Re: [代數] ring

[yasfun (耶死放)]

let $ denote √(-3)

Z[$]不包含i,-i  Unit裡面沒有i,-i

Unit={1,-1}



(2) 5 is a prime element in Z[$]

if 5 | (a+b$)(c+d$), a,b,c,d in Z

then (a+b$)(c+d$)=5(s+t$)

so (a^2+3b^2)(c^2+3d^2)=25(s^2+3t^2)

so (a^2+3b^2)=5k or (c^2+3d^2)=5k

but x^2=0 or 1 or -1 (mod 5) for every x in Z

so if 5不整除(a^2+3b^2), 5也不整除(c^2+3d^2)

then (a^2+3b^2) cannot be 5k, (c^2+3d^2) cannot be 5k

so 5 is a prime element



PS 5 is not prime <=> 5不整除(a+b$) 且 5不整除(c+d$)

(你的原文有點錯)



(3)

(1+$)*(1-$)=4=2*2

並且(1+$)與2並非差unit倍數  =>  Z[$] is not a UFD



(4)

Z[$]/(5) is a field

(0-2$)(0+1$)=1
(0-1$)(0+2$)=1
(0+1$)(0-2$)=1
(0+2$)(0-1$)=1

(1-2$)(2-1$)=1
(1-1$)(-1-1$)=1
(1+0$)(1+0$)=1
(1+1$)(-1+1$)=1
(1+2$)(2+1$)=1

(2-2$)(2+2$)=1
(2-1$)(1-2$)=1
(2+0$)(-2+0$)=1
(2+1$)(1+2$)=1
(2+2$)(2-2$)=1

hence Z[$]/(5) is a field



其實我覺得這系列題目設計的目的應該不是這樣的

設計者應該是在(2)的地方數字設計錯之類的...(?)

純屬臆測

希望有解答到你的問題~XD

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推 abenton     :感謝大大，那第4題這種題目都要把全部的元素算一遍嗎
→ willydp     :不需要, 因為5是prime, 所以(5)是prime ideal
→ willydp     :因dim Z[$]=1,且(5) ≠(0),故(5)是maximal ideal
→ willydp     :  ^^^^^^^^^^ 是因為Z⊂Z[$]是個integral extension