This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
In the next post we study regular local rings (§Dimension, ⁋Definition 9). Before that, we need to define a few concepts.
Invertible Modules
We first make the following definition.
Definition 1 For a ring \(A\), an \(A\)-module \(M\) is called invertible if it is finitely generated and \(M_\mathfrak{p}\cong A_\mathfrak{p}\) holds for every prime ideal \(\mathfrak{p}\) of \(A\).
For a prime ideal \(\mathfrak{p}\) and a maximal ideal \(\mathfrak{m}\) containing \(\mathfrak{p}\), 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 define \(M^\ast=\Hom_A(M,A)\). Since \(A\) is commutative, \(\Hom_A(M, A)\) is also an \(A\)-module, and furthermore the trace map \(M^\ast\otimes M \rightarrow A\) exists. ([Multilinear Algebra] §Hom and Tensor Products, ⁋Definition 6)
Definition 2 Let \(A\) be a ring and let \(K\) be its total ring of fractions. Then an \(A\)-submodule \(\mathfrak{A}\) of \(K\) is called a fractional ideal of \(A\) if there exists a non-zero element \(a\in A\) such that \(a \mathfrak{A}\subseteq A\).
That \(a \mathfrak{A}\) is an ideal of \(A\) is obvious from the definition. Intuitively, if \(\mathfrak{A}\) is finitely generated, the condition \(a \mathfrak{A}\subseteq A\) amounts to multiplying the generators of \(\mathfrak{A}\) by a common denominator to view them as a subset of \(A\). In particular, any finitely generated \(A\)-submodule of \(K\) of the form
\[\left(\frac{a_1}{s_1},\ldots, \frac{a_n}{s_n}\right)A\]is always a fractional ideal via the element \(s_1\cdots s_n\in A\), and the converse also holds if \(A\) is Noetherian. 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 \(K\) we set
\[X^{-1}=(A:_KX)=\{y\in K\mid yX\subseteq A\},\]and by 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.
- An \(A\)-module \(M\) is invertible if and only if the trace map \(M^\ast\otimes_A M \rightarrow A\) is an isomorphism.
- Every invertible module is isomorphic to some fractional ideal of \(A\), and hence 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 \(A\).
-
For any 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.
- For any \(A\)-submodule \(M\subseteq K\), \(M\) is invertible if and only if \(M^{-1}M=A\).
Proof
-
One direction is immediate from §Properties of Localization, ⁋Proposition 4.
\[\tr\left(\sum_{i=1}^n\xi_i\otimes x_i\right)=1\]
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 thatand we show that \(M_\mathfrak{p}\) is generated by the \(x_i\). 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 \(\tr_\mathfrak{p}\) is an isomorphism, i.e. 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 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.
\[(\xi_i)_\mathfrak{p}(a_i x_i)=1\]
Now for a fixed \(\mathfrak{p}\), since \(1\not\in \mathfrak{p}\), among the chosen \(\xi_i\otimes x_i\) there must be one satisfying \(\tr(\xi_i\otimes x_i)=\xi_i(x_i)\not\in \mathfrak{p}\). Then the inverse \(\xi_i(x_i)^{-1}\) exists in \(A_\mathfrak{p}\); write this as \(a_i\) for convenience. From the equationwe see that \((\xi_i)_{\mathfrak{p}}: M_\mathfrak{p} \rightarrow A_\mathfrak{p}\) sends \(a_i x_i\) to \(1\) in \(A_\mathfrak{p}\). Define \(A_\mathfrak{p} \rightarrow M_\mathfrak{p}\) by \(1\mapsto a_ix_i\); this is a section of \((\xi_i)_\mathfrak{p}\), and hence 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 know \(M_\mathfrak{p}\cong A_\mathfrak{p}x_i\oplus\ker(\xi_i)_\mathfrak{p}\). Similarly, viewing \(x_i\) as a map \(M^\ast_\mathfrak{p} \rightarrow A_\mathfrak{p}\) via \(M_\mathfrak{p}\hookrightarrow M^{\ast\ast}_\mathfrak{p}\), we obtain \(M_\mathfrak{p}^\ast\cong A_\mathfrak{p}\xi_i\oplus \ker(x_i)\), and now
\[M^\ast_\mathfrak{p}\otimes M_\mathfrak{p}\cong (A_\mathfrak{p}\xi_i\otimes A_\mathfrak{p}x_i)\oplus ( A_\mathfrak{p}\xi_i\otimes\ker (\xi_i)_\mathfrak{p})\oplus(\ker(x_i)_\mathfrak{p}\otimes A_\mathfrak{p}x_i)\oplus (\ker(x_i)_\mathfrak{p}\otimes \ker(\xi_i)_\mathfrak{p})\]But the first summand on the right recovers all of \(A_\mathfrak{p}\), so when mapped by \(\tr_\mathfrak{p}\) the remaining summands must go to \(0\); in particular, from the second summand mapping to \(0\) we deduce \((\ker\xi_i)_\mathfrak{p}=\ker(\xi_i)_\mathfrak{p}=0\). Thus \(\xi_i\) is an isomorphism from \(M_\mathfrak{p}\) to \(A_\mathfrak{p}\) as claimed, and via this we may view \(M_\mathfrak{p}\) as the free \(A_\mathfrak{p}\)-module generated by \(x_i\).
-
Now let \(M\) be an invertible module. To compare it with a fractional ideal of \(A\), we must first embed \(M\) into \(K\). First, one checks 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. Hence 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 \(M\otimes K\cong K\). Composing this with the localization map \(\epsilon: M \rightarrow S^{-1}M=K\otimes M\) gives the desired embedding. 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
\[\epsilon_\mathfrak{m}: M_\mathfrak{m}\cong A_\mathfrak{m} \rightarrow K\otimes_{A_\mathfrak{m}} A_\mathfrak{m}=S^{-1}A_\mathfrak{m},\]which is injective because the elements of \(S\) (i.e. the non-zerodivisors of \(A\)) are non-zerodivisors in \(A_\mathfrak{m}\).
Now let a fractional ideal \(\mathfrak{A}\) be given, and suppose \(\mathfrak{A}\cap A\) consists only of zerodivisors. Since \(\mathfrak{A}\) is a (finitely generated) fractional ideal, we can find a common denominator \(a\) so that \(a \mathfrak{A}\subseteq A\) is an ideal of \(A\), and applying §Associated Primes, ⁋Theorem 7 we see that \(a\mathfrak{A}\) is an ideal of \(A\) consisting only of non-zerodivisors; hence it lies in the union of the associated prime ideals, and applying §Associated Primes, ⁋Lemma 2 thereto we find that \(a\mathfrak{A}\) is annihilated by some \(b\in A\). Thus \(ab\) annihilates \(\mathfrak{A}\), and localizing at a prime ideal \(\mathfrak{p}\) containing \(\ann(ab)\) we see that \(M_\mathfrak{p}\not\cong A_\mathfrak{p}\). - Let \(M,N\) be two invertible modules. Then by the second result we may regard them as fractional ideals inside \(K\), and the maps given in the statement are defined in this way. Since the given morphisms can be shown to be isomorphisms through §Properties of Localization, ⁋Proposition 4, we may assume from the outset that \(A\) is local; then by the argument in the second result and the definition of an invertible module, both \(M\) and \(N\) are isomorphic to \(A\). Now let \(s,t\) be non-zerodivisors in \(K\) generating \(M,N\) respectively. The first morphism is an epimorphism from the start; moreover, viewing \(M\otimes_A N\) as \(A\cong As\otimes_AAt\), the map \(M\otimes N \rightarrow MN\) sends \(1\otimes1\) to \(st\), and since \(st\) is a non-zerodivisor this is also a monomorphism. For the second morphism, first, by the second result we can choose a non-zerodivisor \(a\in A\cap M\). Then for any non-zero \(t\in M^{-1}N\) we have \(ta\neq 0\), so \(u_t\) is not the zero morphism; hence the morphism in the statement is a monomorphism. That it is an epimorphism follows because for any \(u\in \Hom_A(M,N)\), writing \(u(x)=y\) we have \(u=u_{y/x}\). Setting \(N=A\) in particular gives the last claim.
- First, if \(M\) is invertible then by part 3 we may identify \(M^{-1}\otimes M \rightarrow M^{-1}M\) with the trace map \(M^\ast\otimes M \rightarrow A\). Conversely, if an arbitrary \(A\)-submodule \(M\) of \(K\) satisfies \(M^{-1}M=A\), then just as above we may assume by 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 choose \(y\in M^{-1}\) so that \(yM\not\subseteq \mathfrak{m}\); then by the maximality of \(\mathfrak{m}\) we must have \(yM=A\), and from this 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
Let \(R\) be a UFD and let \(I\) be an invertible fractional ideal of \(R\). For sufficiency we assume \(I \subseteq R\) (otherwise choose a suitable \(s\in R\) so that \(sI\subseteq R\); then if \(I\) is principal, so is \(sI\)).
Since \(I\) is invertible, \(I^{-1}I = R\), and by Definition 1 we have \(I_\mathfrak{p} \cong R_\mathfrak{p}\) at every prime \(\mathfrak{p}\). Therefore \(I\) 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 non-zero fractional ideal is principal, so for a 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 a uniformizer of \(R_\mathfrak{p}\) and \(v_\mathfrak{p}\) is the normalized valuation of \(R_\mathfrak{p}\). Since \(I \subseteq R\), we have \(v_\mathfrak{p}(I) \ge 0\).
Only finitely many \(\mathfrak{p}\) satisfy \(v_\mathfrak{p}(I) > 0\) (since \(I\) is finitely generated). Set \(a = \prod \pi_\mathfrak{p}^{v_\mathfrak{p}(I)}\). Then
\[I = \bigcap_{\mathfrak{p}} I_\mathfrak{p} \cap R = \bigcap_{\mathfrak{p}} (\pi_\mathfrak{p}^{v_\mathfrak{p}(I)}) = (a).\]In the first equality, the inclusion \(\subseteq\) is obvious. For the opposite direction, if \(x\in R\) and \(v_\mathfrak{p}(x) \ge v_\mathfrak{p}(I)\) for every height 1 prime \(\mathfrak{p}\), then comparing valuations at each \(\mathfrak{p}\) in the UFD gives \(x/a\in R\), i.e. \(x\in (a)\). In the last equality, \(\bigcap_{\mathfrak{p}} (\pi_\mathfrak{p}^{v_\mathfrak{p}(I)})\) is the set of \(x\in K\) satisfying \(v_\mathfrak{p}(x) \ge v_\mathfrak{p}(I)\) for all \(\mathfrak{p}\), which coincides with \((a)\). Hence \(I\) is a principal ideal.
Consider the set of isomorphism classes of invertible modules defined over a ring \(A\). Then \(\otimes\) preserves these isomorphism classes, so it defines a binary operation on this set; \(\otimes\) is associative and commutative, and \(A\) serves as the identity element. Moreover, by the first result of Theorem 3, every invertible module has an inverse \(M^\ast\) with respect to \(\otimes\). From this we see that this set becomes an abelian group.
Similarly, the invertible \(A\)-submodules of \(K\) (that is, the invertible fractional ideals of \(A\)) also carry a group structure under ideal product; here the fourth condition of Theorem 3 shows that the inverse of \(M\) is \(M^{-1}\). We name these groups as follows.
Definition 5 For a ring \(A\) we define the following.
- The Picard group \(\Pic(A)\) of \(A\) is the group of isomorphism classes of invertible \(A\)-modules with operation given by \(\otimes\).
- The group \(\CaDiv(A)\) of Cartier divisors 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.
- The map \(\CaDiv(A) \rightarrow \Pic(A)\) sending any invertible \(A\)-submodule of \(K\) to its isomorphism class is surjective, and its kernel is isomorphic to the group of units \(K^\times\) of \(K\).
- \(\CaDiv(A)\) is the free abelian group generated by the invertible ideals of \(A\).
Proof
- That the given map is surjective is the second result of Theorem 3. For any unit \(x\) of \(K\), \(Ax\subseteq K\) is an invertible module mapped to \(A\) by this map. Thus if two invertible submodules are isomorphic, we can write \(I=xJ\) for some \(x\in K^\times\), so the claim about the kernel also follows.
- 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; then again by the second result of Theorem 3, \(\mathfrak{A}^{-1}\) contains a non-zerodivisor of \(A\). Denoting this by \(a\), we have \(a \mathfrak{A}\subseteq A\), so \(\mathfrak{A}=a \mathfrak{A}\cdot (a)^{-1}\).
References
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
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