Hints on Proof Regarding Sign of Discriminant of Number Field Going through Schoof's notes on Algebraic Number Theory, I came across the following problem:
Prove that the sign of the discriminant of a number field $F$ is $(-1)^{r_2}$, where $r_2$ is the number of non-conjugate complex embeddings $F \hookrightarrow \mathbf{C}$.
I am confused how to start the problem for the following reasons and was wondering if someone could give me a nudge in the right direction: the discriminant $\Delta_F$ is the determinant of the matrix $(Tr(\omega_i\omega_j))_{1 \leq i, j \leq n}$ where $\{ \omega_1, \dots, \omega_n \}$ form a $\mathbb{Z}$-basis for $\mathcal{O}_F$, the ring of integers. Meanwhile, one might calculate $r_2$ using the minimal polynomial of a primitive element $\alpha \in F$. This led me to consider evaluating $\Delta(1, \alpha, \dots, \alpha^{n-1})$, though generally this is not $\Delta_F$. So now my confusion comes from trying to figure out a connection between $\det (Tr(\omega_i\omega_j))_{1 \leq i, j \leq n}$ and the embeddings $F \hookrightarrow \mathbf{C}$.
I am not looking for a solution, but just a small hint. Thank you!
 A: This is a complete answer for the benefit of others, so if you just want a hint, only read a little bit!!
While $\Delta(1,\alpha,...,\alpha^{n-1})$ is not in general the discriminant of the whole number field, the two differ by a (positive) square, so they have the same sign.
To figure out the sign, look at $\delta = \sqrt{\Delta(1,\alpha,...,\alpha^{n-1})}$ we can extract the sign of $\Delta$ by determining whether or not complex conjugation acts nontrivially on it; a nontrivial action means the square root must be imaginary, hence from a negative real number, and the opposite for a trivial action.
As it happens, there is a formula for this square root (mentioned in the comments), it is the determinant
$$|\sigma_i(\alpha^j)|$$
If we call complex conjugation $\tau$, we know that it ``plays well'' with this determinant because the determinant is ultimately a polynomial expression in the entries of the matrix, so
$$\tau(|\sigma_i(\alpha^j)|) = |\tau\sigma_i(\alpha^j)|$$
This permutes the $\sigma$'s, i.e. the rows of the matrix. So whether or not the permutation is nontrivial depends on the sign of the permutation. By definition, $\tau$ exchanges each pair of complex-conjugate embeddings, and so $\tau$ acts by swapping $r_2$ rows, which from linear algebra corresponds to changing the sign by $(-1)^{r_2}$.
Thus when $r_2$ is even, complex conjugation acts trivially, and so the discriminant was positive, and if $r_2$ is odd the discriminant was negative, which match $(-1)^{r_2}$.
