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I want to find a primitive element of $\mathbb{Q}(\sqrt{3},\sqrt{7})$ over $\mathbb{Q}$. I try with $\sqrt{3} + \sqrt{7}$.

Let $X = \sqrt{3} + \sqrt{7}$ we get

$$ X^4 - 20X^2 + 16 = 0 $$

We can prove that $[\mathbb{Q}(\sqrt{3},\sqrt{7}) : \mathbb{Q}] = 4$ since $\sqrt{3}$ and $\sqrt{7}$ are the roots of $X^4 - 10X^2 + 21$ and it is irreducible over $\mathbb{Q}$ by the Eisenstein criterion with $p = 3$.

Let $P(X) = X^4 - 20X^2 + 16$. Since $\sqrt{3} + \sqrt{7}$ is a root of $P$ then $[\mathbb{Q}(\sqrt{3}+\sqrt{7}) : \mathbb{Q}] \leq 4$. If we prove that $P$ is irreducible over $\mathbb{Q}$ then the degree is equal to 4. But the Eisenstein criterion does not seem to work...

So, I try another thing. Let $\alpha = \sqrt{3} + \sqrt{7}$, we get $\alpha^2 = 10 + 2 \sqrt{21}$ and $\mathbb{Q} \subset \mathbb{Q}(10 + 2 \sqrt{21}) \subset \mathbb{Q}(\sqrt{3} + \sqrt{7})$. Now

$$ [\mathbb{Q}(10 + 2 \sqrt{21}) : \mathbb{Q}] = 2 \Rightarrow 2 \mid [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q} ] $$

The degree must be 1, 2 or 4.

But $\mathbb{Q}(10 + 2 \sqrt{21}) \subset \mathbb{Q}(\sqrt{3} + \sqrt{7})$ and $ [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q}(10 + 2 \sqrt{21}) ] > 1$ then the degree must be 2 or 4. I don't know how to conclude with the formula

$$ [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q} ] = [\mathbb{Q}(\sqrt{3} + \sqrt{7}) :\mathbb{Q}(10 + 2 \sqrt{21}) ] [\mathbb{Q}(10 + 2 \sqrt{21}) : \mathbb{Q}] $$

Thank you. I think I did a mistake somewhere.

Tohiea
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    For an alternative approach, see https://math.stackexchange.com/a/3922059/589 – lhf Dec 10 '20 at 18:16
  • With this approach, since $[ \mathbb{Q}( \sqrt{3},\sqrt{7} ) : \mathbb{Q} ] = 4$ then there are 4 elements in the basis gave by : $\left{ 1 , \sqrt{3}, \sqrt{7}, \sqrt{21} \right}$. Let $\gamma = \sqrt{3} + \sqrt{7}$ then $\gamma^2 = 10 + 2\sqrt{21}$, $\gamma^3 = 24 \sqrt{3} + 16 \sqrt{7}$. We get : $$ \begin{pmatrix} 1\ \gamma \ \gamma^2 \ \gamma^3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 1 & 0 \ 10 & 0 & 0 & 2 \ 0 & 24 & 16 & 0\ \end{pmatrix} \begin{pmatrix} 1\ \sqrt{3}\ \sqrt{7} \ \sqrt{21} \end{pmatrix} $$ – Tohiea Dec 11 '20 at 11:37
  • then $\left{ 1, \gamma, \gamma^2 , \gamma^3 \right}$ is a basis of $\mathbb{Q}( \sqrt{3} + \sqrt{7} )$. Since this is a basis with 4 elements then $[ \mathbb{Q}( \sqrt{3} + \sqrt{7} ) : \mathbb{Q} ] = 4$. But, $[ \mathbb{Q}( \sqrt{3} , \sqrt{7} ) : \mathbb{Q} ] = 4$ then $$ \mathbb{Q}( \sqrt{3} + \sqrt{7} ) = \mathbb{Q}( \sqrt{3} , \sqrt{7} )$$ as $\mathbb{Q}$-vector space. i.e it is the same extension and therefore, $\sqrt{3} + \sqrt{7}$ is a primitive element. It is correct ? – Tohiea Dec 11 '20 at 11:40

2 Answers2

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Cubing $\alpha=\sqrt 3+\sqrt 7$, you obtain the linear system \begin{cases} \alpha=\sqrt 3+\sqrt 7, \\[1ex] \alpha^3=24\sqrt 3+16\sqrt7, \end{cases} from which can easily deduce $\sqrt 3$ and $\sqrt 7$ as a linear combination of $\alpha$ and $\alpha^3$.

Bernard
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    Should $z$ be $7$ (with no $=7$) and $16\sqrt{16}$ be $16\sqrt{7}?$ – Chris Leary Dec 10 '20 at 19:32
  • @ChrisLeary: Yes. I combined a copy-paste accident and a typo. 'Tis fixed. Thanks for pointing it! – Bernard Dec 10 '20 at 20:08
  • It is simpler to consider $4/\alpha=\sqrt{7}-\sqrt{3}$ and then deduce that $\sqrt{3},\sqrt{7}$ both lie in $\mathbb {Q} (\alpha) $. – Paramanand Singh Dec 13 '20 at 09:49
  • @ParamanandSingh: A bit simpler, you're right. My first reflex is actually to show $\sqrt 3$ and $\sqrt 7$ belong to $\mathbf Q[\alpha]$ (which is the same set, but thinking in terms of polynomials).. – Bernard Dec 13 '20 at 11:53
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I don't know what is your background in Galois theory, but essentially, this theory was conceived to replace tedious (and most of the time blind) polynomial calculations by natural conceptual manipulations of groups. Here we deal with quadratic and biquadratic fields. General quadratic fields are of the form $\mathbf Q(\sqrt a)$, with $a \in \mathbf Q^*, \notin \mathbf {Q^*}^2$, and it is straightforward to show that to quadratic fields $\mathbf Q(\sqrt a),\mathbf Q(\sqrt b)$ are equal iff $ab \in \mathbf {Q^*}^2$. It follows that, if $ab \notin \mathbf {Q^*}^2$, then the extension $K=\mathbf Q(\sqrt a,\sqrt b)$, which is the compositum of two distinct quadratic fields, is actually biquadratic, i.e. galois with group $G\cong C_2 \times C_2$. Obviously $G$ contains exactly three subgroups of order two, isomorphic to $C_2$. Then the first fundamental theorem of Galois theory gives us all the subfields of $K$, which are $\mathbf Q, K$ and three quadratic subfields, $\mathbf Q(\sqrt a),\mathbf Q(\sqrt b),\mathbf Q(\sqrt ab)$ exactly. It follows that the subfield $\mathbf Q(\sqrt a +\sqrt b)$ must coincide with $K$.

Of course we can apply this to $a=3, b=7$, but the galois approach gives much more. For instance it answers one recurrent question in this thread, which is the determination of all the subextensions of $\mathbf Q(\sqrt p_1),...,(\sqrt p_n)$, where the $p_i$ are $n$ distinct primes, see e.g. https://math.stackexchange.com/a/3444776/300700.

nguyen quang do
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