Re: [代數] quotient ring

看板Math作者 (阿鄉)時間14年前 (2011/12/24 11:13), 編輯推噓4(402)
留言6則, 3人參與, 最新討論串4/5 (看更多)
※ 引述《jacky7987 (憶)》之銘言: : 昨天上數論的時候 大概把代數都還給老師了 : 所以來請教大家 : Let R=Z[sqrt(-26)] : show that I=(3,1+sqrt(-26)) is a prime ideal. : 老師提到的提示是用 R/I 是個integral domain下手. : 可是似乎會遇到兩次除法(就是把R也換成quotient ring的寫法) : 然後我就掛了 : 懇請大家幫忙 by 3rd isomorphism theorem, R/I = [R/(3)]/[I/(3)] Note that I/(3)=(1+sqrt(-26)) (generated over R/(3)) and R/(3) has only 9 elements. ({0,1,2,x,1+x,2+x,2x,1+2x,2+2x}, where x^2=1) Indeed, R/(3)=R[x]/(x^2+26,3)=Z3[x]/(x^2+2)={ax+b:0=<a,b=<2} Under this identification, I/(3)=(3,x^2+26,x+1)/(3,x^2+26)=(x+1) (generated over R/(3)) So it can be shown easily that [R/(3)]/(1+x) has only 3 elements. As an additive group, it is isomorphic to Z3. Now it's straightforward to check that this is a ring isomorphism. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.112.51.97

12/25 00:35, , 1F
謝謝:)
12/25 00:35, 1F

12/25 02:24, , 2F
推indeed
12/25 02:24, 2F

12/25 10:53, , 3F
被樓上看出來了
12/25 10:53, 3F

12/25 11:20, , 4F
Indeed那行是Z_3[x]/(x^2+26)?
12/25 11:20, 4F

12/26 10:26, , 5F
26=2 (mod 3)
12/26 10:26, 5F

12/26 10:30, , 6F
原來如此
12/26 10:30, 6F
文章代碼(AID): #1EzKDX-a (Math)
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文章代碼(AID): #1EzKDX-a (Math)