K

Grad. student in mathematics

가환대수학

Fractional ideals, invertible modules, and the Picard group

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In the following we examine regular local rings (§Dimension, ⁋Definition 12). Before doing so, we must define several concepts.

Invertible Modules

First, we define the following.

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

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

Now define \(M^\ast=\Hom_A(M,A)\). Since \(A\) is commutative, \(\Hom_A(M, A)\) is again an \(A\)-module, and moreover the trace map \(M^\ast\otimes M \rightarrow A\) exists. (§Hom and the Tensor Product, ⁋Definition 6)

Definition 2 Let \(A\) be a ring and let \(K\) be the total ring of fractions of \(A\). Then an \(A\)-submodule \(\mathfrak{A}\) of \(K\) is called a fractional ideal of \(A\) if there exists a nonzero element \(a\in A\) such that \(a \mathfrak{A}\subseteq A\).

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

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

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

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

In the following theorem, for any subset \(X\) of \(K\),

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

and, following the above intuition, this can be thought of roughly 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. Any invertible module is isomorphic to some fractional ideal of \(A\), and thus any invertible \(A\)-submodule of \(K\) is a fractional ideal of \(A\). Any 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 the following two formulas

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

    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 obvious by §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 \(M_\mathfrak{p}\cong A_\mathfrak{p}\). 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 the \(a_i\). Then \(M\) must be generated by these \(x_1,\ldots, x_n\) by §Properties of Localization, ⁋Proposition 4.

    Again by this proposition, from the given hypothesis we obtain that for any prime ideal \(\mathfrak{p}\), the localized trace \(\tr_\mathfrak{p}\) is an isomorphism via the following isomorphism

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

    Similarly, we can localize \(\xi_i:M \rightarrow A\) to define \((\xi_i)_\mathfrak{p}: M_\mathfrak{p}\rightarrow A_\mathfrak{p}\), and these will be the desired isomorphisms.
    On the other hand, for a fixed \(\mathfrak{p}\), since \(1\not\in \mathfrak{p}\), among the chosen \(\xi_i\otimes x_i\) above, at least one must satisfy \(\tr(\xi_i\otimes x_i)=\xi_i(x_i)\not\in \mathfrak{p}\). Then the inverse of this element exists in \(A_\mathfrak{p}\); writing it as \(a_i\) for convenience, we know from the equation

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

    that \((\xi_i)_{\mathfrak{p}}: M_\mathfrak{p} \rightarrow A_\mathfrak{p}\) sends \(a_i x_i\) to \(1\in A_\mathfrak{p}\). Now define \(A_\mathfrak{p} \rightarrow M_\mathfrak{p}\) by \(1\mapsto a_ix_i\); this is a section of \((\xi_i)_\mathfrak{p}\), and thus the following 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 know that \(M_\mathfrak{p}\cong A_\mathfrak{p}x_i\oplus\ker(\xi_i)_\mathfrak{p}\). Similarly, via \(M_\mathfrak{p}\hookrightarrow M^{\ast\ast}_\mathfrak{p}\), viewing \(x_i\) as a map \(M^\ast_\mathfrak{p} \rightarrow A_\mathfrak{p}\), we obtain \(M_\mathfrak{p}^\ast\cong A_\mathfrak{p}\xi_i\oplus \ker(x_i)\), and now

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

    However, since the first term on the right-hand side exactly recovers all elements of \(A_\mathfrak{p}\), when mapped by \(\tr_\mathfrak{p}\) the remaining terms must go to \(0\), and in particular from the second term going to \(0\) we know 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 can regard \(M_\mathfrak{p}\) as the free \(A_\mathfrak{p}\)-module generated by \(x_i\).

  2. Now suppose \(M\) is an invertible module. To compare it with a fractional ideal of \(A\), we must first embed \(M\) into \(K\). First, we 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. Thus from §Integral Extensions, ⁋Proposition 13 and the following isomorphisms

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

    we obtain \(M\otimes K\cong K\). Composing this with the localization map \(\epsilon: M \rightarrow S^{-1}M=K\otimes M\), we obtain the desired embedding, and the fact that \(\epsilon\) is injective also follows 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 noting that the elements of \(S\) (i.e., the non-zerodivisors of \(A\)) are non-zerodivisors in \(A_\mathfrak{m}\).
    Now suppose a fractional ideal \(\mathfrak{A}\) is given and assume that \(\mathfrak{A}\cap A\) consists only of zerodivisors. Then, since \(\mathfrak{A}\) is a (finitely generated) fractional ideal, we can find a common denominator \(a\) such that \(a \mathfrak{A}\subseteq A\) becomes an ideal of \(A\), and now applying §Associated Primes, ⁋Theorem 7 we see that \(a\mathfrak{A}\) is an ideal of \(A\) consisting only of non-zerodivisors, so it belongs to the union of associated prime ideals, and applying §Associated Primes, ⁋Lemma 2 (Prime avoidance lemma) here we know that \(a\mathfrak{A}\) actually annihilates some \(b\in A\). That is, \(ab\) annihilates \(\mathfrak{A}\), and thus localizing at a prime ideal \(\mathfrak{p}\) containing \(\ann(ab)\), we know that \(M_\mathfrak{p}\not\cong A_\mathfrak{p}\).

  3. Suppose two invertible modules \(M,N\) are given. Then by the second result we can regard them as fractional ideals inside \(K\), and the maps given in the claim are defined in this way. Since the fact that the given morphisms are isomorphisms will be shown via §Properties of Localization, ⁋Proposition 4, we may assume from the beginning that \(A\) is local, and then by the argument in the second result and the definition of invertible module, both \(M\) and \(N\) are isomorphic to \(A\). Now letting \(s\) and \(t\) be non-zerodivisors of \(K\) generating \(M\) and \(N\) respectively, the first morphism is already an epimorphism, and moreover 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 know 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 \(t\in M^{-1}N\) we have \(ta\neq 0\), so \(u_t\) is not the zero morphism, and thus the morphism in the claim is a monomorphism. That it is an epimorphism follows because for any \(u\in \Hom_A(M,N)\), if \(u(x)=y\) then \(u=u_{y/x}\) holds. In particular, setting \(N=A\) we obtain the last claim.
  4. First, if \(M\) is invertible then by result 3 we can regard \(M^{-1}\otimes M \rightarrow M^{-1}M\) and the trace map \(M^\ast\otimes M \rightarrow A\) as the same. Conversely, if an arbitrary \(A\)-submodule \(M\) of \(K\) satisfies \(M^{-1}M=A\), then as above we may assume via localization that \((A,\mathfrak{m})\) is a local ring and show that \(M\cong A\). But by the condition \(M^{-1}M=A\), we can arrange that for some suitable \(y\in M^{-1}\), \(yM\not\subseteq \mathfrak{m}\), 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 UFD \(R\), we have \(\Pic(R)=0\). That is, every invertible module over a UFD is free.

Proof

Suppose \(R\) is a UFD and \(I\) is an invertible fractional ideal of \(R\). For sufficiency we may assume \(I \subseteq R\) (otherwise we can choose a suitable \(s \in R\) such that \(sI \subseteq R\); if \(I\) is principal then \(sI\) is also principal).

Since \(I\) is invertible we have \(I^{-1}I = R\), and by Definition 1 we have \(I_\mathfrak{p} \cong R_\mathfrak{p}\) at any prime \(\mathfrak{p}\). 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 regular local ring of dimension 1), and in a DVR every nonzero fractional ideal is principal, so for some suitable \(v_\mathfrak{p}(I) \in \mathbb{Z}\) we have \(I_\mathfrak{p} = (\pi_\mathfrak{p}^{v_\mathfrak{p}(I)})\). Here \(\pi_\mathfrak{p}\) is the uniformizer of \(R_\mathfrak{p}\), and \(v_\mathfrak{p}\) is the normalized valuation of \(R_\mathfrak{p}\). Since \(I \subseteq R\), we have \(v_\mathfrak{p}(I) \ge 0\).

Moreover, there are only finitely many \(\mathfrak{p}\) with \(v_\mathfrak{p}(I) > 0\) (since \(I\) is finitely generated). Let \(a = \prod \pi_\mathfrak{p}^{v_\mathfrak{p}(I)}\). Then

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

For the first equality, the \(\subseteq\) direction is obvious, and the converse follows from the fact that if \(x \in R\) and \(v_\mathfrak{p}(x) \ge v_\mathfrak{p}(I)\) for all height 1 primes \(\mathfrak{p}\), then by comparing valuations at each \(\mathfrak{p}\) in the UFD we have \(x/a \in R\), i.e., \(x \in (a)\). In the last equality, \(\bigcap_{\mathfrak{p}} (\pi_\mathfrak{p}^{v_\mathfrak{p}(I)})\) is the set of \(x \in K\) satisfying \(v_\mathfrak{p}(x) \ge v_\mathfrak{p}(I)\) for all \(\mathfrak{p}\), so this coincides with \((a)\). Therefore \(I\) is a principal ideal.

Let us consider the set of isomorphism classes of invertible modules defined over a ring \(A\). Then \(\otimes\) preserves this isomorphism class, so it defines a binary operation on this set; \(\otimes\) satisfies associativity and commutativity, and has identity element \(A\). Moreover, by the first result of Theorem 3, any invertible module has an inverse \(M^\ast\) with respect to \(\otimes\). From this we know that this set becomes an abelian group.

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

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

  1. The Picard group \(\Pic(A)\) of \(A\) is the group of isomorphism classes of invertible \(A\)-modules with operation given by \(\otimes\).
  2. The group of Cartier divisors \(\CaDiv(A)\) of \(A\) is the group of invertible \(A\)-submodules of \(K\), i.e., the group of fractional ideals of \(A\).

Then the following is obvious from Theorem 3.

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

  1. The function \(\CaDiv(A) \rightarrow \Pic(A)\) taking an arbitrary invertible \(A\)-submodule of \(K\) to its isomorphism class is surjective, and its kernel is isomorphic to the group \(K^\times\) of units of \(K\).
  2. \(\CaDiv(A)\) is the free abelian group generated by the invertible ideals of \(A\).
Proof
  1. That the given function is surjective is the second result of Theorem 3, and for any unit \(x\) of \(K\), \(Ax\subseteq K\) is an invertible module mapped to \(A\) by this function. Thus if arbitrary invertible submodules \(M,N\) are isomorphic, we can write \(I=xJ\) for some suitable \(x\in K^\times\), so the claim about the kernel also follows.
  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 by the second result of Theorem 3 again, \(\mathfrak{A}^{-1}\) contains a non-zerodivisor of \(A\). Letting this be \(a\), since \(a \mathfrak{A}\subseteq A\) we have \(\mathfrak{A}=a \mathfrak{A}\cdot (a)^{-1}\).

References

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

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