What is $[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}]$?

Question

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?

Answer 1

8 sounds right, and your reasoning is right as well. $\mathbb{Q}[i,\sqrt{2},\sqrt{3}]$ is the splitting field of the polynomial you gave. The only thing you might be missing is the degree of the splitting field is not necessarily the degree of the polynomial itself.

Answer 2

The polynomial you give is not irreducible and is not the minimal polynomial of a single $\alpha$ with $\mathbb Q(\alpha)=\mathbb Q(i,\sqrt 2,\sqrt 3)$. Out of the blue I suggest that $\alpha=i+\sqrt 2+\sqrt 3$ has the desired property - try to find its minimal polynomial.

Answer 3

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 $1$).