Re: [代數] quotient ring

[yusd24 (阿鄉)]

※ 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.


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推 jacky7987   :謝謝:)
推 LimSinE     :推indeed
→ yusd24      :被樓上看出來了
推 jacky7987   :Indeed那行是Z_3[x]/(x^2+26)?
→ yusd24      :26=2 (mod 3)
推 jacky7987   :原來如此