ring isomorphism


by math-chick_41
Tags: isomorphism, ring
math-chick_41
math-chick_41 is offline
#1
Nov25-05, 11:24 AM
P: 34
what is the easiest way to show that
Q[x]/<x^2-2> is ring isomorphic to
Q[sqrt2]={a+b(sqrt2)|a,b in Q}

just give me a hint how to start
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math-chick_41
math-chick_41 is offline
#2
Nov25-05, 12:29 PM
P: 34
anyone?

do I have to show that x^2 - 2 is in the kernal?
matt grime
matt grime is offline
#3
Nov25-05, 01:48 PM
Sci Advisor
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^2-2) is of the form a+bx, isn't it....)

math-chick_41
math-chick_41 is offline
#4
Nov25-05, 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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