We want to find the $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]$.

My first thought is to find the minimal polynomial of $\sqrt5$ over $\Bbb{Q}(\sqrt2+\sqrt3)$. And from this, to say that $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]=\deg m_{\sqrt5,\Bbb{Q}(\sqrt2+\sqrt3)}(x).$

We take the polynomial $m(x)=x^2-5\in\Bbb{Q}(\sqrt2+\sqrt3)[x].$ This is a monic polynomial which has $\sqrt5\in \Bbb{R}$ as a root. We have to show that this is irreducible over $\Bbb{Q}(\sqrt2+\sqrt3)$, in order to say that $m(x)=m_{\sqrt5,\Bbb{Q}(\sqrt2+\sqrt3)}(x)$. The roots of $m(x)$ are $\pm \sqrt5\in \Bbb{R}.$ So,

$$m(x) \text{ is irreducible over } \Bbb{Q}(\sqrt2+\sqrt3) \iff \pm \sqrt5 \notin \Bbb{Q}(\sqrt2+\sqrt3)=\Bbb{Q}(\sqrt2,\sqrt3)$$

because $\deg m(x) =2.$

And this is the point I stack. I tried with the use of the basis of the $\Bbb{Q}$-vector space $\Bbb{Q}(\sqrt2+\sqrt3)$: $$A:= \{1,\sqrt2,\sqrt3,\sqrt6 \}$$

in order to claim that $\nexists a,b,c,d\in \Bbb{Q}:\sqrt5=a+b\sqrt2+c\sqrt3+d\sqrt6$ but this doesn't help.

Any ideas please?

Thank you in advance.


marked as duplicate by anomaly, Leucippus, Claude Leibovici, Lord Shark the Unknown, Paramanand Singh Aug 19 '17 at 6:56

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  • $\begingroup$ See the question and answers here $\endgroup$ – sharding4 Aug 16 '17 at 22:47
  • $\begingroup$ Thank you for your comment. I saw it but I think this doesn't help. Any other ideas? $\endgroup$ – Chris Aug 16 '17 at 22:53
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    $\begingroup$ My answer to the referenced question showed that $x^2-5$ remains irreducible over $\Bbb{Q}[\sqrt{2}, \sqrt{3}]$. Did you have a question about that? $\endgroup$ – sharding4 Aug 16 '17 at 22:58
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    $\begingroup$ This was one of the more than five answers referenced in the previous question. It's very straightforward. $\endgroup$ – sharding4 Aug 16 '17 at 23:04
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    $\begingroup$ Since this is tagged Galois theory,and you know $\sqrt{5}$ lies in a quadratic extension of $\mathbb{Q}$, you can restrict to showing $\sqrt{5}$ is not in any quadratic subfield of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. $\endgroup$ – Steve D Aug 16 '17 at 23:13

Here is an idea that avoids most of the theory.

The correspondence:
$$a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6} \mapsto a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6}$$ preserves sums and products. So if $$\sqrt{5} = a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6}$$ then $$(a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6})^2= 5$$ so $$(a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6})^2 = 5$$ so $$a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6} = \pm (a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6})$$ so either $a = b = 0$ or $c = d = 0$. Now it should be easy, but really, the same trick applies: If $\sqrt{5} = c\sqrt{3} + d\sqrt{6}$ then $c\sqrt{3} - d\sqrt{6}= \pm( c\sqrt{3} + d \sqrt{6})$, so again, $c= 0$ or $d=0$. In the end we get a contradiction.

Note that this procedure can be generalized, for more radicals, provided we have the uniqueness of the writing. We haven't really used $5$, it really shows which square roots are in such a field.


Here’s another elementary proof, which I used rather more advanced ideas to come up with:

Let’s look at $2\big/(\sqrt2+\sqrt3+1)=1+\frac12(\sqrt2-\sqrt6\,)$, which happens to be a root of $f(X)=X^4-4X^3+2X^2+4X-2$, Eisenstein and thus irreducible. This shows that $\sqrt2+\sqrt3$ is quartic over $\Bbb Q$, and, by degree considerations, that $\Bbb Q(\sqrt2+\sqrt3\,)=\Bbb Q(\sqrt2,\sqrt3\,)$.

So what, you say. But consider $f$ as a polynomial over $k=\Bbb Q(\sqrt5\,)$. The ring of integers of $k$ is $\Bbb Z\bigl[\frac{1+\sqrt5}2\bigr]$ , well known to be a Principal Ideal Domain, and you see easily that $2$ is an irreducible (prime) element there. Thus $f$ remains Eisenstein, and so irreducible, over $k$. As a result, the extension $\Bbb Q(\sqrt5,\sqrt2+\sqrt3\,)\supset\Bbb Q(\sqrt5\,)$ is of degree four, and the big field is of degree eight over $\Bbb Q$. Since $\sqrt2+\sqrt3$ is only quartic over $\Bbb Q$, it follows that $\sqrt5$ is quadratic over that quartic field.

  • $\begingroup$ Thank you for your answer. Could you please explain what do you mean by "The ring of integers of k is $\Bbb Z\bigl[\frac{1+\sqrt5}2\bigr]$ " ? $\endgroup$ – Chris Aug 19 '17 at 1:48
  • $\begingroup$ Every element of $k$ that is root of a monic polynomial with $\Bbb Z$-coefficients may be written $m+n\frac{1+\sqrt5}2$. When we say that there is or isn’t unique factorization in a number field $K$, we are really saying that the ring of integers of $K$ does or doesn’t have unique factorization. $\endgroup$ – Lubin Aug 19 '17 at 18:35

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