Volume computation for an integral basis Let $k$ be a number field, and $\omega_1, ... , \omega_n$ an integral basis for $\mathcal O_k$.  Then the images $\dot{\omega_i}$ of the $\omega_i$ are a basis in the real vector space 
$$k \otimes_{\mathbb Q} \mathbb R = \prod\limits_{v \mid \infty}k_v$$
Let $\sigma_1, ... , \sigma_r$ be the real embeddings of $k$ into $\mathbb{C}$, and let $\tau_1, ... , \tau_s$ be a choice of nonequivalent complex embeddings, so $\overline{\tau_1}, ... , \overline{\tau_s}$ are the remaining embeddings of $k$ into $\mathbb{C}$.  Then
$$k \otimes_{\mathbb Q}\mathbb{R} = \prod\limits_{i=1}^r \mathbb R \times \prod\limits_{i=1}^s \mathbb{C}$$
Let $D = \{ c_1\dot \omega_1 + \cdots + c_n \dot \omega_n : 0 \leq c_i < 1 \}$.  If $\mu$ is the Lebesgue measure on $k \otimes_{\mathbb Q}\mathbb R$, coming from the product of the Lebesgue measures on $\mathbb R$ and $\mathbb C$, then a standard result from measure theory is that 
$$\mu(D) = |\det(\dot \omega_1, ... , \dot \omega_n)|$$
I have read also in Tate's thesis that
$$\mu(D) = 2^{-s} \sqrt{\textrm{disc}(k/\mathbb Q)} \tag{1}$$
where $\textrm{disc}(k/\mathbb Q) = |\det(\phi_i(\omega_j)|^2$, for $\phi_1, ... , \phi_n$ the embeddings of $k$ into $\mathbb{C}$.
I want to verify the equality (1).  I have come up with a proof, which I will give in an answer below, but I don't like it.  Is there any easier proof for this result?
 A: Let me write $\sqrt{-1}$ instead of $i$, since I am using the letter $i$ as an index.  Let $\sigma_i(\omega_j) = c_i^j$, and $\tau_i(\omega_j) = a_i^j + \sqrt{-1}b_i^j$, so that $\overline{\tau_i}(\omega_j) = a_i^j - \sqrt{-1} b_i^j$.
Let $$p_j = \begin{pmatrix} c_1^j \\ \vdots \\ c_n^j \end{pmatrix} ; v_j = \begin{pmatrix} a_1^j \\ \vdots \\ a_n^j \end{pmatrix}; w_j = \begin{pmatrix} \sqrt{-1}b^j_1 \\ \vdots \\ \sqrt{-1} b_n^j\end{pmatrix}$$
So that $$\sqrt{\textrm{disc}(k/\mathbb Q)} = |\det(p_1, ... , p_r,v_1+w_1, v_1-w_1, ... , v_s+w_s,v_s - w_s)|$$
For any alternating function $T(-,-)$, linear in each variable, we have
$$T(v_i + w_i,v_i-w_i) = T(v_i,v_i) + T(w_i,v_i) + T(v_i,-w_i) + T(w_i,-w_i)$$
$$ =0 - T(v_i,w_i) - T(v_i,w_i) + 0 =  -2 \cdot T(v_i,w_i)$$
Repeatedly applying this  to the above determinant, we get 
$$2^s|  \det(p_1, ... , p_r, v_1, w_1, ... , v_s, w_s)|$$
Now interpreting $\dot{\omega_i}$ as a vector in $\mathbb R^r \mathbb C^s = \mathbb R^{r + 2s}$, where we identify each $\mathbb C$ with $\mathbb R^2$ in the usual way (so $a + \sqrt{-1}b$ is identified with $(a,b)$), we have
$$\dot{\omega_i} = (\sigma_1(\omega_i), ... , \sigma_r(\omega_i), \tau_1(\omega_i), ... , \tau_s(\omega_i))$$
$$ =  (c_i^1, ... , c_i^r, a_i^1, b_i^1, ... , a_i^s, b_i^s)$$
so we see that 
$$|\det(p_1, ... , p_r,v_1,w_1, ... , v_s,w_s)| = | \sqrt{-1}^s\det(\dot \omega_1, ... , \dot \omega_s)| = \mu(D)$$
