Field question Divisibility

• Jan 8th 2011, 07:30 AM
kathrynmath
Field question Divisibility
Let f(x), g(x) be in F(x). Show that if g(x)|f(x) and f(x)|g(x), then f(x)=kg(x) for some k in F.

Since g(x)|f(x), then f(x)=g(x)r(x) for some r(x) in F(x).
Similarily, since f(x)|g(x), then g(x)=f(x)s(x)
So f(x)=f(x)s(x)r(x)
I don't know where to go from here.
• Jan 8th 2011, 07:48 AM
TheEmptySet
Quote:

Originally Posted by kathrynmath
Let f(x), g(x) be in F(x). Show that if g(x)|f(x) and f(x)|g(x), then f(x)=kg(x) for some k in F.

Since g(x)|f(x), then f(x)=g(x)r(x) for some r(x) in F(x).
Similarily, since f(x)|g(x), then g(x)=f(x)s(x)
So f(x)=f(x)s(x)r(x)
I don't know where to go from here.

You are almost there :) Notice that

$\displaystyle \displaystyle f(x)=f(x)s(x)r(x) \iff f(x)[1-s(x)r(x)]=0$

Since $\displaystyle \mathbb{F}[x]$ is an integral domain the zero product principle applies so you get

$\displaystyle \displaystyle 1-s(x)r(x)=0 \iff s(x)r(x)=1$ this tells us that $\displaystyle s(x), r(x)$ must be units, and that they are inverses of each other. What are the only invertible elements of $\displaystyle \mathbb{F}[x]$?
• Jan 8th 2011, 08:43 AM
kathrynmath
s(x) and r(x) are the only invertible elements?
• Jan 8th 2011, 11:12 AM
TheEmptySet
Quote:

Originally Posted by kathrynmath
s(x) and r(x) are the only invertible elements?

No. Given any field $\displaystyle \mathbb{F}$ polynomials over the field $\displaystyle \mathbb{F}[x]$ what are the invertible elements?

This may help. Does $\displaystyle 2x$ have an inverse is $\displaystyle \mathbb{R}[x]$. Does there exist an $\displaystyle f(x)$ suchthat
$\displaystyle 2xf(x)=1$? Think about this and good luck.
• Jan 8th 2011, 11:23 AM
kathrynmath
Well for that, we have 2x(1/2x^-1)=1
• Jan 8th 2011, 12:08 PM
Drexel28
Quote:

Originally Posted by kathrynmath
Well for that, we have 2x(1/2x^-1)=1

Is $\displaystyle x^{-1}\in\mathbb{R}[x]$?