[proofplan] The proof splits into two independent arguments, one for each direction. For part (1), we induct on the number of generators $n$. The base case $n = 1$ is the [Structure of Simple Algebraic Extensions](/theorems/1251), which gives $[K(\alpha_1) : K] = \deg P_{\alpha_1} < \infty$. For the inductive step, we set $M = K(\alpha_1, \ldots, \alpha_{n-1})$, observe that $\alpha_n$ remains algebraic over the larger field $M$, and apply the [Tower Law](/theorems/1248) to the chain $K \subset M \subset M(\alpha_n) = K(\alpha_1, \ldots, \alpha_n)$. For part (2), we take a $K$-basis of $L$ and invoke the [Finite Implies Algebraic](/theorems/1249) theorem to see that each basis element is algebraic. [/proofplan] [step:Part (1): Induct on the number of generators $n$, with the base case given by the structure of simple extensions] We prove by induction on $n$ that if $\alpha_1, \ldots, \alpha_n$ are all algebraic over $K$, then $[K(\alpha_1, \ldots, \alpha_n) : K] < \infty$. **Base case ($n = 1$).** The element $\alpha_1$ is algebraic over $K$, so it has a minimal polynomial $P_{\alpha_1} \in K[t]$ of some finite degree $d_1 := \deg P_{\alpha_1} \ge 1$. By the [Structure of Simple Algebraic Extensions](/theorems/1251), the set $\{1, \alpha_1, \alpha_1^2, \ldots, \alpha_1^{d_1 - 1}\}$ is a $K$-basis of $K(\alpha_1)$, so \begin{align*} [K(\alpha_1) : K] = d_1 = \deg P_{\alpha_1} < \infty. \end{align*} [guided] The base case is the content of the [Structure of Simple Algebraic Extensions](/theorems/1251): for a single algebraic element $\alpha_1$ over $K$, the simple extension $K(\alpha_1)$ is a finite-dimensional $K$-vector space whose dimension equals the degree of the minimal polynomial $P_{\alpha_1}$. Concretely, a $K$-basis is $\{1, \alpha_1, \ldots, \alpha_1^{d_1 - 1}\}$ where $d_1 = \deg P_{\alpha_1}$. The hypothesis that $\alpha_1$ is algebraic over $K$ is essential here: if $\alpha_1$ were transcendental, then $K(\alpha_1) \cong K(t)$ (the field of rational functions), which is infinite-dimensional over $K$. So the base case would fail without algebraicity. [/guided] [/step] [step:Inductive step: reduce $K(\alpha_1, \ldots, \alpha_n)$ to a simple extension over $K(\alpha_1, \ldots, \alpha_{n-1})$] **Inductive step.** Assume the result holds for $n - 1$ generators (where $n \ge 2$). Define the intermediate field \begin{align*} M := K(\alpha_1, \ldots, \alpha_{n-1}). \end{align*} By the induction hypothesis applied to the $n - 1$ elements $\alpha_1, \ldots, \alpha_{n-1}$ (each algebraic over $K$), the extension $M/K$ is finite: \begin{align*} [M : K] < \infty. \end{align*} We claim that $\alpha_n$ is algebraic over $M$. Since $\alpha_n$ is algebraic over $K$, there exists a nonzero polynomial $f \in K[t]$ with $f(\alpha_n) = 0$. Since $K \subset M$, the polynomial $f$ also belongs to $M[t]$, so $\alpha_n$ is a root of a nonzero polynomial in $M[t]$. Hence $\alpha_n$ is algebraic over $M$. Let $P_{\alpha_n, M} \in M[t]$ denote the minimal polynomial of $\alpha_n$ over $M$. By the [Structure of Simple Algebraic Extensions](/theorems/1251) applied to the extension $M(\alpha_n)/M$: \begin{align*} [M(\alpha_n) : M] = \deg P_{\alpha_n, M} < \infty. \end{align*} Note that $\deg P_{\alpha_n, M} \le \deg P_{\alpha_n, K}$, since $P_{\alpha_n, K} \in K[t] \subset M[t]$ is a polynomial in $M[t]$ having $\alpha_n$ as a root, and the minimal polynomial over $M$ divides every polynomial in $M[t]$ that vanishes at $\alpha_n$. Now observe that $M(\alpha_n) = K(\alpha_1, \ldots, \alpha_{n-1})(\alpha_n) = K(\alpha_1, \ldots, \alpha_n)$, so we have the tower $K \subset M \subset K(\alpha_1, \ldots, \alpha_n)$ with both steps finite. By the [Tower Law](/theorems/1248): \begin{align*} [K(\alpha_1, \ldots, \alpha_n) : K] = [M(\alpha_n) : M] \cdot [M : K] < \infty. \end{align*} This completes the induction. In particular, since $[K(\alpha_1, \ldots, \alpha_n) : K] < \infty$, the extension $K(\alpha_1, \ldots, \alpha_n)/K$ is finite and hence algebraic by the [Finite Implies Algebraic](/theorems/1249) theorem. [guided] The inductive step reduces the $n$-generator case to the $(n-1)$-generator case by peeling off the last generator. Setting $M = K(\alpha_1, \ldots, \alpha_{n-1})$, the full extension becomes a simple extension $M(\alpha_n)/M$ stacked on top of $M/K$. A natural question arises: we know $\alpha_n$ is algebraic over $K$, but can we use the [Structure of Simple Algebraic Extensions](/theorems/1251) for $M(\alpha_n)/M$? That theorem requires $\alpha_n$ to be algebraic over $M$, not over $K$. The resolution is straightforward but important to state: any polynomial $f \in K[t]$ that witnesses $f(\alpha_n) = 0$ is also a polynomial in $M[t]$ (since $K \subset M$), so algebraicity over $K$ implies algebraicity over any extension of $K$. The converse is false in general — an element can be algebraic over $M$ without being algebraic over $K$ — but we do not need the converse here. With $\alpha_n$ algebraic over $M$, the [Structure of Simple Algebraic Extensions](/theorems/1251) gives $[M(\alpha_n) : M] = \deg P_{\alpha_n, M} < \infty$. We also note that $\deg P_{\alpha_n, M} \le \deg P_{\alpha_n, K}$: the minimal polynomial over $K$ belongs to $M[t]$ and vanishes at $\alpha_n$, so $P_{\alpha_n, M}$ divides it in $M[t]$, forcing the degree inequality. (The degree can strictly decrease: for instance, $\sqrt{2}$ has minimal polynomial $t^2 - 2$ over $\mathbb{Q}$, but minimal polynomial $t - \sqrt{2}$ over $\mathbb{Q}(\sqrt{2})$.) The identity $M(\alpha_n) = K(\alpha_1, \ldots, \alpha_n)$ holds because $K(\alpha_1, \ldots, \alpha_n)$ is defined as the smallest subfield of the ambient field containing $K$ and all of $\alpha_1, \ldots, \alpha_n$. Since $M = K(\alpha_1, \ldots, \alpha_{n-1})$ already contains $K, \alpha_1, \ldots, \alpha_{n-1}$, adjoining $\alpha_n$ to $M$ yields exactly $K(\alpha_1, \ldots, \alpha_n)$. Finally, the [Tower Law](/theorems/1248) applies to the tower $K \subset M \subset M(\alpha_n)$ because both $[M : K]$ and $[M(\alpha_n) : M]$ are finite. The product of two finite numbers is finite, completing the induction. The final clause — that the finite extension is algebraic — follows from the [Finite Implies Algebraic](/theorems/1249) theorem: every element of a finite extension satisfies a nonzero polynomial over the base field, because the powers $1, \alpha, \alpha^2, \ldots$ must eventually be linearly dependent in a finite-dimensional space. [/guided] [/step] [step:Part (2): Express a finite extension as finitely generated by algebraic elements using a $K$-basis] Suppose $[L : K] = d < \infty$. Let $\{e_1, \ldots, e_d\}$ be a $K$-basis of $L$. Since every element of $L$ is a $K$-linear combination of $e_1, \ldots, e_d$, the smallest subfield of $L$ containing $K$ and the elements $e_1, \ldots, e_d$ is $L$ itself. (Any subfield containing $K$ and $e_1, \ldots, e_d$ must contain all $K$-linear combinations of $e_1, \ldots, e_d$, which comprise all of $L$.) Therefore \begin{align*} L = K(e_1, \ldots, e_d). \end{align*} Each $e_i$ is an element of the finite extension $L/K$, so by the [Finite Implies Algebraic](/theorems/1249) theorem, every $e_i$ is algebraic over $K$. (The theorem applies because $[L : K] = d < \infty$, and each $e_i \in L$.) Hence $L = K(e_1, \ldots, e_d)$ with each $e_i$ algebraic over $K$, which is the desired conclusion. [guided] The argument for part (2) is more direct than part (1). Given a finite extension $L/K$ of degree $d$, we need to find finitely many algebraic elements that generate $L$ over $K$. The natural choice is a $K$-basis $\{e_1, \ldots, e_d\}$ of $L$. Why does $L = K(e_1, \ldots, e_d)$? The field $K(e_1, \ldots, e_d)$ is defined as the smallest subfield of $L$ containing $K$ and $e_1, \ldots, e_d$. Since $K(e_1, \ldots, e_d)$ is a subfield containing $K$ and all basis elements, it is a $K$-subspace of $L$ that contains the basis, so it contains every $K$-linear combination of $e_1, \ldots, e_d$. Since $\{e_1, \ldots, e_d\}$ spans $L$ over $K$, every element of $L$ is such a linear combination, giving $L \subset K(e_1, \ldots, e_d)$. The reverse inclusion $K(e_1, \ldots, e_d) \subset L$ holds because $K, e_1, \ldots, e_d$ all lie in $L$ and $L$ is a field. The algebraicity of each $e_i$ follows from the [Finite Implies Algebraic](/theorems/1249) theorem. This theorem states that if $[L : K] < \infty$ and $\alpha \in L$, then the $[L : K] + 1$ elements $1, \alpha, \alpha^2, \ldots, \alpha^{[L:K]}$ are linearly dependent over $K$ (since they are $d + 1$ elements in a $d$-dimensional space), and a nontrivial dependence relation yields a nonzero polynomial in $K[t]$ vanishing at $\alpha$. Since $e_i \in L$ and $[L : K] = d < \infty$, this applies to each $e_i$. Note that the choice of generators is not unique. Any generating set works — a basis is simply the most natural choice because it simultaneously witnesses both the generation and the finiteness. One could also take $L = K(\beta)$ for a single element $\beta$ (a primitive element) when $K$ has characteristic zero or $L/K$ is separable, but the existence of a primitive element requires additional machinery (the [Primitive Element Theorem](/theorems/1267)) and is not needed here. [/guided] [/step]