[proofplan] We show that every element of $K$ already lies in $k$. Given any $\alpha \in K$, its minimal polynomial over $k$ is irreducible and has a root in $k$ because $k$ is algebraically closed. Since the minimal polynomial is irreducible and has a root in $k$, it must be linear, which forces $\alpha \in k$. [/proofplan]

[step:Show that every element of $K$ has a linear minimal polynomial over $k$]Let $\alpha \in K$. Since $K/k$ is algebraic, $\alpha$ is algebraic over $k$. Let $m_{\alpha, k} \in k[x]$ denote the minimal polynomial of $\alpha$ over $k$. By definition, $m_{\alpha, k}$ is the unique monic irreducible polynomial in $k[x]$ satisfying $m_{\alpha, k}(\alpha) = 0$. Since $k$ is algebraically closed, $m_{\alpha, k}$ has a root $\beta \in k$. By the [Factor Theorem](/theorems/???), $(x - \beta)$ divides $m_{\alpha, k}$ in $k[x]$. Since $m_{\alpha, k}$ is irreducible over $k$ and $x - \beta$ is a monic polynomial of degree $1$ dividing it, we must have $m_{\alpha, k}(x) = x - \beta$. (If $\deg m_{\alpha, k} \geq 2$, the factorisation $m_{\alpha, k}(x) = (x - \beta) g(x)$ with $\deg g \geq 1$ would contradict irreducibility.) Therefore $m_{\alpha, k}(x) = x - \beta$, and since $m_{\alpha, k}(\alpha) = 0$, we obtain $\alpha = \beta \in k$.[/step]

[guided]We need to show $K = k$, i.e., that every element of $K$ already belongs to $k$. The strategy is to use the algebraicity assumption to produce a minimal polynomial, and then use the algebraic closure of $k$ to force that polynomial to be linear. Let $\alpha \in K$ be arbitrary. Since $K/k$ is an algebraic extension, $\alpha$ is algebraic over $k$, meaning there exists a non-zero polynomial in $k[x]$ that $\alpha$ satisfies. The minimal polynomial $m_{\alpha, k} \in k[x]$ is the unique monic polynomial of smallest degree such that $m_{\alpha, k}(\alpha) = 0$. A standard result in field theory guarantees that $m_{\alpha, k}$ is irreducible over $k$. Now, $m_{\alpha, k}$ is a non-constant polynomial in $k[x]$ (it has degree $\geq 1$ since $\alpha$ satisfies it and it is monic). Because $k$ is algebraically closed, every non-constant polynomial over $k$ has a root in $k$. Let $\beta \in k$ satisfy $m_{\alpha, k}(\beta) = 0$. By the [Factor Theorem](/theorems/???), $(x - \beta)$ divides $m_{\alpha, k}(x)$ in $k[x]$. But $m_{\alpha, k}$ is irreducible, so it cannot be written as a product of two polynomials of positive degree. Since $x - \beta$ has degree $1$ and divides $m_{\alpha, k}$, the only possibility is $m_{\alpha, k}(x) = x - \beta$ (both are monic, and the quotient must be a unit, hence a non-zero constant, hence $1$ by monicity). From $m_{\alpha, k}(\alpha) = 0$ we get $\alpha - \beta = 0$, so $\alpha = \beta \in k$. Since $\alpha \in K$ was arbitrary, $K \subset k$. Combined with $k \subset K$ (since $K/k$ is an extension), we conclude $K = k$.[/guided]