# The Unapologetic Mathematician

## Roots of Polynomials II

This one might have to suffice for two days. The movers are almost done loading the truck here in New Orleans, and I’ll be taking today and tomorrow to drive to Maryland.

We can actually tease out more information from the factorization we constructed yesterday. Bur first we need a little definition.

Remember that when we set up the algebra of polynomials we noted that the coefficients have to be all zero after some finite number of them. Thus there must be a greatest nonzero coefficient $c_n$. The index $n$ corresponding to this coefficient we call the “degree” of the polynomial. If all the coefficients are zero — giving the zero polynomial — then a common convention is to assign it degree $-\infty$. This actually isn’t completely arbitrary, but the details won’t concern us until later.

So, armed with this information, look at how we constructed the factorization $p=(X-x)q$. We replaced each term $c_kX^k$ in $p$ with the term $c_k(X^k-x^k)$. Then we factored out $(X-x)$ from this term, giving $c_k(X^{k-1}+X^{k-2}x+...+Xx^{k-2}+x^{k-1})$. So the highest power of $X$ that shows up in this term (with a nonzero coefficient) is $c_kX^{k-1}$. And the highest power coming from all the terms of $p$ will be $c_nX^{n-1}$. The power $X^{n-1}$ shows up only once in the expression for $q$, so there’s no way for two such terms to add together and make its coefficient turn out to be zero, and no higher power of $X$ ever shows up at all. Thus the degree of $q$ is one less than that of $p$.

So what does this gain us? Well, each time we find a root we can factor out a term like $(X-x)$, which reduces the degree by ${1}$. So if $p$ has degree $n$ there can only be at most $n$ roots!

A nonzero constant polynomial $p=c_0$ has degree ${0}$, but it also has no roots! Perfect.

A linear polynomial $p=c_0+c_1X$ has degree ${1}$, and it has exactly one root: $-\frac{c_0}{c_1}$.

Now let’s assume that our statement is true for all polynomials of degree $n$: they have $n$ or fewer roots. Then given a polynomial $p$ of degree $n+1$ either $p$ has a root $x$ or it doesn’t. If it doesn’t, then we’re already done. If it does, then we can factor $p=(X-x)q$, where $p$ has degree $n$. But then $q$ can have at most $n$ roots, and thus $p$ can have at most $n+1$ roots!

A nice little corollary of this is that if our base field $\mathbb{F}$ is infinite (like it is for the most familiar examples) then only the zero polynomial can give us the zero function when we evaluate it at various field elements. That is, if $p(x)=0$ for all $x\in\mathbb{F}$, then $p=0$. This must be true because $p$ has an infinite number of roots, and so no finite degree polynomial can possibly have that many roots. The only possibility left is the zero polynomial.

Just to be clear, though, let’s look at this one counterexample. Think about the field $\mathbb{Z}_3$ we used when we talked about Set. The polynomial $p=X^3-X$ is not the zero polynomial, but $p(x)$ is the zero function. Indeed $p(0)=0^3-0=0$, $p(1)=1^3-1=0$, and $p(2)=2^3-2=8-2=6$, which is divisible by $3$, and so is the same as ${0}$ in this field.

July 31, 2008 - Posted by | Algebra, Polynomials, Ring theory