Written by Colin+ in ask uncle colin.

Dear Uncle Colin,

I need to show that $\sqrt{7}$ is in $\mathbb{Q}[\sqrt{2}+\sqrt{3}+\sqrt{7}]$ and I don't really know where to start.

We Haven't Approached Tackling Such Questions

Hi, WHATSQ, and thanks for your message!

I am absolutely *not* a number theorist, although I must admit to getting a bit curious about it recently.

In number theory, you can extend a *field*1 by throwing extra numbers or symbols into it. For example, the complex numbers can be written as $\mathbb{R} [i]$ - the real numbers, with the extra number $i$ added in such that $i^2 = -1$. It's all of the numbers $z = a + bi$, with $a$ and $b$ in the real numbers.

Another example: $\mathbb{Q}[\sqrt{2}]$ is the extension of the rational numbers to include the number $\sqrt{2}$ - that is, all of the numbers $z$ that can be written as $a + b\sqrt{2}$, with $a$ and $b$ both rational.

We have $\mathbb{Q}[\sqrt{2} + \sqrt{3}+\sqrt{7}]$, which contains all of the numbers that can be written as $z = a + b\br{\sqrt{2} + \sqrt{3}+\sqrt{7}}$ with $a$ and $b$ being rational.

This extended field is also a field -- which means (in particular) that any power of the extended number also belongs to the field, and that's how we're going to attack this problem.

We know that $s = \sqrt{2} + \sqrt{3}+\sqrt{7}$ is in the field $\mathbb{Q}[\sqrt{2} + \sqrt{3}+\sqrt{7}]$, by definition - and I'm going to call the field $F$ to save me typing it out every time.

Because $F$ is a field, $s^2 \in F$ as well; that is $s^2 = 12 + 2\br{\sqrt{6}+\sqrt{14}+\sqrt{21}}$, which doesn't look like it helps us much. However, we can use algebraic operations to say $\frac{s^2 - 12}{2} = \sqrt{6}+\sqrt{14}+\sqrt{21}$ is also in $F$. Let's call that number $t$.

Now, $t^2 \in F$ as well, and that works out to be $41 + 2\br{\sqrt{84} + \sqrt{126} +\sqrt{294}}$, which is $41 + 2\sqrt{42}\br{\sqrt{2}+\sqrt{3}+\sqrt{7}}$.

Now we're getting somewhere: this tells us that $\sqrt{42}\br{\sqrt{2}+\sqrt{3}+\sqrt{7}}$ is in the field. Dividing by $\sqrt{2}+\sqrt{3}+\sqrt{7}$ (which is a non-zero element of the field) tells us that $\sqrt{42} \in F$.

If we multiply $t$ by $\sqrt{42}$, we get $21 \sqrt{2} + 14\sqrt{3} + 6 \sqrt{7}$.

If we subtract $6s$ from this, we find that $15\sqrt{2} + 8 \sqrt{3}$ is in $F$.

Squaring this gives $642 + 240\sqrt{6}$ - which means that $\sqrt{6} \in F$.

And finally, since $\sqrt{42}$ and $\sqrt{6}$ are both in $F$, so is $\frac{\sqrt{42}}{\sqrt{6}} = \sqrt{7} \blacksquare$

I don't know that that's the simplest way to do it - I'd be delighted to hear of a less convoluted method! - but I hope it helps all the same.

- Uncle Colin

- A set of numbers where you have the four basic operators defined and following the usual rules of arithmetic [↩]

## Barney Maunder-Taylor

Doesn’t look convoluted to me. Thanks Colin, I did not know about the extensions like C=R(i) but I do now!

## Tom

Thanks for this – I enjoyed that and I understood it, which always makes me feel a little bit cleverer than I often do.

One question that did pop into my head though – I remember bits of number theory (and the notation) from my studies long ago, but I couldn’t remember how to “say” some of this in my head.

For example, I totally get the notation for the rationals being extended by some values as a blackboard Q followed by those values in square brackets, but I’m not sure what a recognisable verbal/ mental shorthand for this is (as “a blackboard Q followed by some values in square brackets” is a bit cumbersome.)

What should my brain be saying to me when I read it?

Thanks!

## Colin

No idea, I’m afraid! I shall ask the Twitters.

## Evelyn

I think people usually use the term adjoin. It’s still a bit cumbersome, but “Q adjoin root 2 plus root 3 plus root 7” is probably what I’d say. I’ve more often seen Q adjoin just one root, so “Q adjoin the square root of 2” is a phrase that I have heard more frequently.

## Colin

Thanks, Evelyn!

## Tom

That’s great, thank you Evelyn!