※ 引述《jacky7987 (忆)》之铭言： : 3.8最後一题习题的最後一小提 : Prove, for all n≧1 and finite field k, : there is an irreducible polynomial in k[x] : Proof: : Let F_p be the prime of k, and so that |k|=p^r, for some r>0. : Let F_p^{nr} be the field having p^{nr} elements. : By thm, for all z in F_p^{nr},z is a root fo f(x)=x^(p^{nr})-x : and since p^r-1 | p^{nr}-1, so k is a subfield of F_p^{nr} : and [F_p^{nr},k][k,F_p]=[F_p^{nr},F_p] : [F_p^{nr},k]*r=nr : [F_p^{nr},k]=n : Therefore, there exists a irreducible polynomial g(x) of degree n : in k[x] such that F_p^{nr} has a root of g(x) http://planetmath.org/encyclopedia/IrreduciblePolynomialsOverFiniteField.html 基於上面的 我应该把 F_p^{nr}写成F_p^(r)(a),a is a primitive element 然後我解释一下里面的minimal polynomial Let K/F be a field extension and a in K be algebraic over F. The minimal polynomial for over F is a monic polynomial m(x) in F[x] such that m(a)=0 and, for any other polynomial f(x) in F[x] with f(a)=0,m divides f Note that, for any element a that is algebraic over F , a minimal polynomial exists; moreover, because of the monic condition, it exists uniquely. Given a in K , a polynomial m is the minimal polynomial of a if and only if m(a)=0 and m is both monic and irreducible. 所以g(x)是a的minimal polynomial,且 F_p^rn isom. to F_p^r/(g(x)) 所以deg(g(x))=n=[F_p^rn,F_p^r] since g(x) is irred. and a is a root. --

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