What does the discriminant of an algebraic number field mean intuitively? If $E/F$ is a finite extension of fields and $\alpha_1,\ldots, \alpha_n$ is a basis
of $E/F$, the discriminant of $\{\alpha_1,\ldots, \alpha_n\}$ is $$\det(\operatorname{Tr}_{E/F}(\alpha_i\alpha_j))$$
I want to know if there is any other interpetration of this quantity? What information is actually conveyed by this quantity?
 A: The discriminant has a geometric interpretation as squared volume. The analogy to have in mind is that the pairing $(x,y) \mapsto {\rm Tr}_{E/F}(xy)$ in a finite extension of fields $E/F$ is analogous to the pairing $(v,w) \mapsto v \cdot w$ for vectors $v$ and $w$ in ${\mathbf R}^n$.  The trace pairing is an $F$-bilinear mapping $E \times E \rightarrow F$ while the dot product is an ${\mathbf R}$-bilinear mapping ${\mathbf R}^n \times {\mathbf R}^n \rightarrow {\mathbf R}$.
If an $n$-dimensional box in ${\mathbf R}^n$ has a vertex at the origin and its edges coming out of the origin are $v_1, \dots, v_n$, then the traditional way of describing the volume uses the matrix with the $v$'s as the columns: the volume is $|\det A|$ where $A = [v_1 \cdots v_n]$ is the $n \times n$ matrix whose $j$th column is $v_j$. The product $A^\top{A}$ has $(i,j)$ entry $v_i \cdot v_j$, and its determinant is $\det(A^\top{A}) = \det(A^\top)\det(A)$, which is $\det(A)^2$. Therefore the volume of the box is $\sqrt{\det(A^\top{A})} = \sqrt{\det(v_i \cdot v_j)}$. So we have two formulas for the volume of that box:
$$
{\rm volume} \ = \det[v_1 \cdots v_n] = \sqrt{\det(v_i \cdot v_j)}.
$$
Therefore
$$
{\rm volume}^2 \ = \det(v_i \cdot v_j).
$$
Replacing $n$-dimensional Euclidean space with a finite extension of fields $E/F$ of degree $n$ and a basis $v_1,\dots,v_n$ of ${\mathbf R}^n$ with a basis $e_1,\dots,e_n$ of $E/F$, then the analogue of $\det(v_i \cdot v_j)$ is $\det({\rm Tr}_{E/F}(e_ie_j))$, which is the discriminant of the basis.  
As one example of this analogy at work, any two bases of $E/F$ have their discriminants related by a nonzero square factor (namely the square of the determinant of the change-of-basis matrix), and this is analogous to the squared volume of two boxes in ${\mathbf R}^n$ being related to each other by the square of the determinant of the change-of-basis matrix. The algebraic calculations underlying both results are very similar (some may even say it's the same calculation).
As another example of the analogy at work, $\det(v_i \cdot v_j)$ and $\det({\rm Tr}_{E/F}(e_ie_j))$ both make sense even if the $v_i$'s and $e_i$'s are not bases, and in that case the numbers are $0$. If the $v_i$'s are linearly dependent then the "box" they span is lower than $n$-dimensional, so it makes geometric sense that $\det(v_i \cdot v_j)$ is $0$ because the $n$-dimensional volume of the box is $0$. If $v_1,\dots, v_n$ is a basis then $\det(v_i \cdot v_j)$ is not $0$ since the the box spanned by the $v_i$'s is $n$-dimensional, so the box has nonzero volume. Is $\det({\rm Tr}_{E/F}(e_ie_j))$ nonzero if $e_1,\dots,e_n$ is a basis of $E/F$? This algebraic question is more subtle than the situation with vectors in ${\mathbf R}^n$. If the trace function ${\rm Tr}_{E/F}$ is not identically zero on $E$ then $\det({\rm Tr}_{E/F}(e_ie_j)) \not= 0$ when the $e_i$'s are a basis, but if the trace is identically $0$ then of course $\det({\rm Tr}_{E/F}(e_ie_j)) = 0$ for any $e_i$'s at all. For number fields, or more generally any fields of characteristic $0$, the trace is not identically $0$ since ${\rm Tr}_{E/F}(1) = [E:F] \not= 0$ in $F$. More generally, the trace is not identically $0$ if and only if $E/F$ is a separable field extension.
A: Let $K$ be a number field and $\mathcal{O}_K$ its ring of integers. Let $\sigma_1, ... \sigma_n$ be the complex embeddings of $K$, where $n = r + 2s$, $r$ is the number of real embeddings, and $s$ is the number of conjugate pairs of complex embeddings. The $\sigma_i$ furnish an embedding
$$K \to \mathbb{R}^r \times \mathbb{C}^s$$
and the image of $\mathcal{O}_K$ in $\mathbb{R}^r \times \mathbb{C}^s$ is a lattice. The square root of the discriminant of $K$ measures the volume of a fundamental domain of this lattice (with respect to a suitable volume form). 
In this setting the prime factorization of the discriminant also provides important information about ramification. In particular, the primes ramifying in $\mathcal{O}_K$ are precisely the primes dividing the discriminant. See, for example, Wikipedia. 
