By Robin Chapman

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**Sample text**

Let δ be any unit. Consider log |δ|. There exists an integer j such that j log η ≤ log |δ| < (j + 1) log η and so 0 ≤ log |δ|η −j < log η. Now |δ|η −j is a unit and 1 ≤ |δ|η −j < η. Thus |δ|η −j = 1 and so δ = ±η j . It is easily seen that the unit η in this theorem is uniquely determined by the field K, as it is called the fundamental unit of K. √ √ Example Let K = Q( 2). Then η = 1 + 2 is a unit in OK , and obviously η > 1. I claim that it is √the fundamental unit. If it were not, then there √ would be a unit ξ = a + b 2 of OK with 1 < ξ < η.

In general we cannot always find an integral basis of this form for a given K. But in special cases, for instance the important case of quadratic fields, we can. We will need to factorize polynomials modulo p. Recall that for a prime number p, the set Fp = {0, 1, 2, . . , p − 1} with addition and multiplication modulo p forms a field. Here we have written elements of Fp in the form a to distinguish them from integers, but in practice we are usually sloppy and write a both for an integer and the corresponding element of Fp .

P − 1} with addition and multiplication modulo p forms a field. Here we have written elements of Fp in the form a to distinguish them from integers, but in practice we are usually sloppy and write a both for an integer and the corresponding element of Fp . If f ∈ Z[X] is a polynomial with integer coefficients then f ∈ Fp [X] denotes its reduction modulo p, that is a0 + a1 X + a2 X 2 + · · · + Am X m = a0 + a1 X + a2 X 2 + · · · + am X m . When OK = Z[α] we can classify all ideals containing p .