[proofplan] We analyse the evaluation homomorphism $\operatorname{ev}_\alpha : K[t] \to L$, $f \mapsto f(\alpha)$. Its image is the subring $K[\alpha] = \{g(\alpha) : g \in K[t]\}$, and its kernel is the principal ideal $\langle P_\alpha \rangle$ generated by the minimal polynomial. The first isomorphism theorem for rings gives $K[\alpha] \cong K[t]/\langle P_\alpha \rangle$. Since $P_\alpha$ is irreducible over $K$ and $K[t]$ is a principal ideal domain, $\langle P_\alpha \rangle$ is a maximal ideal, so $K[t]/\langle P_\alpha \rangle$ is a field. Therefore $K[\alpha]$ is a field, which forces $K[\alpha] = K(\alpha)$. This establishes parts (1) and (3). For part (2), the isomorphism $K(\alpha) \cong K[t]/\langle P_\alpha \rangle$ reduces the basis question to the observation that every coset in $K[t]/\langle P_\alpha \rangle$ has a unique representative of degree less than $d = \deg P_\alpha$, so $\{1, \alpha, \ldots, \alpha^{d-1}\}$ is a $K$-basis. [/proofplan]

[step:Identify the image and kernel of the evaluation homomorphism $\operatorname{ev}_\alpha$]Let $L$ be a field extension of $K$ and let $\alpha \in L$ be algebraic over $K$ with minimal polynomial $P_\alpha \in K[t]$ of degree $d$. Define the evaluation homomorphism \begin{align*} \operatorname{ev}_\alpha : K[t] &\to L \\ f &\mapsto f(\alpha). \end{align*} This is a ring homomorphism from the polynomial ring $K[t]$ to the field $L$, since evaluation preserves addition and multiplication: $\operatorname{ev}_\alpha(f + g) = f(\alpha) + g(\alpha) = \operatorname{ev}_\alpha(f) + \operatorname{ev}_\alpha(g)$ and $\operatorname{ev}_\alpha(fg) = f(\alpha)g(\alpha) = \operatorname{ev}_\alpha(f)\operatorname{ev}_\alpha(g)$, and $\operatorname{ev}_\alpha(1) = 1$. **Image.** The image of $\operatorname{ev}_\alpha$ is the set \begin{align*} \operatorname{im}(\operatorname{ev}_\alpha) = \{f(\alpha) : f \in K[t]\} =: K[\alpha]. \end{align*} This is the smallest subring of $L$ containing $K$ and $\alpha$: it contains all polynomial expressions in $\alpha$ with coefficients in $K$. **Kernel.** A polynomial $f \in K[t]$ lies in $\ker(\operatorname{ev}_\alpha)$ if and only if $f(\alpha) = 0$. The minimal polynomial $P_\alpha$ is defined as the unique monic polynomial of least degree in $K[t]$ satisfying $P_\alpha(\alpha) = 0$. By the division algorithm in $K[t]$, for any $f \in K[t]$ there exist unique $q, r \in K[t]$ with $\deg r < \deg P_\alpha$ (or $r = 0$) such that $f = P_\alpha \cdot q + r$. Evaluating at $\alpha$: $f(\alpha) = P_\alpha(\alpha) \cdot q(\alpha) + r(\alpha) = r(\alpha)$. Therefore $f(\alpha) = 0$ if and only if $r(\alpha) = 0$. Since $\deg r < \deg P_\alpha = d$ and $P_\alpha$ has minimal degree among nonzero polynomials in $K[t]$ vanishing at $\alpha$, the condition $r(\alpha) = 0$ forces $r = 0$. Hence $f \in \ker(\operatorname{ev}_\alpha)$ if and only if $P_\alpha$ divides $f$, giving \begin{align*} \ker(\operatorname{ev}_\alpha) = \langle P_\alpha \rangle. \end{align*}[/step]

[guided]The proof begins by studying the natural map that connects the abstract polynomial ring $K[t]$ to the concrete extension field $L$. The evaluation homomorphism $\operatorname{ev}_\alpha$ sends a polynomial $f \in K[t]$ to the element $f(\alpha) \in L$ obtained by substituting $\alpha$ for the formal variable $t$. Why is $\operatorname{ev}_\alpha$ a ring homomorphism? This follows from the ring axioms in $L$: addition and multiplication in $L$ are compatible with the operations in $K[t]$ because substitution distributes over both. Explicitly: if $f = \sum_i a_i t^i$ and $g = \sum_j b_j t^j$, then $f(\alpha) + g(\alpha) = \sum_i a_i \alpha^i + \sum_j b_j \alpha^j = (f+g)(\alpha)$, and a similar computation holds for multiplication. The image $K[\alpha] = \{f(\alpha) : f \in K[t]\}$ consists of all "polynomial expressions" in $\alpha$ with coefficients in $K$. At this stage, $K[\alpha]$ is a subring of $L$ — we do not yet know it is a field. It contains $K$ (by evaluating constant polynomials) and $\alpha$ (by evaluating $t$). For the kernel, we need to determine which polynomials $f \in K[t]$ satisfy $f(\alpha) = 0$. The claim is that these are precisely the multiples of $P_\alpha$. The inclusion $\langle P_\alpha \rangle \subset \ker(\operatorname{ev}_\alpha)$ is immediate: if $f = P_\alpha \cdot q$, then $f(\alpha) = P_\alpha(\alpha) \cdot q(\alpha) = 0 \cdot q(\alpha) = 0$. For the reverse inclusion, let $f \in \ker(\operatorname{ev}_\alpha)$, so $f(\alpha) = 0$. Apply the division algorithm in $K[t]$ (which is a Euclidean domain with the degree function as the Euclidean norm): write $f = P_\alpha \cdot q + r$ where $q, r \in K[t]$ and either $r = 0$ or $\deg r < \deg P_\alpha = d$. Evaluating at $\alpha$: $0 = f(\alpha) = P_\alpha(\alpha) q(\alpha) + r(\alpha) = 0 + r(\alpha) = r(\alpha)$. So $r(\alpha) = 0$ and $\deg r < d$. But $P_\alpha$ is the nonzero polynomial of minimal degree vanishing at $\alpha$. If $r \neq 0$, then $r$ would be a nonzero polynomial of degree less than $d$ with $r(\alpha) = 0$, contradicting the minimality of $P_\alpha$. Therefore $r = 0$, and $f = P_\alpha \cdot q \in \langle P_\alpha \rangle$. This is the point where the definition of $P_\alpha$ as the *minimal* polynomial is essential: the minimality of its degree is what forces $r = 0$.[/guided]

[step:Apply the first isomorphism theorem to obtain $K[\alpha]\cong K[t]/\langle P_\alpha \rangle$] By the first isomorphism theorem for rings, the evaluation homomorphism $\operatorname{ev}_\alpha : K[t] \to L$ induces an isomorphism of rings \begin{align*} \overline{\operatorname{ev}_\alpha} : K[t]/\langle P_\alpha \rangle &\xrightarrow{\;\sim\;} K[\alpha] \\ f + \langle P_\alpha \rangle &\mapsto f(\alpha). \end{align*} This is a well-defined ring isomorphism: the first isomorphism theorem requires that $\operatorname{ev}_\alpha$ is a surjective ring homomorphism onto its image $K[\alpha]$ with kernel $\langle P_\alpha \rangle$, both of which were established in the previous step.[/step]

[guided]The first isomorphism theorem for rings states: if $\varphi : R \to S$ is a ring homomorphism, then $R / \ker(\varphi) \cong \operatorname{im}(\varphi)$ via the map $r + \ker(\varphi) \mapsto \varphi(r)$. We apply this with $R = K[t]$, $S = L$, and $\varphi = \operatorname{ev}_\alpha$. We have $\ker(\operatorname{ev}_\alpha) = \langle P_\alpha \rangle$ (from the previous step) and $\operatorname{im}(\operatorname{ev}_\alpha) = K[\alpha]$ (by definition). The theorem therefore gives \begin{align*} K[t] / \langle P_\alpha \rangle \cong K[\alpha], \end{align*} with the isomorphism sending $f + \langle P_\alpha \rangle$ to $f(\alpha)$. What has this accomplished? We have related the concrete subring $K[\alpha] \subset L$ to the abstract quotient ring $K[t]/\langle P_\alpha \rangle$. The quotient ring is built entirely from $K$ and the polynomial $P_\alpha$, without reference to the ambient field $L$ — so this isomorphism reveals that the algebraic structure of $K[\alpha]$ depends only on $K$ and $P_\alpha$.[/guided]

[step:Show $K[t]/\langle P_\alpha \rangle$ is a field using irreducibility of $P_\alpha$] The minimal polynomial $P_\alpha$ is irreducible over $K$: if $P_\alpha = gh$ with $g, h \in K[t]$, then $0 = P_\alpha(\alpha) = g(\alpha) h(\alpha)$. Since $L$ is a field and hence an integral domain, either $g(\alpha) = 0$ or $h(\alpha) = 0$. If $g(\alpha) = 0$, then $g \in \ker(\operatorname{ev}_\alpha) = \langle P_\alpha \rangle$, so $P_\alpha$ divides $g$, forcing $\deg g \ge d$ and hence $\deg h = 0$. By symmetry, $h(\alpha) = 0$ forces $\deg g = 0$. In either case, one factor is a unit. Hence $P_\alpha$ is irreducible over $K$. Since $K[t]$ is a principal ideal domain (as $K$ is a field), an element $p \in K[t]$ generates a maximal ideal if and only if $p$ is irreducible (and nonzero, non-unit). Since $P_\alpha$ is irreducible with $\deg P_\alpha = d \ge 1$, the ideal $\langle P_\alpha \rangle$ is maximal in $K[t]$. A quotient ring $R/\mathfrak{m}$ of a commutative ring $R$ with identity is a field if and only if $\mathfrak{m}$ is a maximal ideal. Since $\langle P_\alpha \rangle$ is a maximal ideal of $K[t]$, the quotient \begin{align*} K[t]/\langle P_\alpha \rangle \text{ is a field.} \end{align*}[/step]

[guided]This step establishes the crucial algebraic fact: the quotient $K[t]/\langle P_\alpha \rangle$ is not merely a ring but a field. The argument chains together three standard results from commutative algebra. **Step A: $P_\alpha$ is irreducible over $K$.** Why must the minimal polynomial be irreducible? Suppose $P_\alpha = gh$ where $g, h \in K[t]$ with $\deg g \ge 1$ and $\deg h \ge 1$ (so neither is a unit). Evaluating at $\alpha$: $g(\alpha)h(\alpha) = P_\alpha(\alpha) = 0$. Since $\alpha$ lies in the field $L$, and a field has no zero divisors, either $g(\alpha) = 0$ or $h(\alpha) = 0$. But both $g$ and $h$ have degree strictly less than $d = \deg P_\alpha$ (since $\deg g + \deg h = d$ and both degrees are at least 1). This contradicts the minimality of $P_\alpha$: no nonzero polynomial of degree less than $d$ in $K[t]$ can vanish at $\alpha$. The contradiction shows that no such nontrivial factorisation exists, so $P_\alpha$ is irreducible. What would fail if $L$ were not a field? If $L$ were merely a ring with zero divisors, the equation $g(\alpha)h(\alpha) = 0$ would not force one factor to vanish, and the minimal polynomial might not be irreducible. **Step B: Irreducible elements generate maximal ideals in $K[t]$.** Since $K$ is a field, $K[t]$ is a principal ideal domain (every ideal is generated by a single polynomial, which follows from the Euclidean algorithm). In a PID, a nonzero non-unit element $p$ generates a maximal ideal if and only if $p$ is irreducible. (Proof: if $\langle p \rangle \subset \langle q \rangle$ for some $q$, then $p = qr$, so irreducibility forces $q$ or $r$ to be a unit. If $r$ is a unit, $\langle q \rangle = \langle p \rangle$; if $q$ is a unit, $\langle q \rangle = K[t]$.) Since $P_\alpha$ is irreducible and has degree $d \ge 1$ (so it is not a unit), $\langle P_\alpha \rangle$ is a maximal ideal. **Step C: Quotient by a maximal ideal is a field.** If $\mathfrak{m}$ is a maximal ideal of a commutative ring $R$ with identity, then $R/\mathfrak{m}$ is a field. (Proof: let $f + \mathfrak{m} \neq 0 + \mathfrak{m}$ in $R/\mathfrak{m}$, so $f \notin \mathfrak{m}$. The ideal $\langle f, \mathfrak{m} \rangle = \{rf + m : r \in R, m \in \mathfrak{m}\}$ strictly contains $\mathfrak{m}$ and hence equals $R$ by maximality. So $1 = rf + m$ for some $r \in R$, $m \in \mathfrak{m}$, giving $(r + \mathfrak{m})(f + \mathfrak{m}) = 1 + \mathfrak{m}$.) Combining Steps A, B, and C: $K[t]/\langle P_\alpha \rangle$ is a field.[/guided]

[step:Conclude $K[\alpha]= K(\alpha)$ and establish part (1)] From the previous two steps, $K[\alpha] \cong K[t]/\langle P_\alpha \rangle$ as rings, and $K[t]/\langle P_\alpha \rangle$ is a field. Therefore $K[\alpha]$ is a field. Now, $K(\alpha)$ is defined as the smallest subfield of $L$ containing $K$ and $\alpha$. Since $K[\alpha]$ is a subfield of $L$ containing $K$ and $\alpha$ (it contains $K$ as the image of constant polynomials, and $\alpha$ as the image of $t$), the minimality of $K(\alpha)$ gives $K(\alpha) \subset K[\alpha]$. Conversely, every element $g(\alpha) \in K[\alpha]$ is obtained by adding and multiplying elements of $K$ and $\alpha$, so $g(\alpha)$ lies in any subfield containing $K$ and $\alpha$; in particular $K[\alpha] \subset K(\alpha)$. Therefore \begin{align*} K(\alpha) = K[\alpha] = \{g(\alpha) : g \in K[t]\}. \end{align*} This establishes part (1): every element of $K(\alpha)$ is a polynomial expression in $\alpha$ with coefficients in $K$.[/step]

[guided]This step resolves a subtle distinction between two a priori different objects: - $K[\alpha]$, the subring generated by $K$ and $\alpha$ (polynomial expressions in $\alpha$), and - $K(\alpha)$, the subfield generated by $K$ and $\alpha$ (rational expressions in $\alpha$, i.e., quotients of polynomial expressions). In general, $K[\alpha] \subset K(\alpha)$, and the inclusion can be strict. For instance, if $\alpha$ is transcendental over $K$, then $K[\alpha] \cong K[t]$ is not a field (the polynomial ring lacks inverses for non-constant elements), and $K(\alpha) \cong K(t)$ is strictly larger (the field of rational functions). The key point is that for algebraic $\alpha$, the ring $K[\alpha]$ is already a field — so there is no gap between the ring and the field. This is a consequence of the chain: $K[\alpha] \cong K[t]/\langle P_\alpha \rangle$, and $K[t]/\langle P_\alpha \rangle$ is a field because $P_\alpha$ is irreducible. Concretely, we can see *why* $K[\alpha]$ is closed under inversion. Let $\beta = g(\alpha) \in K[\alpha]$ with $\beta \neq 0$, so $g \notin \langle P_\alpha \rangle$. Since $P_\alpha$ is irreducible and $g$ is not a multiple of $P_\alpha$, we have $\gcd(g, P_\alpha) = 1$. By Bezout's identity in the Euclidean domain $K[t]$, there exist $q, r \in K[t]$ such that $g \cdot q + P_\alpha \cdot r = 1$. Evaluating at $\alpha$: $g(\alpha) \cdot q(\alpha) + P_\alpha(\alpha) \cdot r(\alpha) = 1$. Since $P_\alpha(\alpha) = 0$, this simplifies to $g(\alpha) \cdot q(\alpha) = 1$, i.e., $\beta^{-1} = q(\alpha) \in K[\alpha]$. So every nonzero element of $K[\alpha]$ has a multiplicative inverse in $K[\alpha]$.[/guided]

[step:Extract a $K$-basis from the quotient representation to establish part (2)]By the isomorphism $K(\alpha) = K[\alpha] \cong K[t]/\langle P_\alpha \rangle$, the $K$-vector space structure of $K(\alpha)$ is determined by that of $K[t]/\langle P_\alpha \rangle$. Every element of $K[t]/\langle P_\alpha \rangle$ has a unique representative of degree less than $d = \deg P_\alpha$. To see this, let $f \in K[t]$ be arbitrary. By the division algorithm, $f = P_\alpha \cdot q + r$ with $q, r \in K[t]$ and $\deg r < d$ (or $r = 0$). Then $f + \langle P_\alpha \rangle = r + \langle P_\alpha \rangle$, so every coset has a representative of degree less than $d$. For uniqueness: if $r_1 + \langle P_\alpha \rangle = r_2 + \langle P_\alpha \rangle$ with $\deg r_1, \deg r_2 < d$, then $r_1 - r_2 \in \langle P_\alpha \rangle$, meaning $P_\alpha$ divides $r_1 - r_2$. Since $\deg(r_1 - r_2) < d = \deg P_\alpha$, this forces $r_1 - r_2 = 0$. Every polynomial $r \in K[t]$ with $\deg r < d$ can be written uniquely as $r = a_0 + a_1 t + \cdots + a_{d-1} t^{d-1}$ with $a_0, \ldots, a_{d-1} \in K$. Therefore the cosets $\{1 + \langle P_\alpha \rangle, \, t + \langle P_\alpha \rangle, \, \ldots, \, t^{d-1} + \langle P_\alpha \rangle\}$ form a $K$-basis of $K[t]/\langle P_\alpha \rangle$. Transporting this basis through the isomorphism $\overline{\operatorname{ev}_\alpha}$, which sends $t^k + \langle P_\alpha \rangle$ to $\alpha^k$ for each $k$, we conclude that $\{1, \alpha, \alpha^2, \ldots, \alpha^{d-1}\}$ is a $K$-basis of $K(\alpha)$. In particular, \begin{align*} [K(\alpha) : K] = \dim_K K(\alpha) = d = \deg P_\alpha. \end{align*}[/step]

[guided]The question is: what is $K(\alpha)$ as a $K$-vector space? The isomorphism $K(\alpha) \cong K[t]/\langle P_\alpha \rangle$ reduces this to a problem about the quotient ring, where the answer is transparent. **Existence of low-degree representatives.** Let $f \in K[t]$. The division algorithm (valid because $K[t]$ is a Euclidean domain) produces $q, r \in K[t]$ with $f = P_\alpha \cdot q + r$ and $\deg r < d$. In the quotient, $f$ and $r$ differ by a multiple of $P_\alpha$, so they represent the same coset: $f + \langle P_\alpha \rangle = r + \langle P_\alpha \rangle$. This shows every coset has a representative of degree less than $d$. **Uniqueness of low-degree representatives.** Suppose $r_1$ and $r_2$ both have degree less than $d$ and represent the same coset: $r_1 - r_2 \in \langle P_\alpha \rangle$. Then $P_\alpha$ divides $r_1 - r_2$. But $\deg(r_1 - r_2) \le \max(\deg r_1, \deg r_2) < d = \deg P_\alpha$. A nonzero polynomial cannot be divisible by a polynomial of strictly greater degree, so $r_1 - r_2 = 0$. **Constructing the basis.** The unique representatives of degree less than $d$ have the form $a_0 + a_1 t + \cdots + a_{d-1} t^{d-1}$ with $a_i \in K$. In the quotient, this is $\sum_{k=0}^{d-1} a_k (t^k + \langle P_\alpha \rangle)$, and the uniqueness of the $a_k$ shows that $\{t^k + \langle P_\alpha \rangle : 0 \le k \le d-1\}$ is a $K$-basis of the quotient. Applying the isomorphism $\overline{\operatorname{ev}_\alpha} : t^k + \langle P_\alpha \rangle \mapsto \alpha^k$, a basis maps to a basis under a vector space isomorphism, so $\{1, \alpha, \ldots, \alpha^{d-1}\}$ is a $K$-basis of $K(\alpha)$. The dimension count $[K(\alpha) : K] = d = \deg P_\alpha$ connects the degree of the field extension to the degree of the minimal polynomial. This is the fundamental reason why the minimal polynomial controls the "size" of a simple algebraic extension.[/guided]

[step:Record the field isomorphism to establish part (3)] Combining the results of the preceding steps: the isomorphism \begin{align*} \overline{\operatorname{ev}_\alpha} : K[t]/\langle P_\alpha \rangle &\xrightarrow{\;\sim\;} K(\alpha) \\ f + \langle P_\alpha \rangle &\mapsto f(\alpha) \end{align*} is an isomorphism of fields. Both sides are fields — $K[t]/\langle P_\alpha \rangle$ because $\langle P_\alpha \rangle$ is maximal, and $K(\alpha) = K[\alpha]$ as established in the preceding steps — and $\overline{\operatorname{ev}_\alpha}$ is a ring isomorphism (hence automatically a field isomorphism, since ring isomorphisms between fields preserve the field structure). This is part (3), completing the proof. [/step]