Show that no finite field is algebraically closed.
Hint: if \(K = \{a_i\}_{i=0}^n\), consider \[ f(x) = a_1 + \prod_{i=0}^n (x - a_i) \in K[x] \] where \(a_1 \neq 0\).
Solution
Let \[ K=\{a_0,a_1,\ldots,a_n\} \] be finite, with \(a_1\ne0\), and define \[ f(x)=a_1+\prod_{i=0}^n(x-a_i)\in K[x]. \]
1. The polynomial \(f\) is nonconstant.
The product has degree \(n+1\), so adding the constant \(a_1\) does not change its leading term. Hence \(\deg f=n+1>0\).
2. The polynomial \(f\) has no root in \(K\).
Every \(a\in K\) equals \(a_j\) for some \(j\). Substituting \(a_j\) gives \[ f(a_j)=a_1+\prod_{i=0}^n(a_j-a_i)=a_1, \] because the factor with \(i=j\) is zero. Since \(a_1\ne0\), one has \(f(a_j)\ne0\). Thus no element of \(K\) is a root.
3. Therefore \(K\) is not algebraically closed.
An algebraically closed field has a root for every nonconstant polynomial over it. By 1 and 2, \(f\in K[x]\) is nonconstant and has no root in \(K\).