Fractional Ideals

In the next post, we will examine regular local rings of §Krull Dimension, ⁋Definition 12. Before that, we need to define a few concepts.

Invertible Modules

First, we define the following.

Definition 1 For a ring \(A\), an \(A\)-module \(M\) is invertible if \(M\) is finitely generated and, for every prime ideal of \(A\), \(\mathfrak{p}\), \(M_\mathfrak{p}\cong A_\mathfrak{p}\) holds.

For a prime ideal \(\mathfrak{p}\) and a maximal ideal containing \(\mathfrak{p}\), \(\mathfrak{m}\), if \(A_\mathfrak{m}\cong M_\mathfrak{m}\), then \(A_\mathfrak{p}\cong M_\mathfrak{p}\), so it suffices to check the above condition only for maximal ideals.

Now if we define \(M^\ast=\Hom_A(M,A)\), from the fact that \(A\) is commutative we know that \(\Hom_A(M, A)\) is also an \(A\)-module, and furthermore the trace map \(M^\ast\otimes M \rightarrow A\) also exists. ([Multilinear Algebra] §Hom and the Tensor Product, ⁋Definition 6)

Definition 2 Consider a ring \(A\) and the total ring of fractions of \(A\), \(K\). Then, in \(K\), an \(A\)-submodule \(\mathfrak{A}\) is a fractional ideal of \(A\) if there exists a non-zerodivisor of \(A\), \(a\), such that \(a \mathfrak{A}\subseteq A\).

It is clear that \(a \mathfrak{A}\) obtained from the definition above is an ideal of \(A\). Intuitively, if \(\mathfrak{A}\) is finitely generated, the condition \(a \mathfrak{A}\subseteq A\) amounts to multiplying by a common denominator of the generators of \(\mathfrak{A}\) to view this as a subset of \(A\). In particular, in \(K\), any finitely generated \(A\)-submodule

\[\left(\frac{a_1}{s_1},\ldots, \frac{a_n}{s_n}\right)A\]

is always a fractional ideal via the element \(s_1\cdots s_n\in A\), and if \(A\) is Noetherian, the converse also holds. As with ordinary ideals, the product of fractional ideals is also

\[\mathfrak{A}\mathfrak{B}=\left\{\sum_{i=1}^n a_ib_i\mid a_i\in \mathfrak{A}, b_i\in \mathfrak{B}\right\}\]

defined in this way.

In the following theorem, for an arbitrary subset of \(K\), \(X\), we have

\[X^{-1}=(A:_KX)=\{y\in K\mid yX\subseteq A\}\]

and according to the intuition above, this can roughly be thought of as the collection of denominators of \(X\).

Theorem 3 For a Noetherian ring \(A\), the following hold.

1. An \(A\)-module \(M\) is invertible if and only if the trace map \(M^\ast\otimes_A M \rightarrow A\) is an isomorphism.
2. Every invertible module is isomorphic to some fractional ideal of \(A\). Moreover, every invertible \(A\)-submodule of \(K\) is itself a fractional ideal of \(A\), and every invertible fractional ideal obtained in this way contains a non-zerodivisor of \(A\).

3. For two invertible \(A\)-submodules \(M,N\) of \(K\), the morphisms defined by

   \[M\otimes N \rightarrow MN;\quad s\otimes t\mapsto st,\qquad M^{-1}N \rightarrow \Hom_A(M,N);\quad t\mapsto u_t(-)=t-\]

   are isomorphisms. In particular, \(M^{-1}\cong M^\ast\) holds.

4. For any \(A\)-submodule \(M\subseteq K\), \(M\) is invertible if and only if \(M^{-1}M=A\).

Proof

1. First, one direction is trivial by identifying \((M^\ast)_\mathfrak{m}\) with \((M_\mathfrak{m})^\ast\) via §Properties of Localization, ⁋Proposition 5 and then applying §Properties of Localization, ⁋Proposition 4.
   Conversely, suppose the trace map \(\tr:M^\ast\otimes_A M \rightarrow A\) is an isomorphism; we must show that \(M_\mathfrak{p}\cong A_\mathfrak{p}\). Here, since \(\tr\) is an isomorphism, there exists an element of \(M^\ast\otimes M\) such that

   \[\tr\left(\sum_{i=1}^n\xi_i\otimes x_i\right)=1\]

   and we show that \(M_\mathfrak{p}\) is generated by \(x_i\). Then \(M\) must be generated by these \(x_1,\ldots, x_n\) by §Properties of Localization, ⁋Proposition 4.

   Again through this proposition, from the given hypothesis we obtain the isomorphism

   \[\tr_\mathfrak{p}: (M^\ast\otimes_AM)_\mathfrak{p}\cong M^\ast_\mathfrak{p}\otimes_{A_\mathfrak{p}}M_\mathfrak{p} \rightarrow A_\mathfrak{p}\]

   for any prime ideal \(\mathfrak{p}\). Similarly, localizing \(\xi_i:M \rightarrow A\) defines \((\xi_i)_\mathfrak{p}: M_\mathfrak{p}\rightarrow A_\mathfrak{p}\), and these will be the desired isomorphisms.
   Meanwhile, for a fixed \(\mathfrak{p}\), since \(1\not\in \mathfrak{p}\), at least one of the \(\xi_i\otimes x_i\) chosen above must satisfy \(\tr(\xi_i\otimes x_i)=\xi_i(x_i)\not\in \mathfrak{p}\). Then in \(A_\mathfrak{p}\) the inverse \(\xi_i(x_i)^{-1}\) of this element exists, and writing this as \(a_i\) for convenience, from the equation

   \[(\xi_i)_\mathfrak{p}(a_i x_i)=1\]

   we see that \((\xi_i)_{\mathfrak{p}}: M_\mathfrak{p} \rightarrow A_\mathfrak{p}\) sends \(a_i x_i\) to \(1\). Now defining \(A_\mathfrak{p} \rightarrow M_\mathfrak{p}\) by \(1\mapsto a_ix_i\), this is a section of \((\xi_i)_\mathfrak{p}\), and therefore the short exact sequence

   \[0 \longrightarrow \ker (\xi_i)_\mathfrak{p}\longrightarrow M_\mathfrak{p}\overset{(\xi_i)_\mathfrak{p}}{\longrightarrow}A_\mathfrak{p}\longrightarrow 0\]

   splits. From this we see that \(M_\mathfrak{p}\cong A_\mathfrak{p}x_i\oplus\ker(\xi_i)_\mathfrak{p}\). Similarly, viewing \(x_i\) as the evaluation map \(M^\ast_\mathfrak{p} \rightarrow A_\mathfrak{p};\xi\mapsto \xi(x_i)\), we obtain \(M_\mathfrak{p}^\ast\cong A_\mathfrak{p}\xi_i\oplus \ker(x_i)_\mathfrak{p}\), and now

   \[M^\ast_\mathfrak{p}\otimes M_\mathfrak{p}\cong (A_\mathfrak{p}\xi_i\otimes A_\mathfrak{p}x_i)\oplus ( A_\mathfrak{p}\xi_i\otimes\ker (\xi_i)_\mathfrak{p})\oplus(\ker(x_i)_\mathfrak{p}\otimes A_\mathfrak{p}x_i)\oplus (\ker(x_i)_\mathfrak{p}\otimes \ker(\xi_i)_\mathfrak{p})\]

   holds. However, since the first term on the right-hand side recovers all elements of \(A_\mathfrak{p}\) exactly, when mapped under \(\tr_\mathfrak{p}\) the remaining terms must go to \(0\), and in particular from the fact that the second term goes to \(0\) we see that \((\ker\xi_i)_\mathfrak{p}=\ker(\xi_i)_\mathfrak{p}=0\). That is, \(\xi_i\) is an isomorphism from \(M_\mathfrak{p}\) to \(A_\mathfrak{p}\) as claimed above, and through this we may regard \(M_\mathfrak{p}\) as the free \(A_\mathfrak{p}\)-module generated by \(x_i\).

2. Now let \(M\) be an invertible module. To compare it with a fractional ideal of \(A\), we first need to place \(M\) into \(K\). But first, one can check that the maximal ideals of the total ring of fractions \(K\) of \(A\) correspond exactly to the maximal associated prime ideals of \(A\), and since \(\Ass A\) is finite, \(K\) is a semilocal ring. Therefore, from §Integral Extensions, ⁋Proposition 13 and the isomorphisms

   \[M\otimes K_{\mathfrak{m}K}=M_\mathfrak{m}\cong A_\mathfrak{m}\cong K_{\mathfrak{m}K}\]

   we obtain \(M\otimes K\cong K\). Now composing this with the localization map \(\epsilon: M \rightarrow S^{-1}M=K\otimes M\) yields the desired embedding, where the fact that \(\epsilon\) is injective is also obtained by applying §Properties of Localization, ⁋Proposition 4 to any maximal ideal \(\mathfrak{m}\) of \(A\) to obtain the map

   \[\epsilon_\mathfrak{m}: M_\mathfrak{m}\cong A_\mathfrak{m} \rightarrow K\otimes_{A_\mathfrak{m}} A_\mathfrak{m}=S^{-1}A_\mathfrak{m}\]

   and because the elements of \(S\) (that is, the non-zerodivisors of \(A\)) are non-zerodivisors of \(A_\mathfrak{m}\).
   Meanwhile, that an invertible \(A\)-submodule of \(K\) is a fractional ideal of \(A\) is a separate argument: by Definition 1, such a submodule is finitely generated, so writing its generators as \(a_1/s_1,\ldots, a_n/s_n\), each \(s_i\) is a non-zerodivisor of \(A\), and thus their product \(s_1\cdots s_n\) is also a non-zerodivisor; multiplying by this sends each generator into \(A\), so the condition of Definition 2 is satisfied by this element.
   Now suppose a fractional ideal \(\mathfrak{A}\) is given, and suppose \(\mathfrak{A}\cap A\) consists only of zerodivisors. Then from \(\mathfrak{A}\) being a (finitely generated) fractional ideal, we can find a common denominator \(a\) such that \(a \mathfrak{A}\subseteq A\) is an ideal of \(A\). Now applying §Associated Primes, ⁋Theorem 7, since \(a\mathfrak{A}\) is an ideal of \(A\) consisting only of zerodivisors, it is contained in the union of associated prime ideals, and applying §Associated Primes, ⁋Lemma 2 (Prime avoidance lemma) to this, we see that \(a\mathfrak{A}\) actually annihilates some \(b\in A\). That is, \(ab\) annihilates \(\mathfrak{A}\), and therefore, localizing at a prime ideal \(\mathfrak{p}\) containing \(\ann(ab)\), we see that \(\mathfrak{A}_\mathfrak{p}\not\cong A_\mathfrak{p}\).

3. Let two invertible modules \(M,N\) be given. Then by the second result, we may regard them as fractional ideals contained in \(K\), and the maps given in the claim are also defined in this way. Since we will show anyway that the given morphisms are isomorphisms via §Properties of Localization, ⁋Proposition 4, we may assume from the start that \(A\) is local. In this reduction, that \(\Hom_A(M,N)\) commutes with localization is obtained by §Properties of Localization, ⁋Proposition 5 from the fact that \(A\) is Noetherian and \(M\) is finitely presented since it is finitely generated. Also, \(M^{-1}=(A:_KM)\) is the intersection of the \((A:_Km_i)\) for the finitely many generators \(m_i\) of \(M\), and since localization is exact it commutes with finite intersections; and since \(MN\) is generated by products of generators of \(M,N\), these two also commute with localization. Then by the argument in the second result and the definition of an invertible module, \(M,N\) are both isomorphic to \(A\). Now let \(s,t\) be the non-zerodivisors of \(K\) generating \(M,N\), respectively. The first morphism is an epimorphism to begin with, and additionally, viewing \(M\otimes_A N\) as \(A\cong As\otimes_AAt\), the map \(M\otimes N \rightarrow MN\) can be understood as sending \(1\otimes1\) to \(st\), so from the fact that \(st\) is a non-zerodivisor we see that it is also a monomorphism. For the second morphism, first by the second result we can choose a suitable non-zerodivisor \(a\in A\cap M\). Then for any nonzero \(t\in M^{-1}N\), since \(ta\neq 0\), \(u_t\) is not the zero morphism, and thus the morphism in the claim is a monomorphism. That this is an epimorphism holds because for any \(u\in \Hom_A(M,N)\), setting \(u(s)=y\) gives \(u=u_{y/s}\). In particular, setting \(N=A\) yields the last claim.
4. First, if \(M\) is invertible, then by result 3 we may view \(M^{-1}\otimes M \rightarrow M^{-1}M\) and the trace map \(M^\ast\otimes M \rightarrow A\) as the same. Conversely, suppose an arbitrary \(A\)-submodule \(M\) of \(K\) satisfies \(M^{-1}M=A\). Then that \(M\) satisfies the finitely generated condition of Definition 1 follows from this, because choosing \(y_i\in M^{-1}\) and \(m_i\in M\) such that \(1=\sum_{i=1}^ny_im_i\), for any \(m\in M\) we have \(y_im\in A\), so \(m=\sum_{i=1}^n(y_im)m_i\) is an \(A\)-linear combination of \(m_1,\ldots, m_n\). Now, as above, through localization we may assume that \((A,\mathfrak{m})\) is a local ring and show that \(M\cong A\). But by the condition \(M^{-1}M=A\), we can choose some \(y\in M^{-1}\) such that \(yM\not\subseteq \mathfrak{m}\), and then by the maximality of \(\mathfrak{m}\) we must have \(yM=A\), from which we obtain the isomorphism \(y-\) between \(M\) and \(A\).

Theorem 4 For a Noetherian UFD \(R\), we have \(\Pic(R)=0\). That is, every invertible module over \(R\) is free.

Proof

Let \(R\) be a Noetherian UFD. By the second result of Theorem 3, every invertible \(R\)-module is isomorphic to some fractional ideal of \(R\), so it suffices to show that when \(I\) is an invertible fractional ideal of \(R\), \(I\) is principal. Without loss of generality, assume \(I \subseteq R\) (otherwise, choose a suitable \(s \in R\) so that \(sI \subseteq R\); if \(sI\) is principal, then \(I\) is also principal).

Since \(I\) is invertible, by Definition 1, at every prime \(\mathfrak{p}\) we have \(I_\mathfrak{p} \cong R_\mathfrak{p}\). Thus \(I\) is locally principal. For any height 1 prime \(\mathfrak{p}\), the localization \(R_\mathfrak{p}\) is a DVR (since \(R\) is a UFD, \(R_\mathfrak{p}\) is a \(1\)-dimensional normal local ring), and since every nonzero fractional ideal in a DVR is principal, for some \(v_\mathfrak{p}(I) \in \mathbb{Z}\) we have \(I_\mathfrak{p} = (p_\mathfrak{p}^{v_\mathfrak{p}(I)})\). Here, since \(R\) is a UFD, every height 1 prime \(\mathfrak{p}\) is, for a prime element \(p_\mathfrak{p}\in R\), of the form \(\mathfrak{p}=(p_\mathfrak{p})\), and we have chosen this \(p_\mathfrak{p}\) as a uniformizer of \(R_\mathfrak{p}\); \(v_\mathfrak{p}\) is the normalized valuation of \(R_\mathfrak{p}\). Since \(I \subseteq R\), we have \(v_\mathfrak{p}(I) \ge 0\).

However, \(v_\mathfrak{p}(I) > 0\) holds for only finitely many \(\mathfrak{p}\) (since \(I\) is finitely generated). Let \(a = \prod_\mathfrak{p} p_\mathfrak{p}^{v_\mathfrak{p}(I)}\). This is a finite product of prime elements of \(R\), hence an element of \(R\), and each \((p_\mathfrak{p}^{v_\mathfrak{p}(I)})\) is also an ideal of \(R\). Then

\[I = \bigcap_{\mathfrak{p}} I_\mathfrak{p} \cap R = \bigcap_{\mathfrak{p}} (p_\mathfrak{p}^{v_\mathfrak{p}(I)}) = (a)\]

holds. In the first equality, the \(\subseteq\) direction is obvious, and the reverse direction follows from the fact that if \(x \in R\) and for every height 1 prime \(\mathfrak{p}\), \(v_\mathfrak{p}(x) \ge v_\mathfrak{p}(I)\), then comparing valuations at each \(\mathfrak{p}\) in the UFD gives \(x/a \in R\), i.e., \(x \in (a)\). In the last equality, \(\bigcap_{\mathfrak{p}} (p_\mathfrak{p}^{v_\mathfrak{p}(I)})\) is the set of elements satisfying, for all \(\mathfrak{p}\), \(v_\mathfrak{p}(x) \ge v_\mathfrak{p}(I)\) where \(x \in R\), so it coincides with \((a)\). Therefore \(I\) is a principal ideal.

Consider the collection of isomorphism classes of invertible modules defined over a Noetherian ring \(A\). Then \(\otimes\) preserves these isomorphism classes and thus defines a binary operation on this collection; \(\otimes\) satisfies the associative and commutative laws and has the identity element \(A\). Furthermore, by the first result of Theorem 3, every invertible module has with respect to \(\otimes\) an inverse \(M^\ast\). From this, we see that this collection forms an abelian group.

Similarly, within \(K\), the invertible \(A\)-submodules (i.e., the invertible fractional ideals of \(A\)) also carry a group structure via the ideal product, where the fourth condition of Theorem 3 shows that the inverse of \(M\) is \(M^{-1}\). We name these as follows.

Definition 5 For a Noetherian ring \(A\), we define the following.

1. The Picard group of \(A\), \(\Pic(A)\), is the group under \(\otimes\) of isomorphism classes of invertible \(A\)-modules.
2. The group of Cartier divisors of \(A\), \(\CaDiv(A)\), is the group of submodules of \(K\) that are invertible \(A\)-submodules, that is, the group of invertible fractional ideals of \(A\).

Then the following is immediate from Theorem 3.

Corollary 6 For a Noetherian ring \(A\), the following hold.

1. Taking from \(K\) any invertible \(A\)-submodule and returning its isomorphism class, the function \(\CaDiv(A) \rightarrow \Pic(A)\) is surjective, and its kernel is isomorphic to \(K^\times/A^\times\).
2. \(\CaDiv(A)\) is the abelian group generated by the invertible ideals of \(A\).

Proof

1. The surjectivity of the given function is the second result of Theorem 3, and in \(K\), for any unit \(x\), \(Ax\subseteq K\) is an invertible module that is sent to \(A\) by this function. Therefore, since any invertible submodules \(M,N\) being isomorphic implies that for some \(x\in K^\times\) we can write \(M=xN\), the claim about the kernel can also be shown.
2. For any invertible fractional ideal \(\mathfrak{A}\), by the second and fourth results of Theorem 3, \(\mathfrak{A}^{-1}\) is also an invertible fractional ideal, and then again by the second result of Theorem 3, \(\mathfrak{A}^{-1}\) contains a non-zerodivisor of \(A\). Letting this be \(a\), since \(a \mathfrak{A}\subseteq A\), we have \(\mathfrak{A}=a \mathfrak{A}\cdot (a)^{-1}\).

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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