# Thread: quick question on rings of polynomials.

1. ## quick question on rings of polynomials.

If $F$ is a field then the units in $F[x]$ are precisely the units in $F$. True or false?

The answer is not given in the book.

Proof:

let $f(x),g(x) \in F[x]$
where $f(x)=a_0+a_1x+a_2x^2+\ldots+a_nx^n,\, g(x)=b_0+b_1x+b_2x^2+\ldots+b_mx^m$ that is degree of $f \, is \, n>0$ and degree of $g \, is \, m>0$
now, $f(x)g(x)$ can be written as
$a_0b_0+h(x)+a_nb_mx^{m+n}$.

$f(x)g(x)=1 \Rightarrow a_0b_0+h(x)+a_nb_mx^{n+m}=1$
This gives $a_nb_m=0$ (doesnt it?)
since $F$ is a field one of $a_n\, or\, b_m$ is 0 contradicting the fact that the degrees of the polynomials were $n \,and\, m$
is this correct?

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

3. 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.

4. 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??

5. 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).