I need to use the so called Ring Tower Theorem to show that $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=4$, but I'm quite confused with the notation and some concepts. First of all, my book says that if $a$ in some extension $E$ of $\mathbb{Q}$ is algebraic over $\mathbb{Q}$ then the elements of $\mathbb{Q}(a)$ can be written in the form $p+qa$, with $p,q\in\mathbb{Q}$. Hence we know that, since $\mathbb{Q}(\sqrt2,\sqrt3)=\mathbb{Q}(\sqrt2+\sqrt3)$, elements of $\mathbb{Q}(\sqrt2,\sqrt3)$ can be written in the form $p+(\sqrt2+\sqrt3)q$. But doesn't this imply that the the dimension of the basis for $\mathbb{Q}(\sqrt2,\sqrt3)$ is 2, like it is for $\mathbb{Q}(\sqrt2)$?

The Tower Theorem says that if $E$ is a finite extension field of the field $G$ and $G$ is a finite extension of the field $F$, then $E$ is a finite extension of the $F$, and $[E:F]=[E:G][G:F]$.

If I try to apply this theorem to the case in question, I would do as follows:

$$[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)][\mathbb{Q}(\sqrt2):\mathbb{Q}] $$

So I know that $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$, but how do I find $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)]$?

  • 9
    $\begingroup$ It is not true that all elements of $\mathbb Q(a)$ can be written as $p+qa$, you need to include higher powers of $a$ as well. $\endgroup$ – Santiago Canez Mar 24 '17 at 4:11

(In what follows we assume all fields are subfields of a suitable algebraic closure contained in $\Bbb C$).

Note that $\sqrt{3}$ is a root of $x^2 - 3 \in \Bbb Q(\sqrt{2})[x]$, so if $x^2 - 3$ is irreducible over $\Bbb Q(\sqrt{2})$, we have that $[(\Bbb Q(\sqrt{2}))(\sqrt{3}):\Bbb Q(\sqrt{2})] = 2$, and of course it is not hard to show that:

$\Bbb Q(\sqrt{2},\sqrt{3}) = (\Bbb Q(\sqrt{2}))(\sqrt{3})$, by dual inclusion.

If $x^2 - 3$ is reducible over $\Bbb Q(\sqrt{2})$, this implies that $\sqrt{3} \in \Bbb Q(\sqrt{2})$.

If so, we have $\sqrt{3} = a + b\sqrt{2}$ for some $a,b \in \Bbb Q$. Thus:

$3 = (a + b\sqrt{2})^2 = (a^2 + 2b^2) + (2ab)\sqrt{2}$. Since $\{1,\sqrt{2}\}$ is a $\Bbb Q$-basis of $\Bbb Q(\sqrt{2})$ as a $\Bbb Q$-vector space, this in turn implies:

$a^2 + 2b^2 = 3\\2ab = 0.$

If not both $a,b$ are zero, then either $a = 0$, or $b = 0$. The latter leads to $a^2 = 3$ (and thus $\sqrt{3} \in \Bbb Q$), while the former leads to: $b^2 = \frac{3}{2}$ (and thus $\sqrt{\frac{3}{2}} \in \Bbb Q$), neither of which is possible (why?).

But if both $a,b = 0$, then $0 = 3$. Thus $\sqrt{3} \not\in \Bbb Q(\sqrt{2})$, so that $x^2 - 3$ is irreducible over $\Bbb Q(\sqrt{2})$.


Let $x=\sqrt 2+\sqrt 3$. Then $x-\sqrt 3=\sqrt 2$. So $x^2-2\sqrt 3x+1=0$. But this is not the minimum degree polynomial over $\mathbb{Q}$ since $-2\sqrt 3 \notin\mathbb{Q}$. Thus with $x^2+1=2\sqrt 3x$ squaring both sides gives $x^4-10x^2+1=0$. Hence $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=4$.


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