Group isomorphism and Polynomial ring modulo idealby Xil3 Tags: ideal, isomorphism, modulo, polynomial, ring 

#1
Aug1207, 09:46 AM

P: 3

Hi everyone.
I have two questions that I hope you can help me with. First when trying to show isomorphism between groups is it enough to show that the order of each element within the group is the same in the other group? For example the groups (Z/14Z)* and (Z/9Z)*. They are both of order 6 and both have 1 element of order 1 1 element of order 2 2 elements of order 3 2 elements of order 6 And does the binary operator have to be the same in both groups when doing group isomorphism? And the second question(actually the third one). I am trying to write the multiplication table for the following ring Z_2[X]/(x^2 + x +). I know that this ring has 4 elements. Is it correct that the elements are the following 1 x + 1 x^2 + 1 x^2 + x + 1 ? Hope you can help me with this, Xil3 



#2
Aug1207, 10:12 AM

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#3
Aug1207, 10:27 AM

P: 3

0 x x^2 x^2 + x ? Leaving the 1 out from all the elements? Xil3 



#4
Aug1207, 10:58 AM

Sci Advisor
HW Helper
P: 9,398

Group isomorphism and Polynomial ring modulo ideal
Now where's 1 gone? (Again, it is still there in disguise which makes more confused as toy what you're trying to do).




#5
Aug1207, 11:16 AM

P: 3

Since we are working in Z_2 then a = a where a is an element in Z_2. We know that X^2 + x + 1 = 0 = [0]. Moving numbers around now we then get the remaining 3 elements in the ring. x^2 + x = 1 = [1] x^2 + 1 = x = [x] x + 1 = x^2 = [x^2] Xil3 


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