Finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$

Question

I'm having a difficult time finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$ over $\mathbb{Q}$.

I know that $\sqrt{7}$ satisfies $f(x) = x^2 - 7$ and is irreducible over $\mathbb{Q}$. So $[\mathbb{Q}(\sqrt{7}) : \mathbb{Q}] = 2$. I also know that $\sqrt{5}$ satisfies $f(x) = x^2 - 5$ which is also irreducible in $\mathbb{Q}$. So $[\mathbb{Q(\sqrt{5})}: \mathbb{Q}] = 2$.

Then by some corollary which I forget the name of $[\mathbb{Q}(\sqrt{5}, \sqrt{7}) : \mathbb{Q}] \leq 4$.

I am unsure of what to do from here.

After you have absorbed the answers here you can try and generalize and study this question. Adjoining independent square roots to rationals is an oft enough recurring exercise, so no wonder a suitable umbrella thread exists! For more examples look at the list of *RelatedÄ question in the right margin. — Jyrki Lahtonen, Mar 06 '15 at 06:09

score 3 · Answer 1 · answered Mar 02 '15 at 02:20

You know that $[\mathbb{Q}(\sqrt{5}):\mathbb{Q}]=2$. Then $[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}(\sqrt{5})][\mathbb{Q}(\sqrt{5}):\mathbb{Q}]$. If $\sqrt{7}\in\mathbb{Q}(\sqrt{5})$ then $[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}(\sqrt{5})]=1$, if not then $[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}(\sqrt{5})]=2$. You fill in the details.

score 2 · Answer 2 · answered Mar 02 '15 at 02:21

2

$\mathbb{Q}(\sqrt{5},\sqrt{7})$ as a vector space over $\mathbb{Q}$ is spanned by $1$ and products of the generators $\sqrt{5}$ and $\sqrt{7}$. Therefore, $\mathbb{Q}$ is spanned by $\{1,\sqrt{5},\sqrt{7},\sqrt{35}\}$. Notice that I didn't say that this is a basis. The next step is to show that these elements are linearly independent over $\mathbb{Q}$.

One way to see that these are linearly independent is to find an irreducible polynomial (of degree 4) which has $\sqrt{5}+\sqrt{7}$ as a root. (This polynomial will have roots of $\pm\sqrt{5}\pm\sqrt{7}$, none of which are in $\mathbb{Q}$ and none of the products of the factors are in $\mathbb{Q}[x]$, either).

answered Mar 02 '15 at 02:21

Michael Burr

32,867

So, the minimal polynomial of $\sqrt{5} + \sqrt{7}$ is $x^4 - 24x^2 + 4$. Then we say that since those spanning elements are linearly independent, they are basis elements of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$. Does this sound right? – inoHal Mar 02 '15 at 03:21
@inoHal You can show that the field generated by a root of this polynomial is a subfield of $\mathbb{Q}(\sqrt{5},\sqrt{7})$. But this subfield is also a degree 4 extension of $\mathbb{Q}$. Since the degree of the extension of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ is at most 4, the two fields must be equal and the extension is degree 4. Note that it remains to show that the polynomial $x^4-24x+4$ is irreducible and Eisenstein's criterion doesn't work. However, you can factor it and show that no combinations of the factors are in $\mathbb{Q}[x]$. – Michael Burr Mar 02 '15 at 03:31
Er, wait. x^4 - 24x + 4 does not have roots $\pm \sqrt{5}$ or $\sqrt{4}$. Neither does $x^4 - 24x^2 + 4$. – inoHal Mar 02 '15 at 04:04
The roots are supposed to be $\pm\sqrt{5}+\pm\sqrt{7}$ (not $\sqrt{4}$) – Michael Burr Mar 02 '15 at 11:04
Nice. IMO the pedagogically best way of explaining this to someone new to the area. – Jyrki Lahtonen Mar 06 '15 at 06:11

score 1 · Answer 3 · answered Mar 02 '15 at 04:32

Some of the "hard part" done for you: It should be clear that

$\Bbb Q(\sqrt{5} + \sqrt{7}) \subseteq \Bbb Q(\sqrt{5},\sqrt{7})$ since $\sqrt{5} + \sqrt{7} \in \Bbb Q(\sqrt{5},\sqrt{7})$.

The other direction is a bit "trickier", note that:

$(\sqrt{5} + \sqrt{7})^3 = 5\sqrt{5} + 15\sqrt{7} + 21\sqrt{5} + 7\sqrt{7} = 26\sqrt{5} + 22\sqrt{7}$

Hence $(\sqrt{5} + \sqrt{7})^3 - 26(\sqrt{5} + \sqrt{7}) = -4\sqrt{7}$, that is, if we set $\alpha = \sqrt{5} + \sqrt{7}$, then:

$\sqrt{7} = -\frac{1}{4}(\alpha^3 - 26\alpha) \in \Bbb Q(\alpha) = \Bbb Q(\sqrt{5} + \sqrt{7})$.

In turn, this means $\sqrt{5} = \sqrt{5} + \sqrt{7} - \sqrt{7} \in \Bbb Q(\sqrt{5} + \sqrt{7})$, so that $\Bbb Q(\sqrt{5},\sqrt{7}) \subseteq \Bbb Q(\sqrt{5} + \sqrt{7})$

So we conclude $\Bbb Q(\sqrt{5},\sqrt{7}) = \Bbb Q(\sqrt{5} + \sqrt{7})$. This is useful.

Since we know the minimal polynomial of $\alpha$ is of degree at most $4$, it's natural to start with:

$\alpha^4 = 284 + 48\sqrt{35}$. Now $\alpha^2 = 12 + 2\sqrt{35}$, so:

$\alpha^4 - 24\alpha^2 = -4$, that is $\alpha$ is a root of $x^4 - 24x + 4$. Proving this is irreducible over $\Bbb Q$ then shows $[\Bbb Q(\sqrt{5},\sqrt{7}):\Bbb Q] = 4$.

One possible basis from this is $\{1,\alpha,\alpha^2,\alpha^3\}$, the linear independence of which is guaranteed by the minimal degree of our quartic. Spanning should be clear, from the degree of the extension, but also: any rational polynomial in $\alpha$ can be reduced to one of degree $< 4$, and from:

$\alpha^4 - 24\alpha^2 + 4 = 0$ we obtain: $\alpha\left(6\alpha - \dfrac{\alpha^3}{4}\right) = 1$, that is:

$\alpha^{-1} = 6\alpha - \dfrac{\alpha^3}{4}$, so any rational function of $\alpha$ (that is, $\Bbb Q(\alpha)$) can be expressed as a linear combination of our basis.

score 0 · Answer 4 · answered Mar 02 '15 at 02:26

Followin @A. Wong's hints, to show $[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}(\sqrt{5})]=2$, you have to show $x^2-7$ remains irreducible over $\mathbb{Q}(\sqrt{5})$. If not, it has a root in this field, and you can show this implies $\sqrt 5$ is rational.

score 0 · Answer 5 · answered Mar 02 '15 at 06:17

An element $a+b\sqrt5$ has a minimum polynomial $x^2-2ax+a^2-5b^2$. But $\sqrt7$ has minimum polynomial $x^2-7$. These can only be the same if $a=0$ and $5b^2=7$, and this is impossible for rational b. So $\sqrt7=a+b\sqrt5$ is impossible for rational a,b.

Finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$

5 Answers5