Re: [功課] Irreducible polynomial in finite field
※ 引述《jacky7987 (憶)》之銘言：
: 3.8最後一題習題的最後一小提
: 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
我應該把
F_p^{nr}寫成F_p^(r)(a),a is a primitive element
然後我解釋一下裡面的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.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 123.193.89.201
※ 編輯: jacky7987 來自: 123.193.89.201 (05/29 23:28)
※ 編輯: jacky7987 來自: 123.193.89.201 (05/29 23:30)
推
05/29 23:31, , 1F
05/29 23:31, 1F
推
05/29 23:34, , 2F
05/29 23:34, 2F
推
05/29 23:42, , 3F
05/29 23:42, 3F
討論串 (同標題文章)
本文引述了以下文章的的內容：
完整討論串 (本文為第 2 之 2 篇)：