[proofplan]
We establish a bijection between the set of $K$-homomorphisms $\sigma : K(\alpha) \to \bar{K}$ and the set of roots of the minimal polynomial $P_\alpha$ in $\bar{K}$. The map is $\sigma \mapsto \sigma(\alpha)$. We verify it is well-defined, injective, and surjective.
[/proofplan]
[step:Well-defined]
Let $\sigma : K(\alpha) \to \bar{K}$ be a $K$-homomorphism. We must show that $\sigma(\alpha)$ is a root of $P_\alpha$.
Since $P_\alpha(\alpha) = 0$ and $P_\alpha \in K[t]$, we can write $P_\alpha(t) = a_n t^n + \cdots + a_1 t + a_0$ with each $a_i \in K$. Applying $\sigma$:
\begin{align*}
\sigma(P_\alpha(\alpha)) &= \sigma(a_n \alpha^n + \cdots + a_1 \alpha + a_0) \\
&= a_n \, \sigma(\alpha)^n + \cdots + a_1 \, \sigma(\alpha) + a_0 \\
&= P_\alpha(\sigma(\alpha))
\end{align*}
where in the second equality we used that $\sigma$ is a ring homomorphism fixing $K$ pointwise. Since $P_\alpha(\alpha) = 0$, we obtain $P_\alpha(\sigma(\alpha)) = 0$, so $\sigma(\alpha)$ is indeed a root of $P_\alpha$ in $\bar{K}$.
[/step]
[guided]
The key point is that $\sigma$ fixes every coefficient of $P_\alpha$ because those coefficients live in $K$. A $K$-homomorphism is invisible to elements of $K$, so it commutes with evaluating any polynomial in $K[t]$.
[/guided]
[step:Injective]
Let $\sigma, \tau : K(\alpha) \to \bar{K}$ be $K$-homomorphisms with $\sigma(\alpha) = \tau(\alpha)$. We show $\sigma = \tau$.
Since $\alpha$ is algebraic over $K$ with minimal polynomial $P_\alpha$ of degree $n$, every element of $K(\alpha)$ can be written uniquely as
\begin{align*}
c_0 + c_1 \alpha + c_2 \alpha^2 + \cdots + c_{n-1} \alpha^{n-1}, \qquad c_i \in K.
\end{align*}
For any such element, both $\sigma$ and $\tau$ fix each $c_i$ and respect addition and multiplication, so
\begin{align*}
\sigma\!\left(\sum_{i=0}^{n-1} c_i \alpha^i\right) = \sum_{i=0}^{n-1} c_i \, \sigma(\alpha)^i = \sum_{i=0}^{n-1} c_i \, \tau(\alpha)^i = \tau\!\left(\sum_{i=0}^{n-1} c_i \alpha^i\right).
\end{align*}
Since every element of $K(\alpha)$ has this form, $\sigma = \tau$.
[/step]
[guided]
The crucial fact is that $K(\alpha) = K[\alpha]$, i.e., every element is a $K$-polynomial in $\alpha$ of degree less than $n$. Once you know where $\alpha$ goes, the entire homomorphism is forced.
[/guided]
[step:Surjective]
Let $\beta \in \bar{K}$ be a root of $P_\alpha$. We construct a $K$-homomorphism $\sigma : K(\alpha) \to \bar{K}$ with $\sigma(\alpha) = \beta$.
Consider the evaluation homomorphism $\mathrm{ev}_\beta : K[t] \to \bar{K}$ defined by $t \mapsto \beta$. Its kernel contains $P_\alpha$ since $P_\alpha(\beta) = 0$. Because $P_\alpha$ is irreducible over $K$ and hence generates a maximal ideal in $K[t]$, we have $\ker(\mathrm{ev}_\beta) = (P_\alpha)$. By the first isomorphism theorem, $\mathrm{ev}_\beta$ descends to an injective ring homomorphism
\begin{align*}
\bar{\sigma} : K[t]/(P_\alpha) \hookrightarrow \bar{K}, \qquad \overline{t} \mapsto \beta.
\end{align*}
Now the canonical isomorphism $K(\alpha) \cong K[t]/(P_\alpha)$ sending $\alpha \mapsto \overline{t}$ lets us define $\sigma : K(\alpha) \to \bar{K}$ as $\sigma = \bar{\sigma} \circ \varphi$, where $\varphi : K(\alpha) \xrightarrow{\sim} K[t]/(P_\alpha)$ is this isomorphism. Then $\sigma(\alpha) = \bar{\sigma}(\overline{t}) = \beta$, and $\sigma$ fixes $K$ since $\bar{\sigma}$ sends each constant polynomial to itself.
[/step]
[guided]
Surjectivity is the only part that requires a construction. The engine is the universal property of quotient rings: since $P_\alpha(\beta) = 0$, the map $t \mapsto \beta$ kills the ideal $(P_\alpha)$, so it factors through $K[t]/(P_\alpha)$. Irreducibility of $P_\alpha$ ensures the kernel is exactly $(P_\alpha)$, making the induced map injective.
[/guided]
[step:Conclusion]
The map $\sigma \mapsto \sigma(\alpha)$ from $\mathrm{Hom}_K(K(\alpha), \bar{K})$ to the set of roots of $P_\alpha$ in $\bar{K}$ is well-defined, injective, and surjective. Therefore it is a bijection, and in particular
\begin{align*}
\#\,\mathrm{Hom}_K(K(\alpha), \bar{K}) \;=\; \deg P_\alpha.
\end{align*}
[/step]
