# polynomial functions vs polynomials

Let $k$ be a field. Recall that a function

 $f:k\to k$

is called polynomial function, iff there are $a_{0},\ldots,a_{n}\in k$ such that

 $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}$

for any $x\in k$.

The ring of all polynomial functions (together with obvious addition and multiplication) we denote by $k\{x\}$. Also denote by $k[x]$ the ring of polynomials (see this entry (http://planetmath.org/PolynomialRing) for details).

There is a canonical function $T:k[x]\to k\{x\}$ such that for any polynomial

 $W=\sum_{i=1}^{n}a_{i}\cdot x^{i}$

we have that $T(W)$ is a polynomial function given by

 $T(W)(x)=\sum_{i=1}^{n}a_{i}\cdot x^{i}.$

(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 evaluation map $T$ is a ring homomorphisms which is ,,onto”.

The question is: when $T$ is ,,1-1”?

Proposition 2. $T$ is ,,1-1” if and only if $k$ is an infinite field.

Proof. ,,$\Rightarrow$” Assume that $k=\{a_{1},\ldots,a_{n}\}$ is a finite field. Put

 $W=(x-a_{1})\cdots(x-a_{n}).$

Then for any $x\in k$ we have that $x=a_{i}$ for some $i$ and

 $T(W)(x)=(x-a_{1})\cdots(x-a_{n})=(a_{i}-a_{1})\cdots(a_{i}-a_{i})\cdots(a_{i}-% a_{n})=0$

which shows that $W\in\mathrm{Ker}T$ although $W$ is nonzero. Thus $T$ is not ,,1-1”.

,,$\Leftarrow$” Assume, that

 $W=\sum_{i=1}^{n}a_{i}\cdot x^{i}$

is a polynomial with positive degree, i.e. $n\geqslant 1$ and $a_{n}\neq 0$ such that $T(W)$ is a zero function. It follows from the Bezout’s theorem that $W$ has at most $n$ roots (in fact this is true over any integral domain). Thus since $k$ is an infinite field, then there exists $a\in k$ which is not a root of $W$. In particular

 $T(W)(a)\neq 0.$

Contradiction, since $T(W)$ is a zero function. Thus $T$ is ,,1-1”, which completes the proof. $\square$

This shows that the evaluation map $T$ is an isomorphism only when $k$ is infinite. So the interesting question is what is a kernel of $T$, when $k$ is a finite field?

Proposition 3. Assume that $k=\{a_{1},\ldots,a_{n}\}$ is a finite field and

 $W=(x-a_{1})\cdots(x-a_{n}).$

Then $T(W)=0$ and if $T(U)=0$ for some polynomial $U$, then $W$ divides $U$. In particular

 $\mathrm{Ker}T=(W).$

Proof. In the proof of proposition 2 we’ve shown that $T(W)=0$. Now if $T(U)=0$, then every $a_{i}$ is a root of $U$. It follows from the Bezout’s theorem that $(x-a_{i})$ must divide $U$ for any $i$. In particular $W$ divides $U$. This (together with the fact that $T(W)=0$) shows that the ideal $\mathrm{Ker}T$ is generated by $W$. $\square$.

Corollary 4. If $k$ is a finite field of order $q>1$, then $k\{x\}$ has exactly $q^{q}$ elements.

Proof. Let $k=\{a_{1},\ldots,a_{q}\}$ and

 $W=(x-a_{1})\cdots(x-a_{q}).$

By propositions 1 and 3 (and due to First Isomorphism Theorem for rings) we have that

 $k\{x\}\simeq k[x]/(W).$

But the degree of $W$ is equal to $q$. It follows that dimension of $k[x]/(W)$ (as a vector space over $k$) is equal to

 $\mathrm{dim}_{k}k[x]/(W)=q.$

Thus $k\{x\}$ is isomorphic to $q$ copies of $k$ as a vector space

 $k\{x\}\simeq k\times\cdots\times k.$

This completes the proof, since $k$ has $q$ elements. $\square$

Remark. Also all of this hold, if we replace $k$ 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 PolynomialFunctionsVsPolynomials 2013-03-22 19:18:03 2013-03-22 19:18:03 joking (16130) joking (16130) 4 joking (16130) Theorem msc 13A99