# The Unapologetic Mathematician

## Polynomials as Functions

When I set up the algebra of polynomials I was careful to specify that the element $X$ is not a “variable”, as in high school algebra. Why do I have to do that? What is the “variable” thing that we were all taught, then?

We’ve got the algebra of polynomials $\mathbb{F}[X]$ over the base field $\mathbb{F}$. Now I’m going to define a function $\mathrm{ev}:\mathbb{F}[X]\times\mathbb{F}\rightarrow\mathbb{F}$ called the “evaluation map”. We define $\mathrm{ev}(p,x)$ by first writing out $p$ in terms of the standard basis $p=c_0+c_1X+c_2X^2+...+c_nX^n$

Remember here that the sum must terminate after a finite number of basis elements. Then we just stick the field element $x$ in for $X$ to get an expression written out in the field $\mathbb{F}$ itself: $\mathrm{ev}(p,x)=c_0+c_1x+c_2x^2+...+c_nx^n$

Now the superscripts on each $x$ must be read as exponents. This defines a particular element of the field. If we keep the polynomial $p$ fixed and let $x$ range over $\mathbb{F}$ we get a function from $\mathbb{F}$ to itself, which we can abuse notation to write as $p(x)$. This is the notion of polynomial-as-function we were taught in high school.

But it’s actually more interesting to see what happens as we fix $x$ and let $p$ vary over all polynomials. The map $p\mapsto p(x)$ turns out to be a homomorphism of $\mathbb{F}$-algebras! Indeed, given polynomials $p=c_0+c_1X+c_2X^2+...+c_nX^n$ $q=d_0+d_1X+d_2X^2+...+d_nX^n$

(the top coefficients here may be zero, and all higher coefficients definitely are) and a field element $k$ we find \begin{aligned}\left[p+q\right](x)=(c_0+d_0)+(c_1+d_1)x+(c_2+d_2)x^2+...+(c_n+d_n)x^n\\=c_0+d_0+c_1x+d_1x+c_2x^2+d_2x^2+...+c_nx^n+d_nx^n\\=c_0+c_1x+c_2x^2+...+c_nx^n+d_0+d_1x+d_2x^2+...+d_nx^n\\=p(x)+q(x)\end{aligned} \begin{aligned}\left[kp\right](x)=(kc_0)+(kc_1)x+(kc_2)x^2+...+(kc_n)x^n\\=kc_0+kc_1x+kc_2x^2+...+kc_nx^n\\=k(c_0+c_1x+c_2x^2+...+c_nx^n)\\=kp(x)\end{aligned}

I’ll let you write out the verification that it also preserves multiplication.

In practice this “evaluation homomorphism” provides a nice way of extracting information about polynomials. And considering polynomials as functions provides another valuable slice of information. But we must still keep in mind the difference between the abstract polynomial $p=c_0+c_1X+c_2X^2+...+c_nX^n$

and the field element $p(x)=c_0+c_1x+c_2x^2+...+c_nx^n$

July 29, 2008 Posted by | Algebra, Ring theory | 7 Comments