# Algebraic Number Theory – I. On Generalizing the Integers

Given an integral domain ${R}$, one can form its field of fractions ${\text{Quot}(R)}$ consisting of quotients of elements in ${R}$. More precisely, put on ${R\times R}$ the equivalence relation ${(a,b) \sim (c,d)}$ iff ${ad=bc}$. Then ${\text{Quot}(R) = (R\times R)/\sim}$, with the operations

$\displaystyle [(a,b)] + [(c,d)] = [(ad+cb,bd)]$

$\displaystyle [(a,b)] \cdot [(c,d)] = [(ac,bd)]$

where ${[(a,b)]}$ is the equivalence class of the fraction ${(a,b)}$. One easily checks that these operations are well-defined and give ${\text{Quot}(R)}$ a field structure. Observe that the field of rationals ${\mathbb{Q}}$ is isomorphic with ${\text{Quot}(\mathbb{Z})}$.

A number field is a finite extension ${K/\mathbb{Q}}$. Just as one passes from ${\mathbb{Z}}$ to ${\text{Quot}(\mathbb{Z}) \cong \mathbb{Q}}$, one would like to go back from a number field ${K}$ to a natural analogue of ${\mathbb{Z}}$ within ${K}$, which we call the ring of integers of ${K}$, and denote ${\mathcal{O}_{K}}$. This ${\mathcal{O}_{K}}$ should have the property that ${\text{Quot}(\mathcal{O}_{K}) = K}$.

In general, given a field ${F}$, there is no canonical choice of proper subring ${A}$ such that ${\text{Quot}(A) = F}$. For example, in the case of ${\mathbb{Q}}$, any proper subring containing ${\mathbb{Z}}$ (e.g., ${\mathbb{Z}[1/2]}$, ${\mathbb{Z}[3/7]}$, etc.) has field of fractions ${\mathbb{Q}}$, but any ideal in ${\mathbb{Z}}$ (e.g., ${2\mathbb{Z}}$, ${5\mathbb{Z}}$, etc.) also has field of fractions ${\mathbb{Q}}$.

The question thus arises of how to properly choose the subring ${\mathcal{O}_{K}}$ of ${K}$. To do this, one considers another characterization of ${\mathbb{Z}}$. The elements of ${\mathbb{C}}$ that are roots of a monic polynomial with coefficients in ${\mathbb{Z}}$ are called the algebraic integers. Let ${D \subset \mathbb{C}}$ denote the set of algebraic integers. Suppose ${r\in\mathbb{Q}}$ is an algebraic integer, satisfying

$\displaystyle x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0 = 0$

where each ${c_i}$ is an integer. Writing ${r = a/b}$, where ${a}$ and ${b}$ are relatively prime, one may substitute to see that

$\displaystyle a^n + c_{n-1} b a^{n-1} + \cdots + c_1 b^{n-1} a + c_0 b^{n} = 0$

Since ${b}$ divides the right-hand side, ${b}$ must divide ${a^{n-1}}$, so the only possibility is for ${b=1}$. Hence, one deduces ${r\in\mathbb{Z}}$.

The only rational algebraic integers are the integers. In symbols, ${\mathbb{Z}= \mathbb{Q} \cap D }$. This suggests for us to set ${\mathcal{O}_{K} = K \cap D}$, but this is not, a priori, even a ring. However, it is an important and nontrivial fact that the algebraic integers ${D}$ really do form a ring.

Proposition 1.1: The algebraic integers form a ring.

Proof 1: Let ${\alpha}$ and ${\beta}$ be algebraic integers, and consider the (monic) minimal polynomials ${p}$, ${q \in\mathbb{Z}[x]}$ of ${\alpha}$, ${\beta}$ respectively. Working in the bigger field ${\mathbb{C}}$, one may write the roots of ${p}$ as ${\alpha = \alpha_1}$, ${\alpha_2}$, ${\cdots}$, ${\alpha_n}$ and the roots of ${q}$ as ${\beta = \beta_1}$, ${\beta_2}$, ${\cdots}$, ${\beta_m}$. Define

$\displaystyle r(x) = \prod_{i=1}^{n} \prod_{j=1}^{m} (x-\alpha_i - \beta_j)$

With a little effort (exercise!), one verifies that each coefficient of ${r}$ is a polynomial expression in the coefficients of ${p}$ and ${q}$, hence ${r\in \mathbb{Z}[x]}$. It follows that ${\alpha+\beta}$ is an algebraic integer. An analogous argument shows ${\alpha\beta}$ is an algebraic integer. $\Box$

Proof 2: Suppose ${\alpha\in\mathbb{C}}$, and that there is a finitely generated ${\mathbb{Z}}$-module ${M \subset \mathbb{C}}$ such that ${\alpha M \subset M}$. Then ${x\mapsto \alpha x}$ is a linear transformation ${T: M\rightarrow M}$. Clearly, ${\alpha}$ is an eigenvalue of ${T}$, so the characteristic polynomial of ${T}$ has ${\alpha}$ as a root. But ${M}$ is a ${\mathbb{Z}}$-module, so the characteristic polynomial of ${T}$ is monic in ${\mathbb{Z}[x]}$. Hence, ${\alpha}$ is an algebraic integer.

Now suppose ${\alpha}$ and ${\beta}$ are algebraic integers, with monic minimal polynomials ${p}$, ${q \in \mathbb{Z}[x]}$. Assume also that ${\deg(p) = n}$ and ${\deg(q) = m}$. Then ${\{\alpha^i \beta^j\}_{0\le i \le n-1, 0\le j\le m-1}}$ generates a ${\mathbb{Z}}$-module ${M}$. One sees that ${(\alpha + \beta)M \subset M}$ and ${(\alpha\beta) M \subset M}$, and concludes ${\alpha + \beta}$ and ${\alpha\beta}$ are algebraic integers. $\Box$

Hence ${K\cap D}$ is a ring, the ring of algebraic integers in ${K}$. In fact, one also has that ${\text{Quot}(K\cap D) = K}$. As ${K/\mathbb{Q}}$ is a finite extension, it is algebraic, so any ${x \in K}$ satisfies some polynomial ${p}$ with rational coefficients. By clearing denominators, one may assume without loss of generality that ${p}$ actually has integer coefficients. Writing

$\displaystyle p(x) = a x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$

one multiplies both sides by ${a^{n-1}}$ to get

$\displaystyle (ax)^n + a_{n-1} (ax)^{n-1} + \cdots a^{n-2} a_1 (ax) + a^{n-1} a_0 = 0$

Therefore, ${ax}$ is an algebraic integer, that is ${ax \in K\cap D}$. The element ${x}$ may then be written as the quotient of ${ax}$ and ${a}$, both of which are in ${K\cap D}$.

Consequently, we define ${\mathcal{O}_{K} = K\cap D}$ and think of it as a reasonably natural analogue of ${\mathbb{Z}}$ inside ${K}$. 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 ${\mathbb{Z}}$ do not end here.