Given an integral domain , one can form its field of fractions consisting of quotients of elements in . More precisely, put on the equivalence relation iff . Then , with the operations
where is the equivalence class of the fraction . One easily checks that these operations are well-defined and give a field structure. Observe that the field of rationals is isomorphic with .
A number field is a finite extension . Just as one passes from to , one would like to go back from a number field to a natural analogue of within , which we call the ring of integers of , and denote . This should have the property that .
In general, given a field , there is no canonical choice of proper subring such that . For example, in the case of , any proper subring containing (e.g., , , etc.) has field of fractions , but any ideal in (e.g., , , etc.) also has field of fractions .
The question thus arises of how to properly choose the subring of . To do this, one considers another characterization of . The elements of that are roots of a monic polynomial with coefficients in are called the algebraic integers. Let denote the set of algebraic integers. Suppose is an algebraic integer, satisfying
where each is an integer. Writing , where and are relatively prime, one may substitute to see that
Since divides the right-hand side, must divide , so the only possibility is for . Hence, one deduces .
The only rational algebraic integers are the integers. In symbols, . This suggests for us to set , but this is not, a priori, even a ring. However, it is an important and nontrivial fact that the algebraic integers really do form a ring.
Proposition 1.1: The algebraic integers form a ring.
Proof 1: Let and be algebraic integers, and consider the (monic) minimal polynomials , of , respectively. Working in the bigger field , one may write the roots of as , , , and the roots of as , , , . Define
With a little effort (exercise!), one verifies that each coefficient of is a polynomial expression in the coefficients of and , hence . It follows that is an algebraic integer. An analogous argument shows is an algebraic integer.
Proof 2: Suppose , and that there is a finitely generated -module such that . Then is a linear transformation . Clearly, is an eigenvalue of , so the characteristic polynomial of has as a root. But is a -module, so the characteristic polynomial of is monic in . Hence, is an algebraic integer.
Now suppose and are algebraic integers, with monic minimal polynomials , . Assume also that and . Then generates a -module . One sees that and , and concludes and are algebraic integers.
Hence is a ring, the ring of algebraic integers in . In fact, one also has that . As is a finite extension, it is algebraic, so any satisfies some polynomial with rational coefficients. By clearing denominators, one may assume without loss of generality that actually has integer coefficients. Writing
one multiplies both sides by to get
Therefore, is an algebraic integer, that is . The element may then be written as the quotient of and , both of which are in .
Consequently, we define and think of it as a reasonably natural analogue of inside . These rings of integers are fundamental objects of study in algebraic number theory. Over the next few blog posts, we will study their properties in more depth, and see that the similarities with do not end here.