# 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

$p(x)=c_0+c_1x+c_2x^2+...+c_nx^n$

### Like this:

Posted by John Armstrong | Algebra, Ring theory

## 7 Comments	»

1. [...] of Polynomials I When we consider a polynomial as a function, we’re particularly interested in those field elements so that . We call such an a [...]

Pingback by | July 30, 2008 | Reply

2. The finer points of the distinction between polynomials as
indeterminates and polynomials as functions do not seem to be well enough known, despite the fact that it is closely
related to some interesting mathematics. See for instance
the discussion in Section 2.1 (pages 3-7) of

http://math.uga.edu/~pete/4400ChevalleyWarning.pdf

This discussion is adapted (and expanded) from a similar
discussion in Ireland and Rosen’s text.

Comment by Pete L. Clark | July 30, 2008 | Reply

3. Pete, that’s a great reference, thanks. I’m planning on mentioning some of the stuff in that section, but not all of it. It’d be worth an interested reader’s time to check it out.

Comment by | July 30, 2008 | Reply

4. [...] . But there’s one big difference here: we have a relation that must satisfy. When we use the evaluation map we must find . And, of course, any polynomial which includes as a factor must evaluate to as [...]

Pingback by | August 7, 2008 | Reply

5. [...] going to want to talk about “evaluating” a power series like we did when we considered polynomials as functions. But when we try to map into our base field, we get a sequence of values and we ask questions about [...]

Pingback by | August 26, 2008 | Reply

6. [...] evaluation of power series is specified by two conditions: it should agree with evaluation of polynomials when we’ve got a power series that cuts off after a finite number of terms, and it should be [...]

Pingback by | August 27, 2008 | Reply

7. [...] it as a function in its own right. In a way, this is just extending what we did when we considered polynomials as functions and we can do everything algebraically with abstract “variables” as we can with [...]

Pingback by | October 2, 2009 | Reply

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