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

[Boorglar]

Homework Statement

I need to prove that Q[√2,√3] = \{a+b√2+c√3+d√6: a, b, c, d \in Q\} 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 Q[\sqrt{2},\sqrt{3}] (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 Q[√2,√3] is by definition a linear combination of 1,√2,√3,√6 over the rationals, so the dimension is at most 4.

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

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

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

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

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

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

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

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

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 x^2-3y^2 can only be 0 if x = y = 0 since √3 = \frac{a+b√2}{c+d√2} = \frac{(a+b√2)(c-d√2)}{c^2-2d^2} can not happen, which in turn means a=b=c=d=0. Also, since x^2-3y^2 is a linear combination of 1 and √2, it can be rationalized and \frac{x-y√3}{x^2-3y^2} is a l.c of 1, √2, √3 and √6.

Office_Shredder:
I get the following
\frac{a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}}{a^2-2b^2+3c^2-6d^2+2(ac-2bd)\sqrt{3}}
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 $(\mathbb{Q}(\sqrt{3}))(\sqrt{2})$.