What is $d(\mathbb{Q}(\sqrt{2},\sqrt{3}))$?

Question

$K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4$. We also know the conjugates of $\sqrt{2}+\sqrt{3}$ are, $$x_1=\sqrt{2}+\sqrt{3}$$ $$x_2=\sqrt{2}-\sqrt{3}$$ $$x_3=-\sqrt{2}+\sqrt{3}$$ $$x_4=-\sqrt{2}-\sqrt{3}$$ $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+\sqrt{3})+c(\sqrt{2}+\sqrt{3})^2+d(\sqrt{2}+\sqrt{3})^3|a,b,c,d \in \mathbb{Q}\}$. Thus an integral basis of $K$ is, $$S=\{1,\sqrt{2}+\sqrt{3},(\sqrt{2}+\sqrt{3})^2,(\sqrt{2}+\sqrt{3})^3\}$$ $$S=\{1,\sqrt{2}+\sqrt{3},5+2\sqrt{6},(\sqrt{2}+\sqrt{3})^3\}$$ and hence, $$d(K)=d(\mathbb{Q}(\sqrt{2}+\sqrt{3}))=D(S)$$ $$D(S)=\begin{vmatrix} 1 & \sqrt{2}+\sqrt{3} & (\sqrt{2}+\sqrt{3})^2 & (\sqrt{2}+\sqrt{3})^3\\ 1 & \sqrt{2}-\sqrt{3} & (\sqrt{2}-\sqrt{3})^2 & (\sqrt{2}-\sqrt{3})^3\\ 1 & -\sqrt{2}+\sqrt{3} & (-\sqrt{2}+\sqrt{3})^2 & (-\sqrt{2}+\sqrt{3})^3\\ 1 & -\sqrt{2}-\sqrt{3} & (-\sqrt{2}-\sqrt{3})^2 & (-\sqrt{2}-\sqrt{3})^3 \end{vmatrix}^2$$ $$D(S)=\begin{vmatrix} 1 & \sqrt{2}+\sqrt{3} & 5+2\sqrt{6} & 11\sqrt{2}+9\sqrt{3}\\ 1 & \sqrt{2}-\sqrt{3} & 5-2\sqrt{6} & 11\sqrt{2}-9\sqrt{3}\\ 1 & -\sqrt{2}+\sqrt{3} & 5-2\sqrt{6} & -11\sqrt{2}+9\sqrt{3}\\ 1 & -\sqrt{2}-\sqrt{3} & 5+2\sqrt{6} & -11\sqrt{2}-9\sqrt{3} \end{vmatrix}^2$$ Using Mathematica, $$D(S)=2677248 + 1078272\sqrt{6}$$

Why am I not getting an integer answer?

Appararently I don't know how to type, $D(S)= 147456$

Answer 1

There are two issues here: First, you seem to have made a data entry error, which resulted in a non-integer answer. Second, $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ isn't integrally closed, so you're computing the discriminant of some non-maximal order of $\mathbb{Q}(\sqrt{2} + \sqrt{3})$.

The following SageMath code verifies that $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ in fact has index $8$ in the maximal order:

sage: K.<b> = NumberField((sqrt(2) + sqrt(3)).minpoly())
sage: R = K.order(b)
sage: OK = K.maximal_order()
sage: R.index_in(OK)
8

Thus, the discriminant of this order is $8^2$ times the discriminant of the number field. (For reference, this is number field 4.4.2304.1 in the LMFDB.)

Answer 2

Well, my favorite method of defining (and calculating!) the discriminant of a free algebra is as the determinant of the trace pairing. Before I launch into this, however, in case your ring is of the form $\Bbb Q[a]$, so that a basis is the powers $\{a^m\}_{0\le m<n}$, where the rank is $n$, there’s a quick and dirty way of calculating the discriminant. This is just $\text{Norm}\bigl(f'(a)\bigr)$, where $f$ is the minimal polynomial of $a$. Using $a=\sqrt2+\sqrt3$ and its minimal polynomial, you do indeed get $2^{14}\cdot3^2$. But we know that this method is not apposite to your problem.

If it’s true that an integral basis of your field is $\{1,\sqrt2,\sqrt3, (\sqrt3-1)/\sqrt2\}$, we can label these $b_1$ through $b_4$ and write down the matrix whose $(i,j)$-th entry is $\text{Trace}(b_ib_j)$, and calculate its determinant. I’ll let you check my calculation that the matrix in question is $$ \begin{pmatrix} 4&0&0&0\\0&8&0&-4\\0&0&12&0\\0&-4&0&8 \end{pmatrix}\,, $$ whose determinant is $2304=2^8\cdot3^2$. Since this agrees with Daniel’s result, it seems my guess of the basis was correct.