polynomial functions vs polynomials
Let be a field. Recall that a function
is called polynomial function, iff there are such that
for any .
The ring of all polynomial functions (together with obvious addition and multiplication) we denote by . Also denote by the ring of polynomials (see this entry (http://planetmath.org/PolynomialRing) for details).
There is a canonical function such that for any polynomial
we have that is a polynomial function given by
(Although we use the same notation for polynomials and polynomial functions these concepts are not the same). This function is called the evaluation map. As a simple exercise we leave the following to the reader:
The question is: when is ,,1-1”?
Proposition 2. is ,,1-1” if and only if is an infinite field.
Proof. ,,” Assume that is a finite field. Put
Then for any we have that for some and
which shows that although is nonzero. Thus is not ,,1-1”.
,,” Assume, that
is a polynomial with positive degree, i.e. and such that is a zero function. It follows from the Bezout’s theorem that has at most roots (in fact this is true over any integral domain). Thus since is an infinite field, then there exists which is not a root of . In particular
This shows that the evaluation map is an isomorphism only when is infinite. So the interesting question is what is a kernel of , when is a finite field?
Proposition 3. Assume that is a finite field and
Then and if for some polynomial , then divides . In particular
Proof. In the proof of proposition 2 we’ve shown that . Now if , then every is a root of . It follows from the Bezout’s theorem that must divide for any . In particular divides . This (together with the fact that ) shows that the ideal is generated by . .
Corollary 4. If is a finite field of order , then has exactly elements.
Proof. Let and
By propositions 1 and 3 (and due to First Isomorphism Theorem for rings) we have that
But the degree of is equal to . It follows that dimension of (as a vector space over ) is equal to
Thus is isomorphic to copies of as a vector space
This completes the proof, since has elements.
Remark. Also all of this hold, if we replace with an integral domain (we can always pass to its field of fractions). However this is not really interesting, since finite integral domains are exactly fields (Wedderburn’s little theorem).
|Title||polynomial functions vs polynomials|
|Date of creation||2013-03-22 19:18:03|
|Last modified on||2013-03-22 19:18:03|
|Last modified by||joking (16130)|