Finding basis of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ over $\mathbb{Q}$

Question

I'm having trouble finding a basis for $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$. So far I know that $[K=\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$ and $[L=\mathbb{Q}(\sqrt{2}, \sqrt{3}):K] = 2$, but it's harder to extend $L$ to $L(\sqrt{5})$. Namely I'm trying to prove that $x^2-5$ is irreducible over $L$ by showing that $\sqrt{5} \notin L$ as follows:

Assume by contradiction that $\sqrt{5} \in L$. Then we have

$$\sqrt{5} = a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$$ with $a,b,c \in \mathbb{Q}$. Squaring both sides and using independence of basis gives the following nasty system of equations:

$$-5+a^2+2b^2+3c^2+6d^2 = 0$$ $$ab + 3cd = 0$$ $$ac + 2bd = 0$$ $$ad + bc = 0$$

I can't see how these equations contradict the assumption that $\sqrt{5} \in L$.

Answer 1

From your equations

$-5+a^2+2b^2+3c^2+6d^2 = 0.....(1)$

$ab + 3cd = 0.....(2)$

$ac + 2bd = 0.....(3)$

$ad + bc = 0.....(4)$

it follows by elimination of $a$ in $(2),(3),(4)$ $$\begin{cases}3c^2=2b^2\\2d^2=c^2\end{cases}.....(5)$$ which is easily seen to be impossible.

Optionnally, another way is turning back to equation $(1)$ so you get$$a^2+9c^2=5$$ which have rational solutions, for example,$(a,c)=(1, \frac 23)$ so you get $$b^2=\frac 23$$ in which the impossibility is clearer than in $(5)$

Answer 2

It is possible to derive contradiction from your system, although a bit of work is needed.

First of all, let us observe that if any of the variables is $0$, then precisely three of them are $0$, i.e. one of them is non-zero. Assume, for example, that $a=0$. Then we have $cd = bd = bc = 0$, and let, for example $b\neq 0$ - it immediately follows $c=d=0$. Analogously for other cases. But, none of the equations $a^2 = 5, 2b^2 = 5, 3c^2 = 5, 6d^2 = 5$ has rational solution. We conclude that $a,b,c,d$ are all non-zero.

From equations $(3)$ and $(4)$ we can derive $$c = -\frac{2bd}{a},\ d = -\frac{bc}{a}\implies c = \frac{2b}{a}\cdot\frac{bc}{a} = c\cdot\frac{2b^2}{a^2}\implies \left(\frac{a}{b}\right)^2 = 2$$ which is contradiction because $a$ and $b$ are rational.

Of course, one can similarly derive $a^2 = 3c^2$, $c^2 = 2d^2$, etc. which all lead to contradiction with rationality of $a,b,c,d$.

Answer 3

If you have seen Galois correspondence, then an alternative way of seeing this is the following. Namely one of the first examples of Galois theory is to see that $L=\Bbb{Q}(\sqrt2,\sqrt3)$ is a Galois extension of $\Bbb{Q}$. Furthermore, the Galois group is the Klein four group generated by $\tau_1:\sqrt2\mapsto-\sqrt2, \sqrt3\mapsto\sqrt3$ and $\tau_2:\sqrt2\mapsto\sqrt2,\sqrt3\mapsto -\sqrt3$.

Galois correspondence then tells you that that the quadratic fields contained in $L$ are exactly the fixed fields of $\tau_1,\tau_2$ and $\tau_1\circ\tau_2$. These are, respectively, $K_1=\Bbb{Q}(\sqrt3)$, $K_2=\Bbb{Q}(\sqrt2)$ and $K_3=\Bbb{Q}(\sqrt6)$. But $\sqrt5$ clearly belongs to a quadratic field. So if you can show that it does not belong to any of the fields $K_i$, then you can conclude that $\sqrt5\notin L$.

As shown by the other answers, in this case you can still do this using the form of elements of the field $L$. Using Galois theory may be nicer further on, though. From the structure of the Galois group $Gal(L(\sqrt5)/\Bbb{Q})$ it similarly follows that $L(\sqrt5)$ has exactly seven quadratic subfields that can be identified as $\Bbb{Q}(\sqrt n)$ with $n\in\{2,3,5,6,10,15,30\}$. It is then easy to show that $\sqrt7$ is not an element of any of those seven fields (this is as easy as showing that $\sqrt2\notin\Bbb{Q}(\sqrt3)$). Therefore $[L(\sqrt5,\sqrt7):L(\sqrt5)]=2$, and you can keep going.