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

1. Feb 10, 2010

### gotmilk04

1. The problem statement, all variables and given/known data
Prove Q[sqrt 2, sqrt 3] is a field.

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

3. 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.

2. Apr 9, 2011

### 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$$\sqrt{2}$$).

Last edited: Apr 9, 2011