[proofplan]
Both parts exploit the structure of the vanishing ideal $I_\alpha := \{f \in K[t] : f(\alpha) = 0\}$. Since $K[t]$ is a principal ideal domain, $I_\alpha = \langle P_\alpha \rangle$, so a polynomial $f \in K[t]$ satisfies $f(\alpha) = 0$ precisely when $P_\alpha \mid f$ — this gives part (2) immediately. For part (1), we argue by contradiction: if $P_\alpha = gh$ were a nontrivial factorisation, then $g(\alpha)h(\alpha) = 0$ in the field $L$, and the absence of zero divisors forces one factor to vanish at $\alpha$, producing an element of $I_\alpha$ with degree strictly less than $\deg P_\alpha$ — contradicting the minimality of $P_\alpha$.
[/proofplan]
[step:Establish the divisibility characterisation $f(\alpha) = 0 \Leftrightarrow P_\alpha \mid f$]
Let $L/K$ be a field extension and let $\alpha \in L$ be algebraic over $K$. Recall that the vanishing ideal of $\alpha$ over $K$ is the set
\begin{align*}
I_\alpha := \{f \in K[t] : f(\alpha) = 0\}.
\end{align*}
Since $K[t]$ is a principal ideal domain (it is a Euclidean domain with the degree function as the Euclidean norm), $I_\alpha$ is a principal ideal. The minimal polynomial $P_\alpha$ is defined as the unique monic generator of $I_\alpha$, so
\begin{align*}
I_\alpha = \langle P_\alpha \rangle = \{P_\alpha \cdot q : q \in K[t]\}.
\end{align*}
For any $f \in K[t]$, we therefore have
\begin{align*}
f(\alpha) = 0 \;\Longleftrightarrow\; f \in I_\alpha \;\Longleftrightarrow\; f \in \langle P_\alpha \rangle \;\Longleftrightarrow\; P_\alpha \mid f.
\end{align*}
This proves part (2).
[/step]
[step:Prove irreducibility of $P_\alpha$ by exploiting the integral domain structure of $L$]
Suppose, for contradiction, that $P_\alpha$ is reducible in $K[t]$. Then there exist polynomials $g, h \in K[t]$ with
\begin{align*}
P_\alpha = g \cdot h, \qquad 1 \le \deg g < \deg P_\alpha, \qquad 1 \le \deg h < \deg P_\alpha.
\end{align*}
Evaluating at $\alpha$ gives
\begin{align*}
g(\alpha) \cdot h(\alpha) = P_\alpha(\alpha) = 0,
\end{align*}
where the right-hand side vanishes because $P_\alpha \in I_\alpha$.
Since $L$ is a field, it is an integral domain: the product of two elements of $L$ is zero only if at least one factor is zero. Therefore
\begin{align*}
g(\alpha) = 0 \quad \text{or} \quad h(\alpha) = 0.
\end{align*}
By part (2), whichever factor vanishes at $\alpha$ lies in $I_\alpha = \langle P_\alpha \rangle$. Suppose without loss of generality that $g(\alpha) = 0$. Then $P_\alpha \mid g$, which requires $\deg P_\alpha \le \deg g$. This contradicts the strict inequality $\deg g < \deg P_\alpha$ from the factorisation. The same contradiction arises if $h(\alpha) = 0$ instead, since $\deg h < \deg P_\alpha$.
Therefore no such factorisation exists, and $P_\alpha$ is irreducible in $K[t]$.
[guided]
The question is: why should the monic polynomial of least degree in $I_\alpha$ be irreducible? The answer rests entirely on the fact that $L$ has no zero divisors.
Suppose, for contradiction, that $P_\alpha$ factors nontrivially in $K[t]$: there exist $g, h \in K[t]$ with
\begin{align*}
P_\alpha = g \cdot h, \qquad 1 \le \deg g < \deg P_\alpha, \qquad 1 \le \deg h < \deg P_\alpha.
\end{align*}
(Both factors must have degree at least $1$, since a factorisation where one factor is a nonzero constant would not contradict irreducibility — it would make $P_\alpha$ an associate of the other factor, and since $P_\alpha$ is monic, the only constant that can appear is $1$.)
Evaluating at $\alpha$ gives
\begin{align*}
g(\alpha) \cdot h(\alpha) = P_\alpha(\alpha) = 0,
\end{align*}
where $P_\alpha(\alpha) = 0$ because $P_\alpha$ belongs to the vanishing ideal $I_\alpha$ by definition. Now, $g(\alpha)$ and $h(\alpha)$ are elements of the field $L$. A field has no zero divisors: if a product of two elements is zero, at least one of the factors must be zero. This is the step where the field structure of $L$ is consumed — the conclusion would fail in a ring with zero divisors.
So $g(\alpha) = 0$ or $h(\alpha) = 0$. Suppose $g(\alpha) = 0$. Then $g \in I_\alpha = \langle P_\alpha \rangle$, which means $P_\alpha \mid g$. But divisibility forces $\deg P_\alpha \le \deg g$, contradicting $\deg g < \deg P_\alpha$. The argument is identical if $h(\alpha) = 0$ instead.
Why is the "integral domain" hypothesis essential, not merely convenient? Consider the ring $\mathbb{Z}/6\mathbb{Z}$, which has zero divisors ($2 \cdot 3 = 0$). Over this ring, the polynomial $t^2 - 1 = (t-1)(t+1)$ vanishes at $t = 3$ even though neither $t - 1$ nor $t + 1$ vanishes at $t = 3$ (since $3 - 1 = 2 \ne 0$ and $3 + 1 = 4 \ne 0$ in $\mathbb{Z}/6\mathbb{Z}$). The factorisation argument breaks down precisely because the product of two nonzero elements can vanish.
This irreducibility result is the foundation for the [Structure of Simple Algebraic Extensions](/theorems/1251): it guarantees that $\langle P_\alpha \rangle$ is a maximal ideal of $K[t]$ (since in a PID, an ideal generated by an irreducible element is maximal), so $K[t]/\langle P_\alpha \rangle$ is a field, and therefore $K[\alpha] \cong K[t]/\langle P_\alpha \rangle$ is already a field — no localisation or fraction field construction is needed.
[/guided]
[/step]
