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.

Definition.

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

Note

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.

Definition.

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 $N\ne 0$.

Proof

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:

Definition.

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.

Note

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

Examples

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.

Proof

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

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

Proof

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 $(M : N) \subseteq MN^{-1}$. ♦

Note

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 $(M : N)$.

Exercise

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

Definition.

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

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

Note

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.

Proof

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 $xA \subseteq \mathfrak a$. ♦

Localization

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

Note

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

Proof

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 $\mathrm{Pic} A$.

Example

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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