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匿名使用者 發問時間: 科學數學 · 5 年前

請教"體"的一些性質

圖片參考:https://s.yimg.com/rk/AB00265573/o/1401464528.jpg

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  • Lv 5
    5 年前
    最佳解答

    Since E is a finite field, we know that E*=E\{0} is a cyclic group of order 81-1=80.

    Therefore, for all elements x in E* we have x^80-1=0.

    Thus, all the elements in E* have the form

    圖片參考:https://s.yimg.com/rk/AE03766702/o/1231082060.gif

    where 0<=n<80.

    Note that x^40-1 and x^16-1 are the factors of x^80-1.

    [1]Find all (monic) irreducible polynomials of degree 2 over Z3.

    There are nine possible polynomials of degree 2:

    x^2, x^2+1, x^2+2, x^2+x, x^2+x+1, x^2+x+2, x^2+2x, x^2+2x+1, x^2+2x+2.

    Check the roots of above polynomials by using 0, 1, 2. You shall find that x^2+1, x^2+x+2, and x^2+2x+2 are irreducible over Z3.

    [2]Find the number of elements α∈E, such that Z3(α) is a degree 2 extension of Z3.

    We want to find α∈E such that [Z3(α):Z]=2; that is, find an irreducible polynomial f(x) of degree 2 and the root of f(x).

    For example, f(x)=x^2+1 is irreducible and i is one of the roots of f(x). Let α=i then we have [Z3(i):Z]=2.

    [3]Find the number of elements α∈E, such that Z3(α) is a degree 4 extension of Z3.

    Consider ξ16=cos(2π/5)+isin(2π/5), then (ξ16)^5-1=0.

    But (ξ16)^5-1=(ξ16-1)[(ξ16)^4+(ξ16)^3+(ξ16)^2+ξ16+1], and ξ16≠1.

    Therefore we have ξ16 is a root of the polynomial g(x)=x^4+x^3+x^2+x+1.

    From the factorization of x^80-1 over Z3, we know that

    g(x)=x^4+x^3+x^2+x+1 is irreducible, hence g(x) is the minimal polynomial of ξ16.

    Let α=ξ16=cos(2π/5)+isin(2π/5), then we have [Z3(α):Z3]=4.

    [5]Findthe number of elements with order 80 in E*.

    If(n,80)=1, then the order of ξn is80.

    Hence there are φ(80)=80*(1-1/2)*(1-1/5)=32elements of order 80 in E*, where φ(n) is Euler's function.

    2015-01-04 23:17:22 補充:

    [4]Find the number of (monic) irreducible polynomials of degree 4 over Z3.

    There are 3 irreducible polynomials of order 1: x, x+1, x+2

    There are 3 irreducible polynomials of order 2: x^2+1, x^2+x+2, x^2+2x+2

    There are 8 irreducible polynomials of order 3:

    (1)x^3+2x+1

    (2)x^3+2x+2

    (3)x^3+x^2+2

    2015-01-04 23:28:07 補充:

    因為有字數上的限制,所以我把剩下的都放在以下的網頁裡

    https://docs.google.com/document/d/1vquQj2eZvF_bUK...

    2015-01-04 23:28:51 補充:

    有更簡潔的證明歡迎提供,謝謝

    2015-01-09 16:37:32 補充:

    [4]Find the number of (monic) irreducible polynomials of degree 4 over Z3.

    這一題可以不必大費周章去數有幾個irreducible polynomials of degree 4

    下面的定理可以派上用場

    Lemma : If q(x) in Zp[x] is irreducible of degree n, then q(x)|(x^m-x), where m=p^n

    2015-01-09 16:40:43 補充:

    Let p=3, n=4, then q(x)|(x^81-x)

    Since q(x) is of degree 4, q(x)|(x^80-1)

    Hence x^80-1 has all irreducible polynomials of degree 4 over Z3.

    The number of q(x) is the number of the irreducible polynomials of degree 4 in the linear factorization of x^80-1 over Z3; namely, 18.

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