quick question on rings of polynomials.

If http://latex.codecogs.com/png.latex?F is a field then the units in http://latex.codecogs.com/png.latex?F[x] are precisely the units in http://latex.codecogs.com/png.latex?F. True or false?

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

Proof:

let http://latex.codecogs.com/png.latex?f(x),g(x) \in F[x]
where http://latex.codecogs.com/png.latex?...+\ldots+b_mx^m that is degree of http://latex.codecogs.com/png.latex?f \, is \, n>0 and degree of http://latex.codecogs.com/png.latex?g \, is \, m>0
now, http://latex.codecogs.com/png.latex?f(x)g(x) can be written as
http://latex.codecogs.com/png.latex?...+a_nb_mx^{m+n}.

http://latex.codecogs.com/png.latex?..._nb_mx^{n+m}=1
This gives http://latex.codecogs.com/png.latex?a_nb_m=0 (doesnt it?)
since http://latex.codecogs.com/png.latex?F is a field one of http://latex.codecogs.com/png.latex?a_n\, or\, b_m is 0 contradicting the fact that the degrees of the polynomials were http://latex.codecogs.com/png.latex?n \,and\, m
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.

Quote:

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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.

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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).