Algebraic Integers
Inside the rationals, the "integers" are obvious. Inside a
number field, what
should play that role? The answer — the algebraic integers — is subtler and more
beautiful than you might guess, and getting it exactly right is what makes arithmetic in number
fields work.
The definition
A number is an algebraic integer if it is a root of a monic polynomial
(leading coefficient 1) with ordinary integer coefficients:
x^{n} + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0 = 0, \qquad c_i \in \mathbb{Z}.
The word "monic" is the whole subtlety. \sqrt 2 is an algebraic integer
(root of x^2 - 2); so is i. But
\tfrac12 is not — its minimal polynomial
2x - 1 isn't monic over \mathbb Z. Restricted to
\mathbb{Q}, the algebraic integers are exactly the ordinary integers
\mathbb{Z} — so the definition genuinely extends "integer".
A surprise: a hidden half
You might expect the integers of \mathbb{Q}(\sqrt d) to be just
a + b\sqrt d with a, b \in \mathbb{Z}. Usually
yes — but when d \equiv 1 \pmod 4, extra "half-integer" elements sneak in.
For d = 5, the golden ratio
\tfrac{1 + \sqrt 5}{2} is an algebraic integer — it satisfies the monic
x^2 - x - 1 = 0. The full ring of integers is then
\mathbb{Z}\!\left[\tfrac{1 + \sqrt 5}{2}\right],
larger than the naive guess. Getting this ring right is essential — guess wrong and the arithmetic breaks.
They form a ring
The algebraic integers of a number field are closed under addition and multiplication — they form
the ring of integers \mathcal{O}_K, the proper stage for
all arithmetic in K. In \mathbb{Q}(i) it is the
Gaussian integers.
The looming question — does
unique factorisation
still hold in \mathcal{O}_K? — is the subject of the next page, and the
answer is a shock.