In commutative algebra, the norm of an ideal is a generalization of a norm of an element in the field extension. It is particularly important in number theory since it measures the size of an ideal of a complicated number ring in terms of an ideal in a less complicated ring. When the less complicated number ring is taken to be the ring of integers, Z, then the norm of a nonzero ideal I of a number ring R is simply the size of the finite quotient ring R/I.
Relative norm Let A be a Dedekind domain with field of fractions K and integral closure of B in a finite separable extension L of K. (this implies that B is also a Dedekind domain.) Let I A {\displaystyle {\mathcal {I}}_{A}} and I B {\displaystyle {\mathcal {I}}_{B}} be the ideal groups of A and B, respectively (i.e., the sets of nonzero fractional ideals.) Following the technique developed by Jean-Pierre Serre, the norm map
N B / A : I B → I A {\displaystyle N_{B/A}\colon {\mathcal {I}}_{B}\to {\mathcal {I}}_{A}}
is the unique group homomorphism that satisfies
N B / A ( q ) = p [ B / q : A / p ] {\displaystyle N_{B/A}({\mathfrak {q}})={\mathfrak {p}}^{[B/{\mathfrak {q}}:A/{\mathfrak {p}}]}}
for all nonzero prime ideals q {\displaystyle {\mathfrak {q}}} of B, where p = q ∩ A {\displaystyle {\mathfrak {p}}={\mathfrak {q}}\cap A} is the prime ideal of A lying below q {\displaystyle {\mathfrak {q}}} .
Alternatively, for any b ∈ I B {\displaystyle {\mathfrak {b}}\in {\mathcal {I}}_{B}} one can equivalently define N B / A ( b ) {\displaystyle N_{B/A}({\mathfrak {b}})} to be the fractional ideal of A generated by the set { N L / K ( x ) | x ∈ b } {\displaystyle \{N_{L/K}(x)|x\in {\mathfrak {b}}\}} of field norms of elements of B. For a ∈ I A {\displaystyle {\mathfrak {a}}\in {\mathcal {I}}_{A}} , one has N B / A ( a B ) = a n {\displaystyle N_{B/A}({\mathfrak {a}}B)={\mathfrak {a}}^{n}} , where n = [ L : K ] {\displaystyle n=[L:K]} . The ideal norm of a principal ideal is thus compatible with the field norm of an element:
N B / A ( x B ) = N L / K ( x ) A . {\displaystyle N_{B/A}(xB)=N_{L/K}(x)A.}
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