Discussion
If F is a finite field of characteristic p, a prime, then for some integer \(n \ge 1\) :
Answer: C
F is the splitting field of \(x^{p^{n}} - x\) over \(\mathbf{F}_{p}\).
A finite field of characteristic p contains \(\mathbf{F}_{p}\) and is a vector space over it of some dimension n, so \(|F| = p^{n}\). Its multiplicative group has order \(p^{n} - 1\), so every non-zero element satisfies \(a^{p^{n} - 1} = 1\), and every element satisfies \(a^{p^{n}} = a\). So the \(p^{n}\) elements of F are exactly the roots of \(x^{p^{n}} - x\), and F is its splitting field.
F is not \(\mathbf{Z}_{p^{n}}\), which is not a field for \(n \gt 1\), nor the product ring \(\mathbf{F}_{p} \times \ldots \times \mathbf{F}_{p}\), which has zero divisors. The roots of \(x^{p^{n}} - 1\) are all equal to 1 in characteristic p, since \(x^{p^{n}} - 1 = (x - 1)^{p^{n}}\).
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