6

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.

user938272
  • 983

2 Answers2

6

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?)

Zev Chonoles
  • 132,937
  • 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 06 '11 at 23:30
  • @user938272: Quite right. Glad to help :) – Zev Chonoles May 06 '11 at 23:31
  • The factor theorem says that if $a$ is a root of $f$ then $f=(x-a)q$. But then if $b$ is another root of $f$ I know that $(x-b)r=(x-a)q$, how do I infer that $(x-b)$ divides $q$ and not $(x-a)$? I guess here we need some primality of affine polynomials and precisely where the integral domain assumption comes in. – Oskar Limka Mar 31 '23 at 11:29
  • Could you please clarify what does it mean for f not to be in an integral domain? Example please? Thanks (All I know about abstract algebra is that I have no idea what it is!). – NoChance May 04 '23 at 08:37
6

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

Hence $\ 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 $\ n > \deg f$.

Remark $ $ When $\rm\:R\:$ is not a domain this argument fails since then $\: x\! -\! c\:$ is no longer prime, e.g. $\ (x\!-\!1)(x\!+\!1) = (x\!-\!3)(x\!+\!3)\ $ over $\rm\:\mathbb Z/8,\:$ so none of the factors are prime. Generally we have: $\, D\,$ is a domain $\!\iff\!$ every polynomial $\,f(x)\neq 0\in D[x]\, $ has at most $\, \deg f $ roots in $\,D.\,$ For the simple proof see here, where I illustrate it constructively in $\,\Bbb Z/m\, $ by showing that given any $\,f(x)\,$ with more roots than its degree,$\:$ we can quickly compute a nontrivial factor of $\,m\,$ via a simple gcd computation. The quadratic case of this result is at the heart of many integer factorization algorithms, which try to factor $\:m\:$ by searching for a square root of $1$ that's $\not\equiv \pm1\,$ in $\: \mathbb Z/m$.

Bill Dubuque
  • 282,220