TITLE: What is $[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}]$?
QUESTION [5 upvotes]: What is $[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}]$?
On the one hand, we have $[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}(i,\sqrt{2})]\cdot[\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(i)]\cdot[\mathbb{Q}(i):\mathbb{Q}]=2^3=8.$
On the other hand, the minimum polynomial in $\mathbb{Q}[x]$ containing $i,\sqrt{2},\sqrt{3}$ as roots is $(x^2+1)(x^2-2)(x^2-3)$, which is of degree $6$.
What am I misunderstanding?

REPLY [4 votes]: You're correctly applying the degree theorem, but the equality
$$
[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}(i,\sqrt{2})]=2
$$
is not obvious. I'd use a different chain of subfields, recalling that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$, which has degree $4$ over $\mathbb{Q}$.
It is indeed clear that $\mathbb{Q}(i,\sqrt{2}+\sqrt{3})\ne \mathbb{Q}(\sqrt{2}+\sqrt{3})$, so
$$
[\mathbb{Q}(i,\sqrt{2}+\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]=2
$$
because $i$ satisfies $X^2+1$. Then you can conclude that [$\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}]=8$.
Why isn't it $6$? When you factor $\mathbb{Q}[X]$ by the ideal generated by a non irreducible polynomial, the ring you get is not a field. So the ring
$$
\mathbb{Q}[X]/((X^2+1)(X^2-2)(X^2-3))
$$
has dimension $6$ over $\mathbb{Q}$, but it is not the splitting field of the polynomial.
Another example in the opposite direction: the field $\mathbb{Q}[X]/(X^3-2)$ has degree $3$ over $\mathbb{Q}$, but it is not the splitting field of the polynomial. So you cannot say that $\mathbb{Q}(\omega\sqrt[3]{2},\sqrt[3]{2})$ has degree $3$ over $\mathbb{Q}$ just because both elements satisfy $X^3-2$ (where $\omega$ is a non real cubic root of $1$).