Classify (with proof) all fields with finitely many elements.
Solution
Let \(F\) be a finite field. Its characteristic is a prime \(p\), so its prime subfield is \(\mathbb F_p\). Since \(F\) is finite-dimensional over \(\mathbb F_p\), say \[ [F:\mathbb F_p]=n, \] we have \[ |F|=p^n. \] Thus every finite field has prime-power order.
Conversely, fix a prime \(p\) and \(n\ge1\), and work in an algebraic closure \(\overline{\mathbb F_p}\). Let \[ E=\{a\in\overline{\mathbb F_p}:a^{p^n}=a\}. \] If \(a,b\in E\), then in characteristic \(p\), \[ (a+b)^{p^n}=a^{p^n}+b^{p^n}=a+b, \] and similarly \((ab)^{p^n}=ab\). Also, if \(a\ne0\), then \(a^{p^n-1}=1\), so \[ (a^{-1})^{p^n}=a^{-1}. \] Hence \(E\) is a field. It is exactly the set of roots of \[ X^{p^n}-X. \] The derivative of this polynomial is \(-1\), so all roots are simple. Therefore it has exactly \(p^n\) roots in the algebraic closure, and \[ |E|=p^n. \] Thus a field of order \(p^n\) exists for every prime power.
Finally, let \(F\) be any field with \(p^n\) elements. Its multiplicative group \(F^\times\) has order \(p^n-1\), so for every nonzero \(a\in F\), \[ a^{p^n-1}=1, \] and hence \(a^{p^n}=a\); this is also true for \(a=0\). Thus every element of \(F\) is a root of \(X^{p^n}-X\). Since this polynomial has exactly \(p^n\) roots in an algebraic closure, \(F\) is identified with the full root field \(E\) above.
Therefore, up to isomorphism, there is exactly one finite field of order \(p^n\) for each prime power \(p^n\), and there are no finite fields of any other order.