Thread: 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.

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



Since is an integral domain the zero product principle applies so you get

this tells us that must be units, and that they are inverses of each other. What are the only invertible elements of ?

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

Originally Posted by kathrynmath

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

No. Given any field polynomials over the field what are the invertible elements?

This may help. Does have an inverse is . Does there exist an suchthat
? Think about this and good luck.

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

Originally Posted by kathrynmath

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

Is ?