The evaluation map

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Let $R$ be a commutative ring.


  • Definition of the evaluation map

Let $a \in R$. The evaluation map at $a \in R$ is the map $\varphi_a : R[x] \to R$ given by $f(x) \mapsto f(a)$.


  • Theorem

The evaluation map $\varphi_a$ at $a \in R$ is a ring homomorphism with Im$\,\varphi = R$ and Ker$\,\varphi_a = \{f(x) \in R[x] : f(a) = 0\}$.

  • Consequences

1.) Let $a \in R$. Using the First Isomorphism Theorem, we have that $R[x]/\mbox{Ker}\,\varphi_a \cong R$.

2.) Let $F$ be a field and let $a \in F$. Then $F[x]/\mbox{Ker}\,\varphi_a \cong F$. Hence Ker$\,\varphi_a$ is always a maximal ideal of $F[x]$.


  • Remainder Theorem

Let $F$ be a field. Let $a \in F$ and let $f(x) \in F[x]$. Then there are polynomials $q(x), r(x)$ in $F[x]$ with

$f(x) = q(x)(x-a) + r(x)$ and $r(x) = f(a).$


  • Factor Theorem

Let $F$ be a field. Let $a \in F$ and let $f(x) \in F[x]$. Then $f(a) = 0$ if and only if $(x-a)$ is a factor of $f(x)$.

  • Consequence

Let $F$ be a field and let $\varphi_a$ be the evaluation map at $a \in F$. Then

$$f(x) \in \mbox{Ker}\,\varphi_a \iff f(a) = 0 \iff a \mbox{ is a root of } f(x) \iff (x-a) \mbox{ is a factor of } f(x).$$


  • Theorem

Let $F$ be a field. Let $f(x) \in F[x]$ with degree $n$ and suppose $n \geq 1$. Then $f(x)$ has at most $n$ roots in $F$.




Next section on Irreducible polynomials

Back to Chapter 2: Polynomials