Discussion
Let \(p\) be a prime number and \(n\), \(m\) natural numbers with \(m\) divides \(n\). Then the finite extension \(\mathbf{F}_{p}n/\mathbf{F}_{p}m\) of finite fields is a :
Answer: B
It is a Galois extension with cyclic Galois group of order \(n/m\).
The printed \(\mathbf{F}_{p}n/\mathbf{F}_{p}m\) is the extension of the field with \(p^{m}\) elements by the field with \(p^{n}\) elements. Every finite extension of finite fields is Galois (the larger field is the splitting field of \(x^{p^{n}} - x\)), and its degree is \(n/m\).
The Galois group is generated by the Frobenius map \(a \mapsto a^{p^{m}}\), so it is cyclic of order equal to the degree, \(n/m\).
Non-cyclic groups are wrong, since the group is generated by one element, and order \(nm\) confuses the product with the quotient of the degrees.
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