Prove Q[sqrt 2, sqrt 3] is a field.

[gotmilk04]

Homework Statement

Prove Q[sqrt 2, sqrt 3] is a field.

Homework Equations

Q[sqrt 2, sqrt 3]= {r + s\sqrt{2} + t\sqrt{3} + u\sqrt{6}| r,s,t,u\in Q}

The Attempt at a Solution

I know I have to show each element has an inverse, but I don't know how on these elements.

 

[cmbeelby]

To find the inverse of something in Q[\sqrt{2}, \sqrt{3}] it might be helpfull to first notice the inverse of something in Q[\sqrt{2}].

(a + b\sqrt{2})^-1 = (a - b\sqrt{2})/(a^2 - 2*b^2)

In general (a + b*sqrt n)^-1 = (a - b*sqrt n)/(a^2 - n*b^2)

Next notice that an element of Q[\sqrt{2},\sqrt{3}] is also of the form a + b\sqrt{3} where a and b are elements of Q[\sqrt{2}] (i.e. they are each a + b*\sqrt{2}) so an element of Q[\sqrt{2}, \sqrt{3}] is like ((a + b\sqrt{2}) + (c + d\sqrt{2})\sqrt{3}) = a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}.

So the inverse of such an element is still (a - b\sqrt{3})/(a^2 - 3*b^2) where the a and b in this case are elements of Q[\sqrt{2}] (and so of the form a + b\sqrt{2}).