Let me now prove this Theorem 1. So recall that we have seen that the map F_(p^n) which takes x to x^(p^n) is a homomorphism. Then it follows easily that the set of x such that F_(p^n) of x is equal to x is a subfield such that F_(p^n) of x is equal to x is a subfield. And it contains of course F_p And it contains of course F_p But these are exactly roots of x^(p^n) - x. Let me call this polynomial in what follows Qn of x. Let me call this polynomial in what follows Qn of x. Of course this subfield is exactly the splitting field of Qn, Of course this subfield is exactly the splitting field of Qn. And since Qn does not have multiple roots this can be seen for instance by verifying that Qn is relatively prime with its derivative. verifying that Qn is relatively prime with its derivative. Indeed, this derivative is just equal to 1. So they're exactly p^n of the roots. So we have the subfield of p^n elements and this is a splitting field of Qn of Qn, is exactly the field of roots of Qn and this is a field of p^n elements. So conversely, we want to show that any field of p^n elements is a splitting field of Qn. Let the cardinality of K be p^n, and alpha an element of K. Then alpha^(p^n - 1) is equal to 1 provided that alpha is nonzero. Indeed, the multiplicative group of K has cardinality p^n - 1. So any element of K* to the power of (p^n - 1) must be equal to the unit element. So alpha is a root of x^(p^n) - x and 0 is of course also a root. So, K consist of roots of Qn, and this also answers the question why there is a unique subextension of F_p bar, which has p^n elements. This is just because these subextension consists exactly of the roots of Qn. The unicity of the subextension, of the embedding, well, not exactly of the embedding, but of the image of the embedding of K into F_p bar also follows. Let me now formulate and prove a few other properties of finite field. These are very much in the spirit of Theorem 1, so let me probably call them Theorem 1.2, 1.3, and so on. So Theorem 1.2 claims that F_(p^n) contains F_(p^d) if and only if d divides n. One direction is the multiplicativity of degrees in towers. Indeed, in this case we have that the degree of F_(p^n) over F_p is equal to the degree of F_(p^n) over F_(p^d) times the degree of F_(p^d) over F_p. So this is n, and this is d. So we have d divides n. Conversely, suppose that d divides n, then if x^(p^d) is equal to x, then if x^(p^d) is equal to x, the same is true for x^(p^n), since taking x to the (p to the power n)'th power is just iterating taking x to the power (p^d). is just iterating taking x to the power (p^d). So F_(p^d) is So F_(p^d) is contained in F_(p^n). So these are very small theorems. Theorem 1.3, it's better maybe to consider all of them as parts of the Theorem 1. So theorem 1.3 says that F_(p^n) is a stem field and a splitting field of any irreducible polynomial P from F_p[x] of degree n. So the part about stem field is clear, so indeed a stem field of P has degree n over F_p. So this is F_(p^n). Then the part about the splitting field can be shown for instant as follows: let alpha be a root of P. So alpha is an element of F_(p^n), then Qn of alpha is 0 since Qn has as roots exactly all elements of F_(p^n). So P divides Qn and so P splits in F_(p^n). P splits in F_(p^n). This has a simple corollary, so Qn is the product over all d dividing n of all P, where P is irreducible monic of degree d. Now here's the proof. Indeed, we have already seen that all such P divide Qn. Since, the stem field is Since, the stem field is F_(p^d) and this is contained in F_(p^n), so Qn of alpha is 0 if alpha is a root. So they are relatively prime of course, and their product divides Qn. Also, Qn has no multiple roots, so there are no multiple factors either and then what remains to prove is that there are no other irreducible factors of Qn. So let R be an irreducible factor of Qn. Then if alpha is a root of R Qn of alpha is 0, of course. So this means that F_p of alpha is contained in F_(p^n), and this means that F_p of alpha is F_(p^d), where d divides n. And so this means that the degree of R divides n. So there are no other irreducible factors.