Let \(F\) be a field of characteristic \(p>0\), with \(p\ne3\). If \(\alpha\) is a zero of \[ f(x)=x^p-x+3 \] in an extension field of \(F\), prove that \(f(x)\) has \(p\) distinct zeros in \(F(\alpha)\).
Solution
Let \(\FF_p\subset F\) denote the prime subfield.
1. For every \(c\in\FF_p\),\[ f(\alpha+c)=0. \]
Since the characteristic is \(p\), the Frobenius identity gives \[ (\alpha+c)^p=\alpha^p+c^p. \] Also \(c^p=c\) for every \(c\in\FF_p\). Therefore \[ \begin{aligned} f(\alpha+c) &= (\alpha+c)^p-(\alpha+c)+3\\ &= \alpha^p+c^p-\alpha-c+3\\ &= \alpha^p-\alpha+3\\ &= f(\alpha)\\ &= 0. \end{aligned} \]
2. The elements\[ \alpha+c, \qquad c\in\FF_p, \]all lie in \(F(\alpha)\).
The field \(F(\alpha)\) contains both \(F\) and \(\alpha\). Since \(\FF_p\subset F\), it therefore contains \(\alpha+c\) for every \(c\in\FF_p\).
3. The \(p\) elements \(\alpha+c\), with \(c\in\FF_p\), arepairwise distinct.
If \[ \alpha+c=\alpha+d, \] then cancellation gives \(c=d\). Since \(\FF_p\) has exactly \(p\) elements, the displayed family has exactly \(p\) distinct elements.
4. The polynomial \(f(x)=x^p-x+3\) has\[ \boxed{p\text{ distinct zeros in }F(\alpha)}. \]
By steps <1>1–<1>3, the degree-\(p\) polynomial \(f\) has at least \(p\) distinct zeros in \(F(\alpha)\). A nonzero polynomial of degree \(p\) over a field has at most \(p\) roots. Hence these are exactly its \(p\) distinct zeros.
5. Q.E.D.
Step 4 is the required conclusion.