If $R$ is a (commutative) ring, we have for $f(x), g(x) \in R[x]$ and $b_m$ the leading coefficient of $g$ that there is $t \in \mathbb N$ and $q(x),r(x) \in R[x]$ such that $$b_m^t f(x) = q(x) g(x) + r(x)$$
This can be seen by induction on $\deg f$ and writing:
$$f_1(x) = b_m f(x) - a_n x^{n-m} g(x)$$
when $\deg f \ge m=\deg g$.
A corollary of this is:
If $R$ is a (commutative) ring and $f(x) \in R[x]$ and $a \in R$, then there exists $q(x)$ such that
$$f(x) = (x-a)q(x) + f(a)$$
by taking $g(x) = x - a$.
A corollary to this corollary is:
If $R$ is a ring and $f(x) \in R[x]$ and $a \in R$, then $(x-a) \mid f(x)$ iff $f(a) = 0$.
Now my question is: doesn't this have the following corollary?
If $D$ is a domain and $f(x) \in D[x]$ has degree $n$, then $f(x)$ has at most $n$ roots in $D$.
Suggested proof:
We prove by induction on $t$ that if $a_1,\ldots,a_t$ are distinct roots of $f(x)$, then $f(x)$ is divisible by $(x-a_1)(\cdots)(x-a_t)$. For $t=1$ this follows by the above corollary. Suppose now $a_1,\ldots,a_t$ are roots. By induction hypothesis, $f(x) = g(x) \prod_{i=1}^{t-1} (x-a_i)$.
Now $0= f(a_t) = g(a_t)(a_t - a_1)(\cdots)(a_t - a_{t-1})$. Since $D$ is a domain and each $a_t - a_i \neq 0$, we get $g(a_t) = 0$ and so $x-a_t$ divides $g(x)$. So $f(x) = h(x) \prod (x - a_i)$. So we are done.
But this shows any such $t$ is $\le n$. So this completes the proof.
