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

## 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.

## 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?

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Office_Shredder
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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.

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
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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?

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.

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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
Homework Helper
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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
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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.

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!

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jbunniii
Homework Helper
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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
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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 $(\mathbb{Q}(\sqrt{3}))(\sqrt{2})$.