Conductor (ring theory)

Last updated June 28, 2022

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.

Definition

Let A and B be commutative rings, and assume $A \subseteq B$ . The conductor^[1] of A in B is the ideal

{\mathfrak {f}}(B/A)=\operatorname {Ann} _{A}(B/A).

Here $B / A$ is viewed as a quotient of A-modules, and $Ann$ denotes the annihilator. More concretely, the conductor is the set

{\mathfrak {f}}(B/A)=\{a\in A\colon aB\subseteq A\}.

Because the conductor is defined as an annihilator, it is an ideal of A.

If B is an integral domain, then the conductor may be rewritten as

\{0\}\cup \left\{a\in A\setminus \{0\}:B\subseteq \textstyle {\frac {1}{a}}A\right\},

where $\textstyle {\frac {1}{a}}A$ is considered as a subset of the fraction field of B. That is, if a is non-zero and in the conductor, then every element of B may be written as a fraction whose numerator is in A and whose denominator is a. Therefore the non-zero elements of the conductor are those that suffice as common denominators when writing elements of B as quotients of elements of A.

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 $1 \in B$ , the conductor is also equal to

\{r\in R:rB\subseteq A\}.

Elementary properties

The conductor is the whole ring A if and only if it contains $1 \in A$ and, therefore, if and only if $A = B$ . Otherwise, the conductor is a proper ideal of A.

If the index $m = [B : A]$ is finite, then $mB \subseteq A$ , so $m\in {\mathfrak {f}}(B/A)$ . In this case, the conductor is non-zero. This applies in particular when B is the ring of integers in an algebraic number field and A is an order (a subring for which $B / A$ is finite).

The conductor is also an ideal of B, because, for any $b$ in $B$ and any $a$ in ${\mathfrak {f}}(B/A)$ , $baB \subseteq aB \subseteq A$ . In fact, an ideal J of B is contained in A if and only if J is contained in the conductor. Indeed, for such a J, $JB \subseteq J \subseteq A$ , so by definition J is contained in ${\mathfrak {f}}(B/A)$ . Conversely, the conductor is an ideal of A, so any ideal contained in it is contained in A. This fact implies that ${\mathfrak {f}}(B/A)$ is the largest ideal of A which is also an ideal of B. (It can happen that there are ideals of A contained in the conductor which are not ideals of B.)

Suppose that S is a multiplicative subset of A. Then

S^{-1}{\mathfrak {f}}(B/A)\subseteq {\mathfrak {f}}(S^{-1}B/S^{-1}A),

with equality in the case that B is a finitely generated A-module.

Conductors of Dedekind domains

Some of the most important applications of the conductor arise when B is a Dedekind domain and $B / A$ 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 ${\mathfrak {f}}(B/A)$ .

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:

The ideals of A that are relatively prime to ${\mathfrak {f}}(B/A)$ have unique factorization into products of invertible prime ideals that are coprime to the conductor. In particular, all such ideals are invertible.
If I is an ideal of B that is relatively prime to ${\mathfrak {f}}(B/A)$ , then $I \cap A$ is an ideal of A that is relatively prime to ${\mathfrak {f}}(B/A)$ and the natural ring homomorphism $A/(I\cap A)\to B/I$ is an isomorphism. In particular, I is prime if and only if $I \cap A$ is prime.
If J is an ideal of A that is relatively prime to ${\mathfrak {f}}(B/A)$ , then $JB$ is an ideal of B that is relatively prime to ${\mathfrak {f}}(B/A)$ and the natural ring homomorphism $A/J\to B/JB$ is an isomorphism. In particular, J is prime if and only if JB is prime.
The functions $I\mapsto I\cap A$ and $J\mapsto JB$ define a bijection between ideals of A relatively prime to ${\mathfrak {f}}(B/A)$ and ideals of B relatively prime to ${\mathfrak {f}}(B/A)$ . This bijection preserves the property of being prime. It is also multiplicative, that is, $(I\cap A)(I'\cap A)=II'\cap A$ and $(JB)(J'B)=JJ'B$ .

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 $A = B$ . Because B is a Dedekind domain, J is invertible in B, and therefore

\{x\in K\colon xJ\subseteq J\}=B,

since we may multiply both sides of the equation $xJ \subseteq J$ by J⁻¹. If J is also invertible in A, then the same reasoning applies. But the left-hand side of the above equation makes no reference to A or B, only to their shared fraction field, and therefore $A = B$ . Therefore being an ideal of both A and B implies non-invertibility in A.

Conductors of quadratic number fields

Let K be a quadratic extension of Q, and let $O K$ be its ring of integers. By extending $1 \in O K$ to a Z-basis, we see that every order O in K has the form $Z + cO K$ for some positive integer c. The conductor of this order equals the ideal cO_K. Indeed, it is clear that cO_K is an ideal of O_K contained in O, so it is contained in the conductor. On the other hand, the ideals of O containing cO_K are the same as ideals of the quotient ring $(Z + cO K) / cO K$ . The latter ring is isomorphic to $Z / c Z$ by the second isomorphism theorem, so all such ideals of O are the sum of cO_K with an ideal of Z. Under this isomorphism, the conductor annihilates $Z / c Z$ , so it must be $c Z$ .

In this case, the index $[O K : O]$ 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.

Reference

↑ Bourbaki, Nicolas (1989). Commutative Algebra. Springer. p. 316. ISBN 0-387-19371-5.

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[1] Bourbaki, Nicolas (1989). Commutative Algebra. Springer. p. 316. ISBN 0-387-19371-5.

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