Flatness

We defined the \(\Tor\) functors in [Homological Algebra] §Ext and Tor, ⁋Definition 2, and we know that if \(B\) is a flat \(A\)-algebra, the following canonical isomorphism

\[B\otimes_A\Tor_i^A(M,N)\cong\Tor_i^B(B\otimes_AM, B\otimes_AN)\]

exists. This applies particularly well to the case \(B=S^{-1}A\).

Proposition 1 Let an \(A\)-module \(M\) and an \(A\)-ideal \(\mathfrak{a}\) be given. Then the multiplication map \(\mathfrak{a}\otimes_AM \rightarrow M\) is an injection if and only if \(\Tor_1^A(A/\mathfrak{a}, M)=0\). \(M\) is flat if and only if for every finitely generated \(\mathfrak{a}\), \(\mathfrak{a}\otimes_AM \rightarrow M\) is an injection.

Proof

Taking the short exact sequence

\[0 \rightarrow \mathfrak{a} \rightarrow A \rightarrow A/\mathfrak{a} \rightarrow 0\]

and applying the \(\Tor\) long exact sequence, we obtain

\[\cdots \rightarrow \Tor_1^A(A, M) \rightarrow \Tor_1^A(A/\mathfrak{a}, M) \rightarrow \mathfrak{a}\otimes_AM \rightarrow A\otimes_AM \rightarrow (A/\mathfrak{a})\otimes_AM \rightarrow 0\]

From this, it is immediate that \(\mathfrak{a}\otimes M \rightarrow M\) being an injection is equivalent to \(\Tor_1^A(A/\mathfrak{a},M)=0\).

We now must show the second claim. For this, we must show that for any injection \(L \rightarrow N\), \(L\otimes_AM \rightarrow N\otimes_AM\) is an injection. However, to show this, it suffices to assume that \(L\) and \(N\) are both finitely generated. Viewing \(L\) as a submodule of \(N\), suppose \(z\in L\otimes_AM\) belongs to the kernel of \(L\otimes_AM \rightarrow N\otimes_AM\). Then we can write \(z\) as a finite sum \(z=\sum_k u_k\otimes m_k\), and taking the finitely generated submodule generated by the \(u_k\), \(L'\subseteq L\), and an element \(z'=\sum_k u_k\otimes m_k\in L'\otimes_AM\), the image of \(z'\) in \(L\otimes_AM\) is \(z\). Meanwhile, \(N\) is the direct limit of the finitely generated submodules containing \(L'\), denoted \(N'\), and since \(\otimes\) commutes with colimits by [Algebraic Structures] §Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Theorem 6 (\(\otimes\dashv\Hom\)), \(N\otimes_AM\) is the direct limit of the \(N'\otimes_AM\). Hence, from the fact that the image of \(z'\) in \(N\otimes_AM\) is \(0\), we know that in some suitable \(N'\), the image of \(z'\) is already \(0\). Now if the claim holds for the finitely generated \(L'\subseteq N'\), then \(z'=0\), and therefore \(z=0\).

Now, between two finitely generated \(A\)-modules \(N\), \(L\), we can choose a sequence of submodules

\[L=N_0 \subseteq N_1\subseteq\cdots\subseteq N_p=N\]

so that each \(N_{i+1}/N_i\) is generated by a single element. Then there exists a suitable \(A\)-ideal \(\mathfrak{a}\) such that \(N_{i+1}/N_i\cong A/\mathfrak{a}\). Moreover, repeatedly composing these inclusions yields \(L\hookrightarrow N\), which is also an injection, so consequently it suffices to assume \(p=1\) and \(N/L\cong A/\mathfrak{a}\). Also, recall that if for every finitely generated ideal \(\mathfrak{a}'\), \(\mathfrak{a}'\otimes_AM \rightarrow M\) is an injection, then for every ideal \(\mathfrak{a}\), \(\mathfrak{a}\otimes_AM \rightarrow M\) is an injection. Indeed, if \(z=\sum_k a_k\otimes m_k\in \mathfrak{a}\otimes_AM\) is sent to \(0\) by the multiplication map, then for \(\mathfrak{a}'=(a_1,\ldots,a_n)\), the element \(z'=\sum_k a_k\otimes m_k\in \mathfrak{a}'\otimes_AM\) is likewise sent to \(\sum_k a_km_k=0\), so by assumption \(z'=0\); since \(z\) is the image under \(\mathfrak{a}'\otimes_AM \rightarrow \mathfrak{a}\otimes_AM\) of \(z'\), we have \(z=0\). Now, taking the following short exact sequence

\[0 \rightarrow L \rightarrow N \rightarrow N/L \rightarrow 0\]

and applying the \(\Tor\) long exact sequence, we obtain

\[\cdots \rightarrow \Tor_1^A(N/L, M) \rightarrow L\otimes_AM \rightarrow N\otimes_AM \rightarrow (N/L)\otimes_AM \rightarrow 0\]

where \(\Tor_1^A(N/L,M)=\Tor_1^A(A/\mathfrak{a},M)\) is \(0\), so we obtain the desired result.

From this, we obtain the following corollaries in certain special cases.

Corollary 2 Let a field \(\mathbb{K}\), a ring \(A=\mathbb{K}[t]/(t^2)\), and an \(A\)-module \(M\) be given. Then \(M\) is a flat \(A\)-module if and only if the multiplication \(\times t: M \rightarrow tM\) induces an isomorphism \(M/tM \rightarrow tM\).

Proof

Since the only ideal of \(A\) other than \(0\) and \(A\) is \((t)\), \(M\) being flat is equivalent to \((t)\otimes_A M \rightarrow M\) being injective. Meanwhile, \(\times t: A \rightarrow (t)\) is an \(A\)-linear map, and its kernel is \((t)\). More explicitly, this \(A\)-linear isomorphism is given by

\[A/(t)\cong \mathbb{K} \rightarrow (t);\qquad a+(t)\mapsto at\]

Now from this we obtain the isomorphism

\[M/tM\cong A/(t)\otimes_A M \cong (t)\otimes_A M\]

Meanwhile, the multiplication map \((t)\otimes_AM \rightarrow M\) is obtained through the \(A\)-bilinear map

\[(t)\times M \rightarrow M;\qquad (ta, x)\mapsto (ta)x\]

and composing this with the above isomorphism, the resulting map \(M/tM \rightarrow M\) is given, for any \(x+tM\in M/tM\), by

\[x+tM \mapsto (1+(t))\otimes x\mapsto t\otimes x\mapsto tx\]

That is, \(M/tM \rightarrow M\) is precisely the map obtained by multiplying by \(t\), and since its image is evidently \(tM\), \(\times t: M/tM \rightarrow tM\) being an isomorphism is equivalent to \((t)\otimes_AM \rightarrow M\) being injective, which in turn is equivalent to \(M\) being flat by Proposition 1.

Corollary 3 If \(M\) is a flat \(A\)-module, then an element \(a\in A\) that is not a zero-divisor is not a zero-divisor on \(M\) either. If \(A\) is a PID, the converse also holds.

Proof

If \(M\) is a flat \(A\)-module, then \((a)\otimes_AM \rightarrow M\) is an injection, so \(a\) is a non-zerodivisor on \(M\). Meanwhile, if \(A\) is a PID, every ideal of \(A\) other than \(0\) is generated by a single non-zerodivisor, and in particular, for every ideal \(\mathfrak{a}\), \(\mathfrak{a}\otimes_AM \rightarrow M\) is injective.

In general, there is no unique way to express an element of a tensor product. The following lemma partially remedies this.

Lemma 4 Fix two \(A\)-modules \(M,N\), and suppose \(N\) is generated by \(\{y_j\}_{j\in J}\). Now, when an arbitrary element of \(M\otimes_AN\) is written as a finite sum

\[\sum_{j\in J} x_j\otimes y_j\]

this element is \(0\) if and only if there exist a suitable finite index set \(I\), elements of \(M\), \(\{x_i'\}_{i\in I}\), and \(a_{ij}\in A\) such that the following equations

\[\sum_{i\in I} a_{ij}x_i'=x_j\quad\text{for all $j$},\qquad \sum_{j\in J} a_{ij}y_j=0\quad\text{for all $i$}\]

hold.

Proof

First, if \(x_j'\) and \(a_{ij}\) satisfying the given conditions all exist, then

\[\sum_{j\in J} x_j\otimes y_j=\sum_{j\in J}\left(\left(\sum_{i\in I}a_{ij}x_i'\right)\otimes y_j\right)=\sum_{i\in I} x_i'\otimes\left(\sum_{j\in J} a_{ij} y_j\right)=0\]

so the reverse direction is easily obtained.

Now, to show the forward direction, let us first consider the case where \(N\) is a free module and \(\{y_j\}_{j\in J}\) is a basis of \(N\). Then by [Algebraic Structures] §Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Theorem 6 (\(\otimes\dashv\Hom\)), \(\otimes\) commutes with colimits, and in particular with direct sums, so we obtain the isomorphism

\[M\otimes_AN\cong\bigoplus_{j\in J} (M\otimes_A Ay_j)\cong \bigoplus_{j\in J} M\]

Considering this, the element \(\sum_j x_j\otimes y_j\) corresponds to \((x_j)_{j\in J}\), so this element being \(0\) is equivalent to all \(x_j\) being \(0\).

Now, for an arbitrary module \(N\), we can choose a suitable free \(A\)-module \(F\) and \(\varepsilon: F \rightarrow N\) such that the basis of \(F\), \(\{f_j\}_{j\in J}\), is sent via \(\varepsilon\) to \(\{y_j\}_{j\in J}\). ([Multilinear Algebra] §Bases, ⁋Definition 1) Then we obtain the short exact sequence

\[0 \longrightarrow\ker\varepsilon \longrightarrow F \overset{\varepsilon}{\longrightarrow} N \longrightarrow 0\]

and, again regarding \(\ker \varepsilon\) as a quotient of a free module \(G\) via [Multilinear Algebra] §Bases, ⁋Proposition 2, we obtain the exact sequence

\[G \rightarrow \ker\varepsilon \rightarrow 0\]

From these we obtain a free presentation of \(N\)

\[G \overset{\eta}{\longrightarrow} F \overset{\varepsilon}{\longrightarrow} N \longrightarrow 0\]

Meanwhile, since \(M\otimes_A-\) is right exact, this yields the exact sequence

\[M\otimes_A G \overset{\id_M\otimes\eta}{\longrightarrow} M\otimes_AF \overset{\id_M\otimes \varepsilon}{\longrightarrow} M\otimes_AN \longrightarrow 0\]

and by assumption, \(\sum_{j\in J} x_j\otimes f_j\) is sent via \(\id_M\otimes\varepsilon\) to \(0\). Therefore, from exactness at \(M\otimes_AF\), we can choose suitable \(x_i'\in M\) and \(z_i\in G\) so that

\[\sum_{i\in I} x_i'\otimes\eta(z_i)=(\id_M\otimes\eta)\left(\sum_i x_i'\otimes z_i\right)=\sum_j x_j\otimes f_j\]

Meanwhile, using the basis of \(F\), \(\{f_j\}_{j\in J}\), we can write

\[\eta(z_i)=\sum_{j\in J} a_{ij}f_j,\qquad\text{$(a_{ij})_{j\in J}$ finitely supported for all $i$}\]

Substituting this into the above equation, from

\[\sum_{j\in J} x_j\otimes f_j=\sum_{i\in I} x_i'\otimes\eta(z_i)=\sum_{i\in I}\sum_{j\in J}a_{ij}x_i'\otimes f_j\]

we obtain

\[0=\sum_{j\in J} x_j \otimes f_j-\sum_{i\in I}\sum_{j\in J} a_{ij}x_i'\otimes f_j=\sum_{j\in J} \left(x_j-\sum_{i\in I} a_{ij}x_i'\right)\otimes f_j\]

and therefore, by the result shown above for the case of a free module, we know that \(x_j=\sum_{i\in I} a_{ij}x_i'\). Now, considering the image of \(\eta(z_i)\) under \(\varepsilon\),

\[0=(\varepsilon\circ\eta)(z_i)=\varepsilon\left(\sum_{j\in J} a_{ij}f_j\right)=\sum_{j\in J} a_{ij}y_j\]

so we obtain the desired result.

More generally, the following holds.

Corollary 5 An \(A\)-module \(M\) is flat if and only if the following condition holds.

Whenever \(x_i\in M\) and \(a_i\in A\) satisfying \(0=\sum_{i\in I} a_ix_i\) are given, there exist an index set \(J\), \(x_j'\in M\), and \(b_{ij}\in A\) such that the following two equations hold:

\[\sum_{j\in J} b_{ij} x_j'=x_i\quad\text{for all $i$},\qquad \sum_{i\in I} b_{ij} a_i=0\quad\text{for all $j$}\]

Proof

By the result of Proposition 1, \(M\) being flat is equivalent to the multiplication map

\[\mathfrak{a}\otimes_A M \rightarrow M\]

being injective for every finitely generated ideal \(\mathfrak{a}\). This is equivalent to saying that if

\[\sum_i a_i\otimes x_i\in \mathfrak{a}\otimes_AM\]

belongs to the kernel of the above multiplication map, then this element must necessarily be \(0\); now, examining when this element is \(0\) using Lemma 4 yields the desired result.

Meanwhile, flatness can also be formulated in the language of diagrams, in a form similar to [Homological Algebra] §Resolutions, ⁋Definition 1.

Corollary 6 For an \(A\)-module \(M\), the following two conditions are equivalent.

1. \(M\) is a flat \(A\)-module.

2. For every finitely generated free module \(F\), morphism \(u:F \rightarrow M\), and monogeneous submodule of \(\ker u\), \(K\), there exists a diagram

   such that \(G\) is a finitely generated free \(A\)-module and \(K\subseteq \ker v\).

Moreover, the second condition also holds if \(K\) is taken to be an arbitrary finitely generated submodule of \(\ker u\).

Proof

The equivalence of the first and second conditions is a diagrammatic transcription of the condition of Corollary 5. First, suppose \(M\) is flat, and let a finitely generated free module \(F\) with basis \(\{e_i\}_{i\in I}\), a morphism \(u:F \rightarrow M\), and a monogeneous submodule of \(\ker u\), \(K=(\sum_{i\in I} a_ie_i)\), be given. Setting \(x_i=u(e_i)\), we have \(\sum_{i\in I} a_ix_i=u\left(\sum_{i\in I} a_ie_i\right)=0\), so from Corollary 5 we obtain \(x_j'\in M\) and \(b_{ij}\in A\) such that \(x_i=\sum_{j\in J} b_{ij}x_j'\) and \(\sum_{i\in I} b_{ij}a_i=0\) hold; the \(J\) here is finite, since it comes from the finite index set of Lemma 4. Now let the basis of \(G=A^J\) be \(\{f_j\}_{j\in J}\), and define \(v:F \rightarrow G\) by \(v(e_i)=\sum_{j\in J} b_{ij}f_j\) and \(G \rightarrow M\) by \(f_j\mapsto x_j'\); then \(e_i\) is carried to \(x_i=u(e_i)\), so the above diagram commutes, and since \(v\left(\sum_{i\in I} a_ie_i\right)=\sum_{j\in J}\left(\sum_{i\in I} b_{ij}a_i\right)f_j=0\), we have \(K\subseteq \ker v\).

Conversely, assume the second condition, and suppose that \(\sum_{i\in I} a_ix_i=0\) holds for given \(x_i\in M\) and \(a_i\in A\); since this sum is finite, we may take \(I\) to be finite, and thus we can apply the second condition to \(F=A^I\), with \(u(e_i)=x_i\) defining \(u:F \rightarrow M\), and to \(K=(\sum_{i\in I} a_ie_i)\subseteq \ker u\). For the resulting \(v:F \rightarrow G\), choose for \(G\) a basis \(\{f_j\}_{j\in J}\), write \(v(e_i)=\sum_{j\in J} b_{ij}f_j\), and let \(x_j'\) be the image of \(f_j\) under \(G \rightarrow M\); then from the commutativity of the diagram we obtain \(x_i=\sum_{j\in J} b_{ij}x_j'\), and from \(K\subseteq \ker v\) we obtain \(\sum_{i\in I} b_{ij}a_i=0\), so by Corollary 5, \(M\) is flat.

Finally, the case where \(K\) is a finitely generated submodule of \(\ker u\) is obtained by repeatedly applying the second condition. Letting the generators of \(K\) be \(x_1,\ldots, x_n\), first, to kill the monogeneous submodule \((x_1)\), choose \(v_1:F \rightarrow G_1\); then, applying the same process to \(G_1 \rightarrow M\) and the monogeneous submodule of \(\ker(G_1 \rightarrow M)\), \((v_1(x_2))\), and repeating this up through \(x_n\), the composite \(v_n\circ\cdots\circ v_1\) is the desired morphism.

If \(M\) is finitely presented, then there exists an exact sequence

\[0 \rightarrow K \rightarrow F \rightarrow M \rightarrow 0\]

Here \(F\) is a finitely generated free \(A\)-module, and \(K\) is also finitely generated. If \(M\) is flat, then in the above corollary, \(\im v\subseteq G\) is carried isomorphically by \(G \rightarrow M\) onto \(M\). Therefore \(G \rightarrow M\) splits, and from this we know that \(M\) is a direct summand of \(G\). In other words, a finitely presented flat module is the same as a finitely presented projective module. ([Multilinear Algebra] §Projective, Injective, and Flat Modules, ⁋Proposition 4)

Corollary 7 Fix a field \(\mathbb{K}\) and \(A=\mathbb{K}[\x]\), and let \(E\) be a Noetherian ring that is a flat \(A\)-algebra. If \(E/\x E\) is a domain, and \(S\) is the set of elements that, for \(f\in E\), are of the form \(1-\x f\), then \(S^{-1}E\) is a domain.

Proof

First, let us reduce the given statement to a simpler form. Since taking localization preserves \(\otimes\), replacing \(E\) by \(S^{-1}E\) still leaves \(S^{-1}E\) a flat \(A\)-algebra, and by §Localization, ⁋Corollary 9, \(S^{-1}E\) is also Noetherian. Moreover, if \(E/\x E\) is a domain, then \(S^{-1} E/ \x S^{-1}E\) is also a domain, so we may take \(E\) to be \(S^{-1}E\) from the start. The choice of \(S\) as in the condition can then be replaced by the assumption that every element of \(E\) of the form \(1-\x f\) is a unit.

Now, under the above condition, suppose ideals of \(E\), \(\mathfrak{a}, \mathfrak{b}\), satisfy \(\mathfrak{a}\mathfrak{b}=0\); we must show that \(\mathfrak{a}=0\) or \(\mathfrak{b}=0\). Then, since \(\mathfrak{a}\mathfrak{b}=0\), enlarging \(\mathfrak{a}\) and \(\mathfrak{b}\) if necessary, we may assume they are each other’s annihilators. Now \(\mathfrak{a}\mathfrak{b}=0\) modulo \(\x\), and since \(E/\x E\) is a domain, we may assume \(\mathfrak{b}\subseteq (\x)\). Then \(\x\) is a non-zerodivisor of \(E\) and \(\mathfrak{a}(\mathfrak{b}:(\x))\x=0\), so \((\mathfrak{b}:(\x))\) annihilates \(\mathfrak{a}\). That is, \((\mathfrak{b}:(\x))\subseteq \mathfrak{b}\), and hence \(\mathfrak{b}=\x\mathfrak{b}\). Meanwhile, since \(E\) is Noetherian, \(\mathfrak{b}\) is finitely generated, and since for the ideal \((\x)\subseteq E\) we have \((\x)\mathfrak{b}=\mathfrak{b}\), by §Integral Extensions, ⁋Lemma 7 there exists a suitable \(f\in E\) such that \((1-\x f)\mathfrak{b}=0\). But by the reduction above, \(1-\x f\) is a unit, so \(\mathfrak{b}=0\), and we obtain the desired result.

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References

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

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