# Show that if $R$ is an integral domain then a polynomial in $R[X]$ of degree $d$ can have at most $d$ roots

Show that if $R$ is an integral domain then a polynomial in $R[X]$ of degree $d$ can have at most $d$ roots.

Thoughts so far:

I feel like I might be missing something here. If $R$ is an integral domain, then so is $R[X]$. The factor theorem gives a correspondence between roots and factors. Clearly (is this clear?) a polynomial of degree $d$ can't have more than $d$ irreducible factors.

## 2 Answers

It is only mildly clear; you have to show that, when $R$ (and hence $R[x]$) is an integral domain, the degree function does in fact satisfy $$\text{deg}(fg)=\text{deg}(f)+\text{deg}(g).$$ Then you can argue that (by the factor theorem), if $a_1,\ldots,a_n$ were roots of $f$, then $f=(x-a_1)\cdots(x-a_n)g$ for some $g\in R[x]$, so that $\deg(f)=n+\deg(g)$ and $\deg(g)\geq0$, hence $\deg(f)\geq n$.

When $R$ is not an integral domain, we might have $\text{deg}(fg)<\text{deg}(f)+\text{deg}(g)$ (do you see why?)

• Thanks, that was very helpful. I can see why: take, for example, $\left(\mathbb{Z}/8\mathbb{Z}\right)[X]$, which isn't an integral domain. In this ring, $(4X)(2X) = 0$. – user938272 May 6 '11 at 23:30
• @user938272: Quite right. Glad to help :) – Zev Chonoles May 6 '11 at 23:31

Note $\rm\ a_i\ne a_j\ \Rightarrow\ x-a_i\$ are nonassocate primes in $\rm\:R[x],\:$ since $\rm\: R[x]/(x-a) \cong R\:$ is a domain.

Therefore $\rm\ \ x-a_1\ |\ f(x),\ \ldots\:,\: x-a_n\ |\ f(x)\ \ \Rightarrow\ \ (x-a_1)\cdots (x-a_n)\ |\ f(x)$

since LCM = product for nonassociate primes. But this is contra degree if $\rm\ n > deg\ f\:.$

Remark $\$ In fact a ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\$ has at most $\rm\ deg\ f\$ roots in $\rm\:D\:.\:$ For the simple proof see my post here, where I illustrate it constructively in $\rm\: \mathbb Z/m\:$ by showing that, $\:$ given any $\rm\:f(x)\:$ with more roots than its degree,$\:$ we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which try to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$.