Let $R$ be a commutative ring and $f\in R[x]$ be such that $f(r)=0$ for some $r\in R$. I wonder if $x-r$ divides $f(x)$ in this case.
I know this is true when $R$ is an Euclidean domain, since we can write $f(x)=(x-r)g(x)+c$. But is this true for a more general commutative ring $R$?
Plus, if this is false, is there a condition on $R$ weaker than the Euclidean domain that makes this true (PID, UFD)?
Just realize that $f(x)-f(r)$ is a linear combination of $x^k-r^k$'s and $x-r$ divides every $x^k-r^k$.
Indeed, over any commutative ring, if $m$ is a monic polynomial then for each $f$ there is a unique $q$ with $f-qm$ having degree less than that of $m$. Here for $m(x)=x-r$ you get $f(x)=(x-r)q(x)+c$ and $c$ must be $f(r)$.
Yes, this is true and can be stated slightly more abstractly. Here's a proof that doesn't use the division algorithm.
Lemma. Let $R$ be a unital commutative ring and $R[x]$ be the one-variable polynomial ring over $R$. For $\alpha \in R$, let \begin{align*} \varphi = \text{eval}_\alpha: R[x] &\to R\\ x &\mapsto \alpha \end{align*} be the evaluation homomorphism. Then $\ker(\varphi) = (x-\alpha)$ and the induced map $\overline{\varphi}: R[x]/(x-\alpha) \to R$ is an isomorphism.
Note that $\ker(\varphi) = \{f \in R[x] : f(\alpha) = 0\}$.
Proof. Certainly $(x - \alpha) \subseteq \ker(\varphi)$, so it remains to show the reverse inclusion. Given $f = \sum_{i=0}^n a_i x^i \in \ker(\varphi)$, then $$ 0 = f(\alpha) = \sum_i a_i \alpha^i \, . $$ We can rewrite $f$ as a polynomial in $x-\alpha$ by writing $x = (x - \alpha) + \alpha$ and using the binomial theorem: \begin{align*} f &= \sum_{i=0}^n a_i x^i = \sum_{i=0}^n a_i ((x - \alpha) + \alpha)^i = \sum_{i=0}^n a_i \sum_{j = 0}^i \binom{i}{j} (x-\alpha)^j \alpha^{i-j}\\ &= \sum_{i=0}^n a_i \sum_{j = 0}^n \binom{i}{j} (x-\alpha)^j \alpha^{i-j} = \sum_{j = 0}^n \underbrace{\left(\sum_{i=0}^n a_i \binom{i}{j} \alpha^{i-j}\right)}_{b_j} (x-\alpha)^j = \sum_{j=0}^n b_j (x - \alpha)^j \end{align*}
(Note that $\binom{i}{j} = 0$ for $j > i$, which allows us to change the upper limit of the inner sum from $i$ to $n$.) Since $f(\alpha) = 0$, then $$ b_0 = \sum_{i=0}^n a_i \binom{i}{0} \alpha^{i-0} = \sum_{i=0}^n a_i \alpha^i = f(\alpha) = 0 \, . $$ Since $b_0 = 0$, then every term in $f = \sum_{j=0}^n b_j (x - \alpha)^j$ has a factor of $x - \alpha$, so $f \in (x - \alpha)$.
This idea also works in multivariable polynomial rings; see here for the proof.
The most general formulation of a division algorithm that I know of goes as follows: Let $R$ be an arbitrary commutative ring, and let $R[t]$ be the ring of polynomials in one variable over $R$.
Terminology: For any $h= \sum_{j=0}^m c_j t^j \in R[t]$ where $c_m \neq 0$, we let $\deg(h)=m$ and call $c_mt^m$ the leading term of $h$ and $c_m$ its leading coefficient. If $h=0$ we do not assign it a degree.
Lemma: ("Division algorithm") If $f = \sum_{k=0}^n a_kt^k \in R[t]$ is such that its leading coefficient $a_n \in R^\times$, i.e. there is a $b \in R$ such that $b a_n=1$ then given any polynomial $g \in R[t]$ there are unique polynomials $q,r \in R[t]$ such that $$ g =q.f + r, \text{ where } r=0 \text{ or } \deg(r)<\deg(f) $$ Proof: We will prove the existence and uniqueness of the pair $(q,r)$ for each $g \in R[t]$ by induction on $\deg(g)$.
Step 1: If $h \in R[t]$ is nonzero then we have $\deg(h.f) = \deg(h)+\deg(f)\geq \deg(f)$, since if the leading term of $h$ is $c_mt^m$ then the leading term of $h.f$ is $a_nc_m.t^{m+n}$, since $a_nc_m =0$ implies $0=b(a_nc_m) = (ba_n)c_m = c_m$ while $c_m \neq 0$ by assumption.
Step 2: Now suppose that the existence and uniqueness of the pair $(q,r)$ is known for all $h$ with $\deg(h)<k$ and suppose $g \in R[t]$ has $\deg(g)=k$. If $g= q.f +r$, then if $q=0$ it follows $r=g$ and hence $\deg(g)<\deg(f)$. Conversely, if $q \neq 0$ then (as $r=0$ or $\deg(r)<\deg(f)$) the leading term of $q.f+r$ is $d a_n t^{\deg(q)+n}$ while that of $g$ is, say, $c_mt^m$. It follows that if $m<n$ we must have $q=0$, and hence the only possible pair $(q,r)$ is $(0,g)$, while if $m \geq n$ we must have $d a_n t^{\deg(q)+n} = c_mt^m$ so that the leading term of $q$ must be $c_m b t^{m-n}$
On the other hand, if $m \geq n$ then consider $$ g_1: g-c_mb.t^{m-n}f $$ Then since $\deg(g) =m = \deg(t^{m-n}.f)$ it follows that $\deg(g_1)\leq k$, while the coefficient of $t^k$ in $g_1$ is $c_k - c_kba_n = c_k - c_k =0$. It follows that $\deg(g_1)\leq k-1$, and hence by induction there are unique polynomials $(q_1,r_1)$ with $g_1 = q_1.f+r_1$. But then if we set $q=c_k.bt^{k-n} + q_1$ and $r=r_1$ it follows that $g = q.f+r$, and hence we have established the existence of pair $(q,r)$ for all $g\in R[t]$ of degree at most $k$.
Finally, to see that $(q,r)$ are unique, note that, as observed above, the leading term of any nonzero $q$ for which $g=q.f+r$ is necessarily equal to $c_m b t^{\deg(g)-n}$, and the inductive hypothesis ensures that $r$ and the lower order terms of $q$ are all unique, hence the pair $(q,r)$ is indeed unique.
