Algebraic number theory is the study of rings of algebraic numbers which are sophisticated generalizations of . Let us quickly recall some necessary definitions.
Definition 1 (Algebraic numbers). A complex number is called an algebraic number if it satisfies a polynomial equation with rational coefficients, i.e. there exists and , such that
Definition 2 (Algebraic Integers). An algebraic integer is an algebraic number which satisfies a monic polynomial equation with integer coefficients, i.e., we can take and for .
The Rational Root Theorem implies that any rational number which is not an integer is an algebraic number but not an algebraic integer. On the other hand, a purely imaginary number can be an algebraic integer. For example, is an algebraic integer since .
It turns out that every algebraic number satisfies a unique monic polynomial in , called the minimal polynomial and its degree is defined to be the degree of the algebraic number .
Looking for a criterion which will say that an algebraic number is in fact an algebraic integer is an interesting and nontrivial problem. In this post, we present a sufficient condition for an algebraic number to be an algebraic integer. This requires the definition of what is called the trace of an algebraic number in a number field. A number field is a finite extension field of the rationals .
Definition 3 (Trace in a Number Field).
Let be a number field of index . We know that there are exactly field embeddings of in which fix point-wise. Let be the embeddings. Then, for any algebraic number , we define the trace of , denoted as to be the following quantity
Now we are in a position to state and prove our main theorem. The proof will require some basic module theory, especially modules over a commutative ring with unity. The book Introduction to Commutative Algebra by Atiyah and MacDonald is a good reference if the reader is not familiar with these topics.
Let be an algebraic number. Consider the number field . If for all , then is an algebraic integer.
We need the following lemma for proving this theorem.
Let be a nonzero algebraic number of degree . Let be the -module generated by . Define
Then is a finitely generated -module.
We first prove the following claim:
For any , we have if and only if for .
Let . Then . Now, for . Hence for all .
Conversely, let for . Let . Then there are integers , such that . Since is a -linear functional and then for all we have
Therefore for all . Therefore and hence, . This completes the proof of the claim.
Consider the form defined by for all . Since is a -linear functional, is a symmetric bilinear form. Also, since is a separable extension, is non-degenerate. Now, is a -basis of . Hence non-degeneracy of the trace form implies that the matrix
Let . Then there exists such that . Also, according to the claim above, there exists such that
This system of equations can be written in terms of matrices as follows:
Then . Hence we observe that any can be written as and ‘s are independent of . Take an arbitrary combination for . Then
Since both and are matrices over and are linearly independent over , we have
This implies that for . Hence from a previous claim, we conclude that . Therefore, we have shown
and hence is a finitely generated -module.
Now we apply this to prove that is an algebraic integer if for all .
Let . We know that is an algebraic integer if and only if is finitely generated -module. We will show under our assumption on that is indeed a finitely generated -module. Since for all , the claim in the lemma above tells us for all with is as defined in the lemma. Therefore all finite linear combinations of over are elements of . Hence . In fact, it is easy to see that is a -submodule of the -module . Since is a PID, and is a finitely generated -module, is Noetherian. Therefore being a submodule of , is a finitely generated -module. This shows is an algebraic integer, as required.
The lemma used in the proof can be extended to a more general setting. We recall the definition of the basic structural setup in algebraic number theory:
Let be a Dedekind domain and be its field of fractions. Let be a finite separable extension field of and be the integral closure of in . This diagram is often called an -setup.
The lemma can be stated in the following general form. First, recall the definition of a Dedekind domain.
Definition 4 (Dedekind domain) An integral domain is said to be a Dedekind domain, if it satisfies the following conditions
(1) Every ideal is finitely generated;
(2) Every non-zero prime ideal is maximal;
(3) It is integrally closed in its field of fractions, i.e. The set of elements of its fraction fields which satisfy a monic polynomial in is precisely .
Lemma (A general AKLB-setup).
Let be a Dedekind domain and be its field of fractions. Let be a finite separable field extension with . Let be a basis of as a -vector space. Consider the -module and define . Then is a finitely generated -module.
We can follow the similar method as done in the case when where is the ring of algebraic integers of .
 Number Fields by Daniel Marcus
 Etale algebras, norm and trace, Dedekind extensions Lecture notes by Andrew Sutherland
 Introduction to Commutative Algebra by M. F. Atiyah, I. G. MacDonald