Proving Bijections: Tau_1,2,3,4 on Q(sqrt(2), sqrt(3))

[Gott_ist_tot]

Sorry for no tex. When I previewed it it would just come up as not finding the images so I made it in ascii as well as I could.

Homework Statement

Show that each of the rational linear mappings tau_1 tau_2 tau_3 and tau_4 is a bijection of Q(sqrt(2) sqrt(3))

Homework Equations

These are meant to resemble piecewise functions.

...{ sqrt(2) |--> sqrt(2)
tau_1 : {
...{ sqrt(3) |--> sqrt(3)

...{ sqrt(2) |--> sqrt(2)
tau_2 : {
...{ sqrt(3) |--> (-)sqrt(3)

The other two just makes the mapping of sqrt(2) go to the negative and the last one has both going to the negative

The Attempt at a Solution

I seem to be able to argue it in words. I just do not think it is a mathematically correct method. I am new to the whole abstract aspect of math.

Tau_i where i = 1,2,3,4 are just a series of mappings of alpha |--> beta or alpha |--> (-)beta. In the cases alpha |--> beta alpha = beta. This means that if you map two elements x,y such that x =/= y you will get two elements of beta w,v such that w =/= v. Therefore if tau(x) =/= tau(y) w =/= v and it is injective. To prove surjective every element of beta has a value in alpha since the value in alpha equals the value in beta. Thus it is not restricted to any range of values.

I would use the same argument for the negative I don't know any special things I should look out for since I believe there is a number y such that x+y=0 where y is denoted as -x. The only problem I can see with this is that was for the reals and I am dealing with rationals. Thanks for any advice.

 

[HallsofIvy]

$\tau_2$ takes $\sqrt{2}$ into $\sqrt{2}$ and $\sqrt{3}$ into $-\sqrt{3}$

Any number is $Q(\sqrt{2},\sqrt{3})$ can be written in the form $a+ b\sqrt{2}+ c\sqrt{3}+ d\sqrt{6}$
(I'm going to call $\tau_2$ "f" since it is easier to type!)
Then $f(a+ b\sqrt{2}+ c\sqrt{3}+ d\sqrt{6})$= $f(a)+ f(b)f(\sqrt{2})+ f(c)f(\sqrt{3})+ f(d)f(\sqrt{2})f(\sqrt{3})$= $a+ b\sqrt{2}- c\sqrt{3}- d\sqrt{6}$.