In ring theory, a branch of mathematics, the conductor is a measurement of how far apart a commutative ring and an extension ring are. Most often, the larger ring is a domain integrally closed in its field of fractions, and then the conductor measures the failure of the smaller ring to be integrally closed.
The conductor is of great importance in the study of non-maximal orders in the ring of integers of an algebraic number field. One interpretation of the conductor is that it measures the failure of unique factorization into prime ideals.
Let A and B be commutative rings, and assume . The conductor[1] of A in B is the ideal
ak{f}(B/A)=\operatorname{Ann}A(B/A).
ak{f}(B/A)=\{a\inA\colonaB\subseteqA\}.
If B is an integral domain, then the conductor may be rewritten as
\{0\}\cup\left\{a\inA\setminus\{0\}:B\subseteq
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\right\},
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Suppose R is a ring containing B. For example, R might equal B, or B might be a domain and R its field of fractions. Then, because, the conductor is also equal to
\{r\inR:rB\subseteqA\}.
The conductor is the whole ring A if and only if it contains and, therefore, if and only if . Otherwise, the conductor is a proper ideal of A.
If the index is finite, then, so
m\inak{f}(B/A)
The conductor is also an ideal of B, because, for any in and any in
ak{f}(B/A)
ak{f}(B/A)
ak{f}(B/A)
Suppose that S is a multiplicative subset of A. Then
S-1ak{f}(B/A)\subseteqak{f}(S-1B/S-1A),
Some of the most important applications of the conductor arise when B is a Dedekind domain and is finite. For example, B can be the ring of integers of a number field and A a non-maximal order. Or, B can be the affine coordinate ring of a smooth projective curve over a finite field and A the affine coordinate ring of a singular model. The ring A does not have unique factorization into prime ideals, and the failure of unique factorization is measured by the conductor
ak{f}(B/A)
Ideals coprime to the conductor share many of pleasant properties of ideals in Dedekind domains. Furthermore, for these ideals there is a tight correspondence between ideals of B and ideals of A:
ak{f}(B/A)
ak{f}(B/A)
ak{f}(B/A)
A/(I\capA)\toB/I
ak{f}(B/A)
ak{f}(B/A)
A/J\toB/JB
I\mapstoI\capA
J\mapstoJB
ak{f}(B/A)
ak{f}(B/A)
(I\capA)(I'\capA)=II'\capA
(JB)(J'B)=JJ'B
All of these properties fail in general for ideals not coprime to the conductor. To see some of the difficulties that may arise, assume that J is a non-zero ideal of both A and B (in particular, it is contained in, hence not coprime to, the conductor). Then J cannot be an invertible fractional ideal of A unless . Because B is a Dedekind domain, J is invertible in B, and therefore
\{x\inK\colonxJ\subseteqJ\}=B,
Let K be a quadratic extension of Q, and let be its ring of integers. By extending to a Z-basis, we see that every order O in K has the form for some positive integer c. The conductor of this order equals the ideal cOK. Indeed, it is clear that cOK is an ideal of OK contained in O, so it is contained in the conductor. On the other hand, the ideals of O containing cOK are the same as ideals of the quotient ring . The latter ring is isomorphic to by the second isomorphism theorem, so all such ideals of O are the sum of cOK with an ideal of Z. Under this isomorphism, the conductor annihilates, so it must be .
In this case, the index is also equal to c, so for orders of quadratic number fields, the index may be identified with the conductor. This identification fails for higher degree number fields.