[proofplan]
We construct the evaluation homomorphism $\mathrm{ev}_\alpha \colon K[t] \to L$ and identify its kernel as $(P_\alpha)$ and its image as $K[\alpha]$. The First Isomorphism Theorem then gives $K[\alpha] \cong K[t]/(P_\alpha)$, which is a field because $P_\alpha$ is irreducible over $K$. Since $K[\alpha]$ is a field containing $K$ and $\alpha$, it equals $K(\alpha)$. Finally, the division algorithm in $K[t]$ shows that $\{1, \alpha, \dots, \alpha^{n-1}\}$ is a $K$-basis for $K(\alpha)$, so $[K(\alpha):K] = n$.
[/proofplan]
[step:The evaluation homomorphism and its kernel]
Define the evaluation map
\begin{align*}
\mathrm{ev}_\alpha \colon K[t] &\to L, \\
f(t) &\mapsto f(\alpha).
\end{align*}
This is a ring homomorphism: $\mathrm{ev}_\alpha(f + g) = f(\alpha) + g(\alpha)$ and $\mathrm{ev}_\alpha(fg) = f(\alpha)g(\alpha)$ follow directly from the ring operations in $L$. Its image is
\begin{align*}
\mathrm{im}(\mathrm{ev}_\alpha) = \{ f(\alpha) : f \in K[t] \} = K[\alpha],
\end{align*}
the smallest subring of $L$ containing $K$ and $\alpha$.
Since $\alpha$ is algebraic over $K$, the kernel $\ker(\mathrm{ev}_\alpha) = \{ f \in K[t] : f(\alpha) = 0 \}$ is nonzero. Now $K[t]$ is a principal ideal domain, so $\ker(\mathrm{ev}_\alpha) = (g)$ for some monic polynomial $g \in K[t]$.
[guided]
We claim $g = P_\alpha$. By definition $P_\alpha$ is the monic polynomial of least degree in $K[t]$ satisfying $P_\alpha(\alpha) = 0$, so $P_\alpha \in \ker(\mathrm{ev}_\alpha) = (g)$, which gives $g \mid P_\alpha$. Conversely, $g(\alpha) = 0$ and $g$ is monic, so the minimality of $\deg P_\alpha$ forces $\deg g \geq \deg P_\alpha$. A divisibility relation $g \mid P_\alpha$ together with $\deg g \geq \deg P_\alpha$ (with both polynomials monic) yields $g = P_\alpha$.
[/guided]
Therefore $\ker(\mathrm{ev}_\alpha) = (P_\alpha)$.
[/step]
[step:First Isomorphism Theorem and K(α) is a field]
By the First Isomorphism Theorem for rings,
\begin{align*}
K[\alpha] = \mathrm{im}(\mathrm{ev}_\alpha) \cong K[t] / \ker(\mathrm{ev}_\alpha) = K[t]/(P_\alpha).
\end{align*}
[guided]
We show $K[t]/(P_\alpha)$ is a field. Since $P_\alpha$ is the minimal polynomial of $\alpha$ over $K$, it is irreducible in $K[t]$: if $P_\alpha = gh$ with $\deg g, \deg h < \deg P_\alpha$, then $0 = P_\alpha(\alpha) = g(\alpha)h(\alpha)$, so $g(\alpha) = 0$ or $h(\alpha) = 0$ (as $L$ is a field and hence an integral domain), contradicting the minimality of $\deg P_\alpha$. Now $K[t]$ is a PID, and in a PID an ideal generated by an irreducible element is maximal: if $(P_\alpha) \subseteq (d)$ then $d \mid P_\alpha$, so $d$ is a unit or an associate of $P_\alpha$, giving $(d) = K[t]$ or $(d) = (P_\alpha)$. Therefore $(P_\alpha)$ is a maximal ideal and $K[t]/(P_\alpha)$ is a field.
[/guided]
Since $K[\alpha] \cong K[t]/(P_\alpha)$ is a field, and $K[\alpha]$ is a subring of $L$ containing $K$ and $\alpha$, it is already closed under taking inverses. But $K(\alpha)$ is by definition the smallest subfield of $L$ containing $K$ and $\alpha$, so $K(\alpha) \subseteq K[\alpha]$. The reverse inclusion $K[\alpha] \subseteq K(\alpha)$ is immediate (every polynomial expression in $\alpha$ lies in any field containing $K$ and $\alpha$). Therefore
\begin{align*}
K(\alpha) = K[\alpha] \cong K[t]/(P_\alpha).
\end{align*}
[/step]
[step:Basis from the division algorithm]
Let $n = \deg P_\alpha$. We show that $\{1, \alpha, \dots, \alpha^{n-1}\}$ is a $K$-basis for $K(\alpha)$.
**Spanning.** Let $x \in K(\alpha) = K[\alpha]$. Then $x = f(\alpha)$ for some $f \in K[t]$. By the division algorithm in $K[t]$, write $f = qP_\alpha + r$ with $q, r \in K[t]$ and $\deg r < n$. Evaluating at $\alpha$:
\begin{align*}
f(\alpha) = q(\alpha) P_\alpha(\alpha) + r(\alpha) = r(\alpha),
\end{align*}
since $P_\alpha(\alpha) = 0$. Writing $r = c_0 + c_1 t + \cdots + c_{n-1} t^{n-1}$ with $c_i \in K$, we get
\begin{align*}
x = c_0 + c_1 \alpha + \cdots + c_{n-1} \alpha^{n-1},
\end{align*}
so $\{1, \alpha, \dots, \alpha^{n-1}\}$ spans $K(\alpha)$ over $K$.
**Linear independence.** Suppose $c_0 + c_1 \alpha + \cdots + c_{n-1} \alpha^{n-1} = 0$ for some $c_i \in K$. Then the polynomial $h(t) = c_0 + c_1 t + \cdots + c_{n-1} t^{n-1}$ satisfies $h(\alpha) = 0$, i.e., $h \in \ker(\mathrm{ev}_\alpha) = (P_\alpha)$. So $P_\alpha \mid h$. But $\deg h \leq n - 1 < n = \deg P_\alpha$, which is impossible unless $h = 0$. Therefore $c_0 = c_1 = \cdots = c_{n-1} = 0$.
Thus $\{1, \alpha, \dots, \alpha^{n-1}\}$ is a $K$-basis for $K(\alpha)$, and
\begin{align*}
[K(\alpha) : K] = n = \deg P_\alpha. \qquad \blacksquare
\end{align*}
[/step]
