Commutative Algebra 44

Fractional Ideals

Throughout this article, let A be an integral domain and K its field of fractions. We do not assume the ring to be noetherian. The objective here is to develop the theory of multiplying and dividing certain classes of non-zero ideals so that we get a group structure.


fractional ideal of A is an A-submodule M\subseteq K such that there is an a\in A-\{0\} such that aM\subseteq A.


The condition a M \subseteq A ensures that the elements of M do not have “too many denominators” occurring in them. Thus \mathbb Q is not a fractional ideal of \mathbb Z because all positive integers can appear as denominators among a\in \mathbb Q.

Easy Exercise

1. Let M\subseteq K be an A-submodule; prove that M is a fractional ideal if and only if there exists a\in K - \{0\} such that aM \subseteq A.

2. Find an example of A where K is a fractional ideal of A.

We can copy and paste many of the ideal constructions here.


Let M, N be fractional ideals of A. We have M\cap N, M+N, which are A-submodules of K. We also define the following.

  • MN is the set of finite sums x_1 y_1 + \ldots + x_n y_n, where x_i \in M, y_i \in N.
  • (M : N) is the set of all x\in K such that xN\subseteq M.

As before, M\cap N represents the “lcm” of M and N, M+N represents their “gcd”, while MN, (M : N) represent “product” and “division” respectively.

Be aware that unlike the case of ideals, in the definition of (M : N) we have to consider all x\in K.

Proposition 1.

If M and N are fractional ideals of A, so are M \cap N, M+N, MN and; also (M:N) is a fractional ideal if N\ne 0.


It is easy to show they are all A-submodules of K. Now pick a,b\in A-\{0\} such that aM\subseteq A and bN \subseteq A. Then

\begin{aligned} ab(M+N) &= abM + abN \subseteq bA + aA \subseteq A, \\ a(M\cap N) &\subseteq aM \subseteq A, \\ ab(MN) &= (aM)(bN) \subseteq A\cdot A= A. \end{aligned}

For the last case pick any x\in N - \{0\}; then ax(M:N) \subseteq A because

y \in (M : N) \implies yN \subseteq M \implies yx \in M \implies axy \in A.

Exercise A

Write down and prove as many relations as you can for the above four operations. E.g.

AM = M, \quad (M_1 + M_2)M_3 = M_1 M_3 + M_2 M_3, \quad (M_1 M_2)M_3 = M_1(M_2 M_3).


Invertible Fractional Ideals

To avoid complications, henceforth we will only consider non-zero fractional ideals.

Note that the set of fractional ideals of A forms a commutative monoid under product, with identity A. In other words, it satisfies all axioms of a group except the existence of inverse. Hence we define:


Let M be a fractional ideal of A.

We say M is principal if it is of the form xA for some x\in M; we say it is invertible if MN = A for some fractional ideal N, in which case we write N = M^{-1}.

An ideal of A is said to be invertible if it is invertible as a fractional ideal.


Clearly principal fractional ideals are invertible. The following examples show that the converse is not true, which is why the theory is interesting.


1. Let A = \mathbb Z[\sqrt{-5}], a non-UFD. The prime ideal \mathfrak p = (2, 1 + \sqrt{-5}) is not principal but it is invertible because

(2, 1 + \sqrt{-5})\cdot (2, 1-\sqrt{-5}) = (4, 2(1+\sqrt{-5}), 2(1-\sqrt{-5}), 6) = (2).

2. Let A = \mathbb C[X, Y]/(Y^2 - X^3 + X). The maximal ideal \mathfrak m = (X, Y) \subset A is invertible because

(X, Y)^2 = (X^2, XY, Y^2) = (X^2, XY, X^3 - X) = (X).

Proposition 2.

An invertible fractional ideal of A is finitely generated as an A-module.


Suppose MN = A; write 1 = x_1 y_1 + \ldots + x_n y_n for x_i \in M, y_i \in N. We claim x_1, \ldots, x_n generate M as an A-module. Indeed let x\in M. Then

x = 1\cdot x = (x_1 y_1 + \ldots + x_n y_n)x = x_1 (y_1 x) + \ldots + x_n (y_n x)

and each xy_i \in MN \in A. ♦

Clearly the set of invertible fractional ideals of A forms a group under product, for if M and N are invertible so is MN. The identity of this group is A. The inverse is given by the following.

Proposition 3.

Let N be an invertible fractional ideal of A. For any fractional ideal M, we have

(M : N) = MN^{-1}.


We have (MN^{-1})N = M and thus MN^{-1} \subseteq (M : N) by definition. Conversely (M : N)N \subseteq M and multiplying both sides by N^{-1} gives (M : N) \subseteq MN^{-1}. ♦


Two special cases are important here.

  • Let MA: we have N^{-1} = (A: N) = \{x \in K : xN \subseteq A\}.
  • If M and N are invertible, so is (M : N).


Prove that a fractional ideal is free if and only if it is principal.

Prove that in the ring k[X, Y]/(Y^2 - X^3 + X), where k is a field, the maximal ideal \mathfrak m = (X, Y) is invertible. Prove also that \mathfrak n = (X-1, Y) is invertible if \mathrm{char} k \ne 2.


Picard Group


The set of principal ideals forms a subgroup of the group of invertible fractional ideals. The resulting quotient is called the Picard group of A, i.e.

\mathrm{Pic} A = (\text{invertible fractional ideals of } A) / (\text{principal fractional ideals of } A).


Every element of the Picard group can be represented by an invertible ideal \mathfrak a \subseteq A.

We immediately have:

Proposition 4.

The Picard group of a UFD is trivial.


Let \mathfrak a be an invertible ideal of a UFD A. By proposition 2, we can find a finite set of generators x_1, \ldots, x_n of M. We claim that x = \gcd(x_i) generates \mathfrak a.

Since x | x_i for each i we have x_i \in xA and thus \mathfrak a\subseteq xA.

Conversely let y\in \mathfrak a^{-1} so y\mathfrak a \subseteq A. For each i, x_i y \in A and thus xy = \gcd(x_i) y = \gcd(x_i y) \in A. So y \in x^{-1}A. Thus \mathfrak a^{-1} \subseteq x^{-1}A and so xA \subseteq \mathfrak a. ♦



Now suppose S\subseteq A is a multiplicative subset not including 0.

We get a map from the set of fractional ideals of A to that of S^{-1}A. Indeed if M\subseteq K is an A-submodule satisfying aM\subseteq A then S^{-1}M \subseteq S^{-1}K = K is an (S^{-1}A)-submodule satisfying \frac a 1 (S^{-1}M) \subseteq S^{-1} A. Clearly if M is invertible (resp. principal), so is S^{-1}M.

Localization preserves the four operations we defined:

Proposition 5.

For any fractional ideals M, N of A, we have

  • S^{-1}(M + N) = S^{-1}M + S^{-1}N;
  • S^{-1}(M \cap N) = S^{-1}M \cap S^{-1}N;
  • S^{-1}(MN) = (S^{-1}M)(S^{-1}N);
  • if N\ne 0 is finitely generated, then S^{-1}(M : N) = (S^{-1}M : S^{-1}N).


In particular, if M is an invertible fractional ideal of A, then S^{-1}M is an invertible fractional ideal of S^{-1}A.


The first two properties follow from that of general submodules. The third is easy. For the last, we have

\begin{aligned} &S^{-1}(M:N) S^{-1}N = S^{-1}((M : N)N) \subseteq S^{-1}M \\ \implies &S^{-1}(M:N) \subseteq (S^{-1}M : S^{-1}N).\end{aligned}

For the reverse inclusion suppose \frac y s S^{-1}N \subseteq S^{-1}M for \frac y s \in S^{-1}A-\{0\}. Write N = Ax_1 + \ldots + Ax_n; then \frac{yx_i} 1 \in S^{-1}M for each i. Write \frac {yx_i} 1 = \frac {z_i}{s_i} for z_i \in M and s_i \in S; we see that tyx_i \in M for each i where t = s_1 \ldots s_n. [ Recall that A is an integral domain. ] Hence

tyN\subseteq M \implies ty \in (M : N)\implies \frac y s \in S^{-1}(M:N). ♦

Corollary 1.

We obtain a group homomorphism

\mathrm{Pic} A \longrightarrow \mathrm{Pic} S^{-1}A,\quad [M] \mapsto [S^{-1}M].

where M is an invertible ideal of A representing an element of \mathrm{Pic} A.


Let A = \mathbb Z[\sqrt{-5}] with \mathfrak p =(2, 1 + \sqrt{-5}) \subset A, a non-principal invertible ideal. If S = \{2^n\} then \mathrm{Pic} A \to \mathrm{Pic} S^{-1}A takes \mathfrak p \mapsto 1 since \mathfrak p \cdot S^{-1}A is generated by (1 + \sqrt{-5}).


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5 Responses to Commutative Algebra 44

  1. Vanya says:

    I guess the “M” in ” By proposition 2, we can find a finite set of generators x_1, \ldots, x_n of M” should be \mathfrak{a}.

  2. Vanya says:

    Is it true that $M

  3. Vanya says:

    In the definition of Picard group, should it be “invertible fractional ideals” instead of just “fractional ideals”? Also, can you give an example of a fractional ideal that is not invertible?

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