# polynomial functions vs polynomials

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

$$f:k\to k$$ |

is called polynomial function, iff there are ${a}_{0},\mathrm{\dots},{a}_{n}\in k$ such that

$$f(x)={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\mathrm{\cdots}+{a}_{n}{x}^{n}$$ |

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:

Proposition^{} 1. 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 $k=\{{a}_{1},\mathrm{\dots},{a}_{n}\}$ is a finite field^{}. Put

$$W=(x-{a}_{1})\mathrm{\cdots}(x-{a}_{n}).$$ |

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

$$T(W)(x)=(x-{a}_{1})\mathrm{\cdots}(x-{a}_{n})=({a}_{i}-{a}_{1})\mathrm{\cdots}({a}_{i}-{a}_{i})\mathrm{\cdots}({a}_{i}-{a}_{n})=0$$ |

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. $n\u2a7e1$ and ${a}_{n}\ne 0$ 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

$$T(W)(a)\ne 0.$$ |

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

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 $k=\{{a}_{1},\mathrm{\dots},{a}_{n}\}$ is a finite field and

$$W=(x-{a}_{1})\mathrm{\cdots}(x-{a}_{n}).$$ |

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 $W$. $\mathrm{\square}$.

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

Proof. Let $k=\{{a}_{1},\mathrm{\dots},{a}_{q}\}$ and

$$W=(x-{a}_{1})\mathrm{\cdots}(x-{a}_{q}).$$ |

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

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

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

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 |
---|---|

Canonical name | PolynomialFunctionsVsPolynomials |

Date of creation | 2013-03-22 19:18:03 |

Last modified on | 2013-03-22 19:18:03 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 13A99 |