[proofplan]
The Remainder Theorem is a one-line consequence of the [Polynomial Division Algorithm](/theorems/1706) applied to the monic linear divisor $g(X) = X - \alpha$. Division produces $f = (X - \alpha)f_1 + r$ with $\deg(r) < 1$, forcing $r$ to be a constant $c \in R$. Evaluating at $X = \alpha$ collapses the first term and identifies the constant as $c = f(\alpha)$. The equivalence "$\alpha$ is a root iff $(X - \alpha) \mid f$" is then immediate from the formula.
[/proofplan]
[step:Apply polynomial division with divisor $X - \alpha$]
Consider the polynomial $g(X) := X - \alpha \in R[X]$. Its leading coefficient is $1 \in R$, which is a unit in any commutative ring with identity. The hypothesis of the [Polynomial Division Algorithm](/theorems/1706) — that the leading coefficient of the divisor is a unit — is therefore satisfied. Applying the algorithm to the pair $(f, g)$ produces unique polynomials $f_1, r \in R[X]$ with
\begin{align*}
f(X) &= (X - \alpha)\,f_1(X) + r(X), & \deg(r) &< \deg(g) = 1.
\end{align*}
Since $\deg(r) < 1$, the polynomial $r$ is a constant: there exists $c \in R$ with $r(X) = c$ (the zero polynomial is included under the convention $\deg(0) < 0 < 1$, corresponding to $c = 0$).
[guided]
We want to divide $f$ by $X - \alpha$, and the natural tool is the [Polynomial Division Algorithm](/theorems/1706). That theorem requires the divisor's leading coefficient to be a unit. For us the divisor is $g(X) = X - \alpha$, whose leading coefficient is $1$ — and $1$ is a unit in any ring with identity (its inverse is itself). So the hypothesis is satisfied automatically; in particular we do **not** need $R$ to be a field or even an integral domain. The theorem gives unique $f_1, r \in R[X]$ with
\begin{align*}
f(X) &= (X - \alpha)\,f_1(X) + r(X), & \deg(r) &< \deg(g) = 1.
\end{align*}
Why does the degree bound force $r$ to be a constant? The polynomials of degree $< 1$ are exactly the constants (together with the zero polynomial, to which we assign degree $-\infty$ or $<0$). So $r(X) = c$ for some $c \in R$. This is the only step where the choice of divisor $X - \alpha$ pays off: a higher-degree divisor would leave a remainder of positive degree, and the next step — evaluation at $\alpha$ — would not isolate a single value of $f$.
[/guided]
[/step]
[step:Evaluate at $X = \alpha$ to identify the constant remainder]
Evaluation at $\alpha$ is a ring homomorphism $\mathrm{ev}_\alpha : R[X] \to R$, $p(X) \mapsto p(\alpha)$. Applying $\mathrm{ev}_\alpha$ to the identity $f(X) = (X - \alpha)f_1(X) + c$ gives
\begin{align*}
f(\alpha) &= (\alpha - \alpha)\,f_1(\alpha) + c = 0 \cdot f_1(\alpha) + c = c.
\end{align*}
Therefore $c = f(\alpha)$, and substituting back:
\begin{align*}
f(X) = (X - \alpha)\,f_1(X) + f(\alpha),
\end{align*}
which is the stated identity.
[/step]
[step:Deduce that $\alpha$ is a root iff $(X - \alpha) \mid f(X)$]
From the identity $f(X) = (X - \alpha)f_1(X) + f(\alpha)$ we read off both directions of the divisibility equivalence.
($\Rightarrow$) Suppose $f(\alpha) = 0$. Then $f(X) = (X - \alpha)f_1(X)$, so $(X - \alpha) \mid f(X)$ in $R[X]$.
($\Leftarrow$) Suppose $(X - \alpha) \mid f(X)$, i.e. $f(X) = (X - \alpha)\,h(X)$ for some $h \in R[X]$. Evaluating at $X = \alpha$ gives $f(\alpha) = (\alpha - \alpha)\,h(\alpha) = 0$, so $\alpha$ is a root of $f$.
[guided]
Both directions drop out of the identity $f(X) = (X - \alpha)f_1(X) + f(\alpha)$.
For the forward direction, assume $\alpha$ is a root, i.e. $f(\alpha) = 0$. Then the constant term in the identity vanishes and we are left with $f(X) = (X - \alpha)f_1(X)$, which is exactly the statement $(X - \alpha) \mid f$ in $R[X]$.
For the reverse direction, assume $(X - \alpha) \mid f(X)$, so $f(X) = (X - \alpha)h(X)$ for some $h \in R[X]$. Evaluating at $X = \alpha$ via the homomorphism $\mathrm{ev}_\alpha$ gives $f(\alpha) = 0 \cdot h(\alpha) = 0$. Note that this direction is completely general — it needs no hypothesis on $R$ at all — because we never cancelled any factor. In the next theorem ([Root Bound for Integral Domains](/theorems/1708)) we will have to cancel, and that is precisely where the integral-domain hypothesis becomes indispensable.
[/guided]
[/step]
