1.
Please cite all of the properties, theorems, corollaries and definitions you use.
Let F be a field. If f(x) and g(x) are associates in F[x], prove that the two polynomials have the
same roots in F.
Solution
For simplicity, we call the multiplicative unit of F 1, and we identify the multiplicative unit of
F[x] as 1(x)
Since f(x) and g(x) are associates in F[x], there exists a unit u(x) in F[x] such that:
f(x) = u(x)*g(x)
First a lemma: u(x) has no zeros in F.
Proof: u(x) a unit implies there exists an element v(x) in F[x] such that u(x)*v(x) = 1(x)
If s were a zero of u(x), then plugging into our uv=1 equation:
u(x)*v(x) = 1(x)
u(s)*v(s) = 1 [Note: 1(x) = 1 for all values of x]
0*v(s) = 1
0 = 1
Contradiction. Thus u has no zeros.

2.
Now, let a be a zero of f:
f(x) = u(x)*g(x)
f(a) = u(a)*g(a)
0 = u(a)*g(a)
Since F is a field, it is an integral domain, so
u(a) = 0 or g(a) = 0
The first is impossible by our lemma, so we have g(a) = 0.
Therefore a is a zero of g.
Now let b be a zero of g:
f(x) = u(x)*g(x)
f(b) = u(b)*g(b)
f(b) = u(b)*0
f(b) = 0
Therefore b is a zero of f.
We have proven that an element of F is a zero of one if and only if it is a zero of the other. Thus f
and g have the same zeros in F.