[proofplan] Let $k$ be a finite field and list its elements. We construct a single polynomial over $k$ that is nonconstant but takes the value $1_k$ at every element of $k$. Hence this polynomial has no root in $k$, which contradicts the defining property of an [algebraically closed field](/page/Algebraically%20Closed%20Field). [/proofplan] [step:Construct a nonconstant polynomial from the list of field elements] Let $k$ be a finite field, and let $0_k$ and $1_k$ denote the additive and multiplicative identities of $k$. Define $n := |k|$, and choose an enumeration of the elements of $k$: \begin{align*} k = \{a_1, \dots, a_n\}. \end{align*} Define the polynomial \begin{align*} p: k \to k,\quad x \mapsto 1_k + \prod_{j=1}^{n}(x - a_j). \end{align*} Equivalently, $p(x) = 1_k + \prod_{j=1}^{n}(x - a_j)$ and $p \in k[x]$. The product $\prod_{j=1}^{n}(x-a_j)$ is monic of degree $n$, so $p$ has leading coefficient $1_k$ and degree $n$. Since a field has at least one element, $n \geq 1$, and therefore $p$ is nonconstant. [/step] [step:Show that the polynomial has no root in the finite field] Let $b \in k$ be arbitrary. Since $k = \{a_1, \dots, a_n\}$, there exists an index $j_b \in \{1,\dots,n\}$ such that $b = a_{j_b}$. Therefore the factor $b - a_{j_b}$ occurs in the product defining $p(b)$, and this factor is $0_k$. Hence \begin{align*} \prod_{j=1}^{n}(b - a_j) = 0_k. \end{align*} It follows that \begin{align*} p(b) = 1_k + 0_k = 1_k. \end{align*} Since $1_k \neq 0_k$ in a field, $b$ is not a root of $p$. Because $b \in k$ was arbitrary, $p$ has no root in $k$. [guided] The point of the construction is that the product $\prod_{j=1}^{n}(x-a_j)$ vanishes at every element of $k$: every possible input is one of the listed elements. Fix an arbitrary element $b \in k$. Since the list $\{a_1,\dots,a_n\}$ contains every element of $k$, there is an index $j_b \in \{1,\dots,n\}$ with $b = a_{j_b}$. Now evaluate the product at $b$. Among the factors \begin{align*} (b-a_1), \dots, (b-a_n), \end{align*} the factor with index $j_b$ is \begin{align*} b-a_{j_b} = b-b = 0_k. \end{align*} A product in a field with one factor equal to $0_k$ is equal to $0_k$, so \begin{align*} \prod_{j=1}^{n}(b-a_j) = 0_k. \end{align*} Therefore the added constant term forces the value of $p$ to be \begin{align*} p(b) = 1_k + \prod_{j=1}^{n}(b-a_j) = 1_k + 0_k = 1_k. \end{align*} Since $1_k \neq 0_k$, the element $b$ is not a root of $p$. Because $b$ was arbitrary, no element of $k$ is a root of $p$. [/guided] [/step] [step:Conclude that the finite field is not algebraically closed] The polynomial $p \in k[x]$ is nonconstant and has no root in $k$. By definition, a field is algebraically closed if every nonconstant polynomial in $k[x]$ has a root in $k$. This contradicts the polynomial $p$ constructed above. Therefore $k$ is not algebraically closed. Since $k$ was an arbitrary finite field, no finite field is algebraically closed. [/step]