Thread: Field question Divisibility

1. 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.

2. 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

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

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

$\displaystyle 1-s(x)r(x)=0 \iff s(x)r(x)=1$ this tells us that $s(x), r(x)$ must be units, and that they are inverses of each other. What are the only invertible elements of $\mathbb{F}[x]$?

3. s(x) and r(x) are the only invertible elements?

4. Originally Posted by kathrynmath
s(x) and r(x) are the only invertible elements?
No. Given any field $\mathbb{F}$ polynomials over the field $\mathbb{F}[x]$ what are the invertible elements?

This may help. Does $2x$ have an inverse is $\mathbb{R}[x]$. Does there exist an $f(x)$ suchthat
$2xf(x)=1$? Think about this and good luck.

5. Well for that, we have 2x(1/2x^-1)=1

6. Originally Posted by kathrynmath
Well for that, we have 2x(1/2x^-1)=1
Is $x^{-1}\in\mathbb{R}[x]$?