Conductor (ring theory)

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

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

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

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

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

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

where ${\displaystyle \textstyle {\frac {1}{a}}A}$ 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 ∈ B, the conductor is also equal to

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

The conductor is the whole ring A if and only if it contains 1 ∈ 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 ⊆ A, so ${\displaystyle m\in {\mathfrak {f}}(B/A)}$. 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 ${\displaystyle {\mathfrak {f}}(B/A)}$, baB ⊆ aB ⊆ 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 ⊆ J ⊆ A, so by definition J is contained in ${\displaystyle {\mathfrak {f}}(B/A)}$. Conversely, the conductor is an ideal of A, so any ideal contained in it is contained in A. This fact implies that ${\displaystyle {\mathfrak {f}}(B/A)}$ 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

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

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

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 ${\displaystyle {\mathfrak {f}}(B/A)}$.

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

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

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

since we may multiply both sides of the equation xJ ⊆ 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.

Let K be a quadratic extension of Q, and let O_K be its ring of integers. By extending 1 ∈ 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 / cZ 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 / cZ, so it must be cZ.

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.

• Integral element