Finding the inverse of an element in Q[√2,√3].

[Boorglar]

Homework Statement

I need to prove that [itex]Q[√2,√3] = \{a+b√2+c√3+d√6: a, b, c, d \in Q\} [/itex] is a field.

Homework Equations

The Attempt at a Solution

I proved all axioms except the existence of inverses for nonzero elements. My problem is that multiplication is quite hairy. I ended up trying to invert a 4x4 matrix with letters, which got tedious very quickly and I couldn't finish it. Actually, proving that the determinant is never 0 would be good enough since this would prove linear independence and the existence of a unique solution.

Is there a much simpler method, or do I really have to deal with ugly algebra?

 

[Office_Shredder]

A totally different approach is to calculate the dimension of [itex] Q[\sqrt{2},\sqrt{3}][/itex] (or at least an upper bound) as a vector space over Q, and then confirm that you have a linearly independent set of the right size that is spanning your field.

 

[Boorglar]

Well, any element in [itex]Q[√2,√3][/itex] is by definition a linear combination of [itex]1,√2,√3,√6[/itex] over the rationals, so the dimension is at most 4.

Now, to show that [itex]1, √2, √3 [/itex] and [itex]√6[/itex] are linearly independent, I guess I can set up the equation [itex]a+b√2+c√3+d√6=0[/itex]. I will divide the proof in 2 parts.

PART 1
First, I will show that [itex]1, √2 [/itex] and [itex]√3[/itex] are linearly independent.
Suppose [itex](*) a + b√2 + c√3 = 0[/itex]. Then:

[itex](b√2+c√3)^2 = a^2[/itex] so [itex]2b^2+3c^2+2bc√6 = a^2[/itex]. Rearranging:
[itex]2bc√6 = a^2 - 2b^2 - 3c^2[/itex]. But the RHS is rational, so [itex]bc = 0[/itex].

Case I: [itex]b = 0[/itex]
[itex]a^2 = 3c^2 [/itex] so [itex]a = 0, c = 0[/itex] (since [itex]√3[/itex] is irrational).

Case II: [itex]c = 0[/itex]
[itex]a^2 = 2b^2[/itex] so [itex]a = 0, b = 0[/itex] (since [itex]√2[/itex] is irrational).

Thus the only solution to [itex](*)[/itex] is [itex]a=b=c=0[/itex], and [itex]1, √2[/itex] and [itex]√3[/itex] are linearly independent over the rationals.

PART 2
Now it remains to be shown that [itex]√6[/itex] is linearly independent from [itex]1, √2[/itex] and [itex]√3[/itex]. Suppose [itex]a+b√2+c√3 = √6[/itex].
Then, move [itex]a[/itex] to the RHS and square both sides:
[itex]2b^2+3c^2+2bc√6 = a^2+6-2a√6[/itex] so
[itex]a^2-2b^2-3c^2 + 6 = 2(a+bc)√6[/itex]. So [itex]a+bc=0[/itex] (√6 is irrational).
Replacing [itex]a = -bc[/itex]: [itex]2b^2-b^2c^2+3c^2-6 = 0[/itex]
Factoring: [itex](c^2-2)(3-b^2) = 0[/itex] which has no solutions in the rationals.

Thus the set [itex]\{1,√2,√3,√6\}[/itex] is linearly independent over the rationals, and the dimension of [itex]Q[√2,√3][/itex] is 4.


Good, this is done! But I don't see how this proves invertibility.

 

[Office_Shredder]

Ah nuts, I realize now that I bungled my field extension notation (Q(...) vs Q[...] - I thought the intent of the question was to prove that the field Q(sqrt(2),sqrt(3)) was representable as the set on the right) - perhaps my answer isn't applicable for where you are in your class/this specific question. Do you know what the tower theorem for field extensions is?

 

[Boorglar]

Unfortunately not. Also, I might have mistaken the () braces with [] braces. Not sure which should be used.
To put you in context, this question comes from the exercise set of a chapter dealing with the definition of rings, integral domains and fields (from an intro to abstract algebra book). Field extensions have not yet been mentioned.

 

[Boorglar]

My guess is that the inverse formula is relatively simple and analogous to that in the field of quaternions. But I don't want to keep guessing how the inverse might look like. A systematic approach would be better.

 

[jbunniii]

Here is a trick that might help to find an explicit formula for the inverse. Note that we can write a general element of ##\mathbb{Q}[\sqrt{2},\sqrt{3}]## as
$$a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} = (a + b\sqrt{2}) + (c + d\sqrt{2})\sqrt{3} = x + y\sqrt{3}$$
where ##x = a + b\sqrt{2}## and ##y = c + d\sqrt{2}##. Since ##(x+y\sqrt{3})(x-y\sqrt{3}) = x^2 - 3y^2##, we have
$$(x + y\sqrt{3})^{-1} = \frac{x-y\sqrt{3}}{x^2 - 3y^2}$$
Now plug in the values for ##x## and ##y## and simplify.

 

[Office_Shredder]

No you got the right notation, [ ] means take the ring generated by sqrt(2) and sqrt(3) (which is what you have), ( ) means take the field generated by them (which would make the question silly, but I deformed it slightly in my head to make it reasonable).

If you do get to field extensions and the tower theorem, I recommend coming back to this problem and seeing how much more systematic it becomes.

Anyway, for the straightforward approach I recommend trying
[tex] \frac{1}{(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}} \frac{a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}}{a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}} [/tex]

I'm trying to rationalize with respect to sqrt(2) by itself, instead of both at the same time, which is generally a good approach to taking multiple smaller steps against these kinds of problems.

 

[Boorglar]

jbunniii, this is a great idea!
Now, the denominator [itex]x^2-3y^2[/itex] can only be 0 if [itex]x = y = 0[/itex] since [itex]√3 = \frac{a+b√2}{c+d√2} = \frac{(a+b√2)(c-d√2)}{c^2-2d^2}[/itex] can not happen, which in turn means [itex]a=b=c=d=0[/itex]. Also, since [itex]x^2-3y^2[/itex] is a linear combination of 1 and √2, it can be rationalized and [itex]\frac{x-y√3}{x^2-3y^2}[/itex] is a l.c of 1, √2, √3 and √6.

Office_Shredder:
I get the following
[tex]\frac{a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}}{a^2-2b^2+3c^2-6d^2+2(ac-2bd)\sqrt{3}}[/tex]
which can also be easily rationalized. Thank you both!

 

[jbunniii]

Yes, that looks good. The other approach you tried, involving the 4x4 matrix, will also work, but as you noted, the calculation is really nasty.

 

[jbunniii]

By the way, the factorization I used illustrates a general principle regarding field extensions.

As you have now shown, ##\mathbb{Q}(\sqrt{2},\sqrt{3})## is a field. In fact, it is the smallest field that contains ##\mathbb{Q}##, ##\sqrt{2}##, and ##\sqrt{3}##, because any field containing these items must also contain all linear combinations over the rationals of ##1##, ##\sqrt{2}##, ##\sqrt{3}##, and ##\sqrt{6}##.

Moreover, this field can be constructed in two steps.

First, extend ##\mathbb{Q}## to ##\mathbb{Q}(\sqrt{2})##, which is the set of all linear combinations of ##1## and ##\sqrt{2}## over the rationals, or equivalently, the smallest field containing ##\mathbb{Q}## and ##\sqrt{2}##. Then extend ##\mathbb{Q}(\sqrt{2})## to ##(\mathbb{Q}(\sqrt{2}))(\sqrt{3})##. This is the set of all linear combinations of ##1## and ##\sqrt{3}## over the field ##\mathbb{Q}(\sqrt{2})##, or equivalently, the smallest field containing both ##\mathbb{Q}(\sqrt{2})## and ##\sqrt{3}##.

Of course, you can also construct it in the reverse order, first extending to include ##\sqrt{3}##, resulting in ##\mathbb{Q}(\sqrt{3})## and then to include ##\sqrt{2}##, resulting in ##(\mathbb{Q}(\sqrt{3}))(\sqrt{2})##.

In both cases you get the same field, which justifies the notation ##\mathbb{Q}(\sqrt{2},\sqrt{3})## instead of the more cumbersome ##(\mathbb{Q}(\sqrt{2}))(\sqrt{3})## or ##(\mathbb{Q}(\sqrt{3}))(\sqrt{2})##.