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Ring isomorphism 
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#1
Nov2505, 11:24 AM

P: 34

what is the easiest way to show that
Q[x]/<x^22> is ring isomorphic to Q[sqrt2]={a+b(sqrt2)a,b in Q} just give me a hint how to start 


#2
Nov2505, 12:29 PM

P: 34

anyone?
do I have to show that x^2  2 is in the kernal? 


#3
Nov2505, 01:48 PM

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HW Helper
P: 9,398

why don't you just write down the (obvious) isomorphism? (obvious in the sense of one side only has x as a special quantity, the other sqrt(2), and anything in Q[x]/(x^22) is of the form a+bx, isn't it....)



#4
Nov2505, 02:06 PM

P: 34

Ring isomorphism
yes! thank you. the example in the book goes into too much detail and I was trying follow that, but yes the function f(a+bx)=a+b(sqrt2) is a ring isomorphism.



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