Number of Roots of a Polynomial is Bounded by its Degree?

Question

Most of the textbooks state that provided a nonzero field $F$, a nonzero polynomial $f\in F[x]$ of degree $n$ has at most $n$ distinct roots. I am wondering whether the word "distinct" can be removed? I guess the answer is yes, but I cannot come up with a nice proof.

Sorry for the confusion. The question above can be rephrased as "if $f$ has roots $\alpha_i$ of multiplicity $n_i$, then is it possible that $\sum n_i>n$?"

Here $\alpha$ is of multiplicity $k$ means $(x-\alpha)^k\mid f$ but $(x-\alpha)^{k+1}\nmid f$.

For commutative ring this is possible, e.g, $x^2-x\in\mathbb{Z}_6[x]$ has roots $\overline{0},\overline{1},\overline{3},\overline{4}$.

Answer 1

Suppose that $\alpha$ were not a root of g.

Then $\alpha$ is a simple root of f, which is a contradiction.

Answer 2

The key thing you need is that these are no non-trivial zero divisors, so that if $ab=0$ then either $a=0$ or $b=0$. So the formula for maximum number of roots works if you are working within a domain (and not just a field - the integers will do too).

Then you can use the polynomial division algorithm with $x-a$ to give, for any polynomial $p(x)$, that $$p(x)=(x-a)q(x)+r$$If you now make $a$ a root of $p$ and set $x=a$ you get $r=p(a)=0$ and $p(x)=(x-a)q(x)$. Then if $q(x)$ has a root $b$ you can take out a factor $x-b$ so you get $$p(x)=(x-a)^h(x-b)^t \dots (x-f)^us(x)$$ where $s$ has no roots, and may be a scalar.

Now, because there are no zero divisors, if $p(y)=0$ one of the factors must be zero. It can't be $s(y)$ ($s$ has no roots) so it must be one of the others, and so $y$ must be one of the roots you already know. Now equate degrees (or work the other way and use induction on degree).

In the case of integers modulo $6$ you have $2\times 3 \equiv 0$ so that $x(x-1)\equiv 0$ can be true even if $x$ is not $1$ or $2$.

Indeed we also have $x(x-1)\equiv (x-4)(x-3)$, which shows that factorisation is not unique.