Total number of roots of a polynomial of degree $n$ over a ring $R$

Question

Let $R$ be a commutative ring with unity and $f$ be a polynomial of degree $n$ over $R$. Under what conditions on $R$, does $f$ has at most $n$ roots ?

I am asking because in $\mathbb{Z}/12\mathbb{Z}$, the polynomial $x^2-4$ has 4 roots $2,4,8,10$ since $X^2-4=(x-8)(x-4)=(x-10)(x-2)$.

The above example shows unique factorization is necessary. So I believe, $R$ should be a UFD as $R[x]$ will also be a UFD then.

Answer 1

Claim - $R$ should be an `integral domain' for $f$ to have at most $n$ roots.

Proof - Let's consider the simplest equation $x^2-m=0$ where $m$ is a non-zero element in $R$. Suppose it has more than 2 roots i.e. $$x^2-m=(x-a)(x-b)=(x-c)(x-d), \enspace a,b,c,d \in R.$$ Let $c \neq a,b. $ Then $(c-a)(c-b)=0.(c-d)=0$ which implies $R$ is not an integral domain.

Answer 2

Let $R$ be a commutative ring where each degree $n$ polynomial has at most $n$ roots ($n>0$).

Suppose $R$ is not a domain; then there are $a,b\in R$, both nonzero, with $ab=0$. Consider now $$ f(x)=(x-a)(x-b)=x^2-(a+b)x+ab=x^2-(a+b)x $$ This degree two polynomial has at least three roots, provided $a\ne b$. Contradiction.

This leaves open the case where in the ring it holds that $ab=0$ if and only if $a=b$, when $a\ne0$ is a zero divisor. But then the polynomial $x^2$ has all zero divisors as roots. If there's more than one nonzero zero divisor we are done. Thus we reduce to the case the ring has a single nonzero zero divisor $a$. Since $2a$ is clearly a zero divisor and $2a\ne a$, we conclude $2a=0$. Therefore $2=0$ in $R$ or $a=2$ (identifying $2$ with $2\cdot 1$).

In the former case the characteristic of $R$ is $2$, in the latter case the characteristic is $4$, because $a^2=0$ by assumption.

Suppose the characteristic of $R$ is $4$. Then $2x-2$ has two distinct roots, namely $1$ and $-1$: contradiction.

Thus we remain with the case when the characteristic is $2$ and the ring has exactly one nilpotent element $a\ne0$ with $a^2=0$ and no other nonzero zero divisor. The nilradical of $R$ is then $\{0,a\}$. Hence, for every $r\in R$, we have $ar=0$ or $ar=a$. By assumption, $ar=0$ can only hold for $r=a$ or $r=0$. Take $r\in R$, $r\ne 0$ and $r\ne a$. Then $ar=a$, so $a(r-1)=0$. Therefore $r-1=0$ or $r-1=a$.

Then the ring is $R=\{0,1,a,1+a\}\cong\mathbb{F}_2[x]/(x^2)$. This should be the only counterexample to $R$ not being a domain.

Of course, for a domain the statement holds.