An $n$ degree polynomial with more than $n$ roots?

Question

I know that a polynomial of degree $n$ over a field has at most $n$ roots (even counted with multiplicity).

My question is how this works with polynomials over integral domains.

For example, $f(x) = x^2 + x + 1 \in \mathbb{Z}_{4}[x]$ has no roots.

$f(x) = x^3 + x - 1\in \mathbb{Z}_9[x]$ does not have more than $3$ roots.

From just playing with example like these it seems that even an $n$ degree polynomial over an integral domain has at most $n$ roots. Is that correct?

Answer 1

An integral domain $D$ embeds in its field of fractions $F$ say, so a polynomial can't have more roots in $D$ than it does in $F$. If you don't require multiplication to be commutative or if you allow zero divisors then a polynomial of degree $n$ can have more than $n$ roots. E.g., in the quaternions (which aren't commutative), the quadratic $x^2 + 1$ has infinitely many roots; in the ring $\Bbb{Z} \times \Bbb{Z}$ (which has zero divisors), the linear equation $(0, 1)x = (0, 1)$ also has infinitely many roots.

Answer 2

That's correct. It follows from the fact that every integral domain can be included in a field (and, of course, from the fact that you stated about fields).