# Degree of Extension

1. Jun 28, 2007

### Kummer

Let $$p_1,p_2,...,p_n$$ be distinct primes.
Show that $$[\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},...,\sqrt{p_n} ) : \mathbb{Q}]= 2^n$$

Last edited: Jun 28, 2007
2. Jun 28, 2007

### StatusX

What have you tried? It's obvious you should use induction, and then the tricky part becomes proving the irreducibility of a certain polynomial in an extension of Q. Have you gotten this far yet?

3. Jun 28, 2007

### Kummer

Yes. I tried induction of course.

Note that, $$[\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}}): \mathbb{Q}] = [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})]:\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})]\cdot [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})]:\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})] \cdot 2^n$$

So it remains to show,
$$[\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})]:\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})] = 2$$

Which can be shown if,
$$\mbox{deg}(\sqrt{p_{n+1}},\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}))=2$$

But the difficult step is the last step. I was reading about this problem and it was solved with "Kummer Theory" but I tried to find my own "elementary" solution.

4. Jun 29, 2007

### StatusX

You know $\sqrt{p_{n+1}}$ is a root of $x^2-p_{n+1}$, so generates an extension of at most degree 2 over any extension of Q. It will generate an extension of degree 2 iff it is not in the original extension. In other words, you need to show there's no element in $\mathbb{Q}[\sqrt{p_1},...,\sqrt{p_n}]$ whose square is $p_{n+1}$. One way to do this is to note (by induction) that every element in $\mathbb{Q}[\sqrt{p_1},...,\sqrt{p_n}]$ can be written in the form $\alpha + \beta \sqrt{p_n}$ for some $\alpha,\beta \in \mathbb{Q}[\sqrt{p_1},...,\sqrt{p_{n-1}}]$, then square this and find conditions on $\alpha$ and $\beta$.

Last edited: Jun 29, 2007
5. Jun 30, 2007

### Kummer

Yes you are correct, it is really not that bad to proof.