Upper bound for the number of number fields with fixed degree and bounded discriminant

You need to know: Field, field {\mathbb Q} of rational numbers, isomorphic fields, vector space, vector space over a field, dimension of a vector space, matrix, determinant of a matrix.

Background: Number field is a field F that contains {\mathbb Q} and has finite dimension n when considered as a vector space over {\mathbb Q}. Number n is called the degree of F. A set e=\{e_1, e_2, \dots, e_n\} of n elements of F is called basis of F if every x\in F can be written as x=\sum_{i=1}^n c_i(x) e_i with coefficients c_i(x) \in {\mathbb Q}. Sum \text{Tr}(x)=\sum_{i=1}^nc_i(x\cdot e_i) does not depend on the choice of basis e and is called trace of x. If, for every x\in F, all c_i(x) are integers, e is called integral basis of F. The determinant of an n\times n matrix with entries \text{Tr}(e_i\cdot e_j), i=1,\dots,n, j=1,\dots,n, does not depend on the choice of integral basis e and is called the discriminant of F. For X>0, let N_n(X) denotes the number of non-isomorphic number fields of degree n with absolute value of the discriminant at most X.

The Theorem: On 8th September 2003, Jordan Ellenberg and Akshay Venkatesh submitted to arxiv a paper in which they proved the existence of constant B_n depending only on n and absolute constant C, such that inequality N_n(X) \leq B_n X^{\exp(C\sqrt{\log n})} holds for all n>2 and all X>0.

Short context: Counting number fields up to isomorphism is a basic and important open problem in the area. There is a conjecture that N_n(X) grows as linear function of X for every fixed n, but, before 2003, this was known only for n\leq 3 (in later works – see here and here – Bhargava proved it for n=4 and n=5). For general n, the best upper bound was N_n(X) \leq B_n X^{(n+2)/4}. The Theorem proves a bound which is significantly better for large n.

Links: Free arxiv version of the original paper is here, journal version is here. See also Section 6.2 of this book for an accessible description of the Theorem.

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