📂Abstract Algebra

Zeros of a Polynomial Function

Definition¹

$$ f(x) : = \sum_{k=0}^{n} a_{k} x^{k} = a_{0} + a_{1} x + \cdots + a_{n} x^{n} $$ For a polynomial function $f \in F [x]$ and fields $F \le E$, define the evaluation $\phi_{\alpha} : F [ x ] \to E$ at $\alpha \in E$ as follows. $$ \phi_{\alpha} ( f(x) ) : = a_{0} + a_{1} \alpha + \cdots + a_{n} \alpha^n = f (\alpha) $$ An $\alpha \in E$ satisfying $f( \alpha ) = 0$ is called a zero of $f(x)$.

Explanation

Evaluation

As a fact, $\phi_{\alpha}$ turns out to be a homomorphism.

The definition may feel too vague, so as a simple example, consider $\phi_{i} : \mathbb{R} [ x ] \to \mathbb{C}$. For instance, $\phi_{i} ( x^2 - 3 x + 2)$ is simply the result of substituting $i$ into $$ f(x) = x^2 - 3 x + 2 $$ which gives $$ i^2 - 3 i + 2= 1 -i2 \in \mathbb{C} $$ Of course, the fact that $\mathbb{R} \le \mathbb{C}$ poses no problem here.

The Motivation for Zeros

Thinking of the kernel, $\ker ( \phi_{\alpha} )$ would be the set of functions satisfying $f(\alpha) = 0$. Continuing the example above, the elements of $\ker ( \phi_{i} ) \subset \mathbb{R} [ x ]$ are the polynomial functions having $(x-i)$ as a factor.

Thus, calling $\alpha$ a zero of $f(x)$ when $f( \alpha ) = 0$ holds is natural beyond any doubt. Equivalently, we said that $\alpha$ is a zero of $f(x)$ if $\phi_{\alpha} ( f(x) ) = 0$.

The reason for going so far as to bring functions into the familiar notions of substitution and roots is precisely to rigorously define the concept of a “solution of an equation.” $$ f(x) = g(x) $$ For example, consider an equation like the one above. There is no reason we could not think of a set collecting such equations, but rather than collecting relations, it is far simpler and clearer to consider each side separately and deal with $f(x)$ and $g(x)$ themselves. If a set of equations $X$ were a collection of equations like the one above, we would have to be able to write something like $$ \left( f(x) = g(x) \right) \in X $$ but since transposing gives $$ f(x) = g(x) \iff f(x) - g(x) = 0 $$ anyway, there is no reason to leave the right-hand side messy and unconstrained. Letting the set $X$ follow a structure holding functions rather than equations is no different from collecting the equations whose right-hand side is $0$ and being interested in when they hold.

Following this expansion of thought, facts such as “a polynomial function with real coefficients can still have imaginary roots” will be abstracted and generalized.