Thread: quick question on rings of polynomials.

If is a field then the units in are precisely the units in . True or false?

The answer is not given in the book.
My answer: True.

Proof:

let
where that is degree of and degree of
now, can be written as
.


This gives (doesnt it?)
since is a field one of is 0 contradicting the fact that the degrees of the polynomials were
is this correct?

F field => F integral domain, so the coefficient of the largest term x^(m + n) will never be zero.
The units are constants.

what you have shown is that if deg f > 0 or deg g > 0, fg ≠ 1.

therefore, if fg = 1, it must be that deg f = deg g = 0, hence f(x) = a0, and g(x) = b0, and neither a0 nor b0 can be 0.

Originally Posted by Deveno

what you have shown is that if deg f > 0 or deg g > 0, fg ≠ 1.

therefore, if fg = 1, it must be that deg f = deg g = 0, hence f(x) = a0, and g(x) = b0, and neither a0 nor b0 can be 0.

so doesn't it prove that if F is a field then F[x] has precisely the units in F??

by the very definition of a field, every non-zero element is a unit, because for any non-zero a, 1/a exists, and we have a(1/a) = 1.

so, yes, the non-zero constant polynomials are the units of F[x], which can be considered "the units of F" through the isomorphism F-->F[x]:

a--->f(x) = a.

(there is a slight distinction between a field element, a, and f(x) = a, which is a "polynomial expression". the constant polynomials of F[x]

"act just like" field elements, and the isomorphism is often understood implicitly).