Integral domain

**james123** · December 15th 2009, 09:04 AM

Originally Posted by The problem

Suppose D is an integral domain with the property that there is a map $v<img src=$ \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

**tonio** · December 15th 2009, 09:15 AM

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)....

**james123** · December 15th 2009, 09:32 AM

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

). Thanks, Tonio!

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