Theorem. Let be a finite field extension and an exponent valuation of the extension field . Then there exists one and only one positive integer such that the function
defined in the base field , is an exponent of .
Proof. The exponent of attains in the set also non-zero values; otherwise would be included in , the ring of the exponent . Since any element of are integral over , it would then be also integral over , which is integrally closed in its quotient field (see theorem 1 in ring of exponent); the situation would mean that and thus the whole would be contained in . This is impossible, because an exponent of attains also negative values. So we infer that does not vanish in the whole . Furthermore, attains in both negative and positive values, since .
Let be such an element of on which attains as its value the least possible positive integer in the field and let be an arbitrary non-zero element of . If
then , and thus on grounds of the choice of . This means that is always divisible by , i.e. that the values of the function in are integers. Because and , the function attains in every integer value. Also the conditions
are in force, whence is an exponent of the field .
Definition. Let be a finite field extension. If the exponent of is tied with the exponent of via the condition (1), one says that induces to and that is the continuation of to . The positive integer , uniquely determined by (1), is the ramification index of with respect to (or with respect to the subfield ).
- 1 S. Borewicz & I. Safarevic: Zahlentheorie. Birkhäuser Verlag. Basel und Stuttgart (1966).