Re: [代數] 五次方不可公式解的多項式
※ 引述《omni1234 (啾啾)》之銘言：
: ※ 引述《snaredrum (好聽木琴)》之銘言：
: : 大家都知道Galois理論是證明五次方以上的多項式沒有公式解
: : 請問有人可以給個例子嗎?
: : 目前初淺的想到 假設有個degree 5的polynomial。首先它的Galois group not solvable
: : 那我猜這個 group 必須含有一個5-cycle.
: : 請問例子是? 還有能否解釋一下如何找到?
: 最近剛好代數課學到這邊就上來獻醜一下了
: 考慮f(x)=x^5-6x+3∈Q[x]，注意到f(x)在Q[x]是irreducible且separable
: 令K = splitting field of f(x) over Q
: 因為f(x)是separable，所以K over Q是一個Galois extension
: 假設f(x)的根分別是a1 a2 a3 a4 a5，則K=Q(a1,a2,a3,a4,a5)
: 對所有σ∈ Gal(K/Q)，σ一定會把每一個K裡面的元素送到他的（在Q裡的）最小多項式的根
: 換句話說，對每個ai，i=1到5，σ(ai)=aj for some j=1到5
: 所以Gal(K/Q)會是S5的一個子群
: 藉由一些高中數學(?)跟微積分的幫助我們可以知道f(x)=0有三個實根兩個虛根
已經忘光了沒辦法像 omni 大寫出詳解QQ
不過剛好有個習題是 omni 大這段的後半段 也可以給原原 PO 參考
(就是說 三個實根兩個虛根的都不可根式解)
Fraleigh 的 A First Course in Abstract Algebra (7/e) Section 56 習題 8:
8. This exercise exhibits a polynomial of degree 5 in Q[x] that is not
solvable by radicals over Q.
a. Show that if a subgroup H of S_5 contains a cycle of length 5
and a transposition τ, then H = S_5. [Hint: Show that H contains
every transposition of S_5 and (後略)].
b. Show that if f(x) is an irreducible polynomial in Q[x] of degree 5
having exactly two complex and three real zeros in C, then the group
of f(x) over Q is isomorphic to S_5. [Hint: Use Sylow theory to show
that the group has an element of order 5. Use the fact that f(x) has
exactly two complex zeros to show that the group has san element of
of order 2. Then apply part (a).]
c. The polynomial f(x) = 2x^5 - 5x^4 + 5 is irreducible in Q[x], by the
Eisenstein criterion, with p = 5. Use the techniques of calculus to
find relative maxima and minima and to ``graph the polynomial function
f'' well enough to see that f(x) must have exactly three real zeros
in C. Conclude form part (b) and Theorem 56.4 that f(x) is not solvable
by radicals over Q.
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06/21 12:09, , 1F
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