[proofplan] Fix an arbitrary $\alpha \in L$ and set $d := [L : K]$. The $d + 1$ elements $1, \alpha, \alpha^2, \ldots, \alpha^d$ all belong to the $K$-vector space $L$, which has dimension $d$. Since $d + 1$ elements in a $d$-dimensional vector space must be linearly dependent, there exist coefficients $c_0, c_1, \ldots, c_d \in K$, not all zero, with $\sum_{i=0}^{d} c_i \alpha^i = 0$. This dependence relation is precisely the statement that the nonzero polynomial $f(t) = \sum_{i=0}^{d} c_i t^i \in K[t]$ satisfies $f(\alpha) = 0$, so $\alpha$ is algebraic over $K$. [/proofplan] [step:Produce $d+1$ powers of $\alpha$ in the $d$-dimensional $K$-vector space $L$] Let $\alpha \in L$ be arbitrary and set $d := [L : K]$, so that $L$ is a $K$-vector space of dimension $d$, with $d < \infty$ by hypothesis. Consider the $d + 1$ elements \begin{align*} 1, \, \alpha, \, \alpha^2, \, \ldots, \, \alpha^d \in L. \end{align*} Each power $\alpha^k$ for $k \in \{0, 1, \ldots, d\}$ belongs to $L$ because $\alpha \in L$ and $L$ is closed under multiplication (being a field). We have $d + 1$ elements in a vector space of dimension $d$. [guided] The starting point is to choose an arbitrary element $\alpha \in L$ — the goal is to show that every element of the extension is algebraic over $K$, so we must make no special assumptions about $\alpha$. Why consider the powers $1, \alpha, \alpha^2, \ldots, \alpha^d$? The definition of "$\alpha$ is algebraic over $K$" requires a nonzero polynomial $f \in K[t]$ with $f(\alpha) = 0$. A polynomial evaluated at $\alpha$ is a $K$-linear combination of powers of $\alpha$, so we need to find coefficients $c_i \in K$ (not all zero) such that $\sum c_i \alpha^i = 0$. This is precisely a $K$-linear dependence relation among powers of $\alpha$. The key insight is that $L$, viewed as a $K$-vector space, has finite dimension $d$. Any collection of more than $d$ elements in a $d$-dimensional vector space must be linearly dependent — this is a fundamental fact of linear algebra. By listing $d + 1$ powers (from $\alpha^0 = 1$ through $\alpha^d$), we guarantee that we have exceeded the dimension. [/guided] [/step] [step:Extract a nontrivial $K$-linear dependence relation and interpret it as a nonzero polynomial vanishing at $\alpha$] Since $\dim_K L = d$ and we have $d + 1$ elements $\{1, \alpha, \alpha^2, \ldots, \alpha^d\} \subset L$, any set of more than $d$ elements in a $d$-dimensional $K$-vector space is linearly dependent over $K$. Therefore, there exist scalars $c_0, c_1, \ldots, c_d \in K$, not all zero, such that \begin{align*} \sum_{i=0}^{d} c_i \alpha^i = 0. \end{align*} Define the polynomial \begin{align*} f: K[t] &\to K \\ f(t) &:= \sum_{i=0}^{d} c_i t^i. \end{align*} Since not all $c_i$ are zero, $f$ is a nonzero element of $K[t]$, with $\deg f \le d$. The dependence relation states exactly that $f(\alpha) = 0$. By definition, $\alpha$ is algebraic over $K$. [guided] The step from "linearly dependent" to "nonzero polynomial" is purely a matter of reinterpretation. A $K$-linear dependence relation among $1, \alpha, \ldots, \alpha^d$ has the form \begin{align*} c_0 \cdot 1 + c_1 \cdot \alpha + c_2 \cdot \alpha^2 + \cdots + c_d \cdot \alpha^d = 0, \quad c_i \in K, \text{ not all zero}. \end{align*} Reading the left-hand side as the evaluation of the polynomial $f(t) = c_0 + c_1 t + c_2 t^2 + \cdots + c_d t^d$ at $t = \alpha$, this equation says $f(\alpha) = 0$. The polynomial $f$ has coefficients in $K$ and is nonzero (since at least one $c_i \neq 0$), which is precisely what is required for $\alpha$ to be algebraic over $K$. Note that the degree of $f$ is at most $d = [L : K]$, so the finite degree of the extension provides an explicit upper bound on the degree of a polynomial annihilating $\alpha$. This bound will be refined in the theory of minimal polynomials: the minimal polynomial of $\alpha$ over $K$ divides $f$ and has degree equal to $[K(\alpha) : K]$, which divides $[L : K]$ by the Tower Law. [/guided] [/step] [step:Conclude that $L/K$ is algebraic] Since $\alpha \in L$ was arbitrary and we produced a nonzero polynomial $f \in K[t]$ with $f(\alpha) = 0$, every element of $L$ is algebraic over $K$. Therefore $L/K$ is an algebraic extension. [/step]