Ring isomorphism

1. Nov 25, 2005

math-chick_41

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

2. Nov 25, 2005

math-chick_41

anyone?

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

3. Nov 25, 2005

matt grime

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....)

Last edited: Nov 25, 2005
4. Nov 25, 2005

math-chick_41

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.