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.

### What does that notation mean?

In number theory, you can extend a field ((A set of numbers where you have the four basic operators defined and following the usual rules of arithmetic)) 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.

### Messing around and seeing what we get

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 $\sqrt{42} \in F$.

### Keep on messing!

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