my map on D[x]\0 V)
$V(a_0 + a_1x + a_2x^2 + \ldots + a_nx^n) = 2^{n}(v(a_0) + v(a_1) + \ldots + v(a_n))$
(the 2^n is to distinguish a_0 being a unit from a_1, say, being a unit for property (ii)), but I'm always hitting a snag for property (i) since there isn't linearity of v. I know the integral domain $\mathbb{Z}[\sqrt{-5}]$ is an example of such a D, with $v(a+b\sqrt{-5}) = a^2 + 5b^2$ so I've been playing with $(\mathbb{Z}[\sqrt{-5}])[x]$ to help come up with a desirable V, but to no avail.

If you could suggest what the map should like I'd be most appreciative.

Many thanks

2. Originally Posted by james123
I've tried all sorts of things, such as (call my map on D[x]\0 V)
$V(a_0 + a_1x + a_2x^2 + \ldots + a_nx^n) = 2^{n}(v(a_0) + v(a_1) + \ldots + v(a_n))$
(the 2^n is to distinguish a_0 being a unit from a_1, say, being a unit for property (ii)), but I'm always hitting a snag for property (i) since there isn't linearity of v. I know the integral domain $\mathbb{Z}[\sqrt{-5}]$ is an example of such a D, with $v(a+b\sqrt{-5}) = a^2 + 5b^2$ so I've been playing with $(\mathbb{Z}[\sqrt{-5}])[x]$ to help come up with a desirable V, but to no avail.

If you could suggest what the map should like I'd be most appreciative.

Many thanks

An idea: if $f(x)=a_0 + a_1x + a_2x^2 + \ldots + a_nx^n\,,\,\,a_n\neq 0$ is a pol. in $D[x]$ of degree n, try $V(f(x)):= v(a_n)$ .

Tonio

Disclaimer: I'm not 100% sure the above works...not even 93.156% sure, in fact.

Hmmm...now I'm almost sure it doesn't work because of condition (2)....

3. Slapping on $2^n$ (ie. $2^{\deg(f)}$) at the front fixes that (which I suspect you knew ). Thanks, Tonio!