Finite field extensions are algebraic

Assume \([L:K]=n<\infty\) and let \(\alpha\in L\). The \(n+1\) vectors \[ 1,\alpha,\alpha^2,\dots,\alpha^n \] lie in the \(n\)-dimensional \(K\)-vector space \(L\), so they are linearly dependent. Hence there exist \(c_0,\dots,c_n\in K\), not all zero, such that \[ c_0+c_1\alpha+\cdots+c_n\alpha^n=0. \] Thus \(\alpha\) is a root of the nonzero polynomial \[ c_0+c_1x+\cdots+c_nx^n\in K[x]. \] Since \(\alpha\) was arbitrary, every element of \(L\) is algebraic over \(K\). Therefore every finite field extension is algebraic.