Flatness

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.

We defined the \(\Tor\) functors in §Ext and Tor, ⁋Definition 2, and we saw that if \(B\) is a flat \(A\)-algebra, then the canonical isomorphism

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

holds. This applies in particular when \(B=S^{-1}A\).

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

Proof

Applying the \(\Tor\) long exact sequence to the short exact sequence

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

yields

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

From this, the equivalence of the injectivity of \(\mathfrak{a}\otimes M \rightarrow M\) and the vanishing of \(\Tor_1^A(A/\mathfrak{a},M)=0\) is immediate.

We now turn to the second claim. It suffices to show that for any injection \(L \rightarrow N\), the induced map \(L\otimes_AM \rightarrow N\otimes_AM\) is injective. To establish this, we may assume that \(N\) is finitely generated: any element \(z\in N\otimes_AM\) is a finite sum of elements of the form \(x\otimes y\), so we may replace \(N\) by the finitely generated submodule \(N'\) spanned by the corresponding \(x\)’s and assume \(z\in N'\otimes_A M\).

Now choose a chain of submodules between two finitely generated \(A\)-modules \(N\) and \(L\)

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

such that each quotient \(N_{i+1}/N_i\) is cyclic. Then there exists an ideal \(\mathfrak{a}\) of \(A\) with \(N_{i+1}/N_i\cong A/\mathfrak{a}\). Since the successive inclusions compose to an injection \(L\hookrightarrow N\), it is enough to treat the case \(p=1\), i.e. \(N/L\cong A/\mathfrak{a}\). Recall also that, by the preceding discussion, it suffices to show injectivity for finitely generated ideals: if \(\mathfrak{a}'\otimes_AM \rightarrow M\) is injective for every finitely generated ideal \(\mathfrak{a}'\), then the same holds for arbitrary ideals \(\mathfrak{a}\).

Applying the \(\Tor\) long exact sequence to

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

gives

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

and since \(\Tor_1^A(N/L,M)=\Tor_1^A(A/\mathfrak{a},M)=0\), the desired result follows.

From this we obtain the following corollaries in several special cases.

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

Proof

Since \((t)\) is the only ideal of \(A\), the flatness of \(M\) is equivalent to the injectivity of \((t)\otimes_A M \rightarrow M\). On the other hand, \(\times t: A \rightarrow (t)\) is an \(A\)-linear map with kernel \((t)\). More explicitly, this map induces the \(A\)-linear isomorphism

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

From this we obtain the isomorphism

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

The multiplication map \((t)\otimes_AM \rightarrow M\) arises from the \(A\)-bilinear map

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

and composing with the isomorphism above yields the map \(M/tM \rightarrow M\) sending

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

Thus the map \(M/tM \rightarrow M\) is precisely multiplication by \(t\), with image equal to \(tM\). Hence \(\times t: M/tM \rightarrow tM\) is an isomorphism if and only if \((t)\otimes_AM \rightarrow M\) is injective, which by Proposition 1 is equivalent to \(M\) being flat.

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

Proof

If \(M\) is a flat \(A\)-module, then \((a)\otimes_AM \rightarrow M\) is injective, so \(a\) is a non-zerodivisor on \(M\). Conversely, if \(A\) is a PID, every ideal of \(A\) is generated by a single non-zerodivisor; hence \(\mathfrak{a}\otimes_AM \rightarrow M\) is injective for every ideal \(\mathfrak{a}\).

In general, the representation of an element in a tensor product is not unique. The following lemma clarifies this to some extent.

Lemma 4 Fix two \(A\)-modules \(M\) and \(N\), and assume that \(N\) is generated by \(\{y_j\}_{j\in J}\). Let an arbitrary element of \(M\otimes_AN\) be expressed as a finite sum

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

Then this element is zero if and only if there exist elements \(\{x_j'\}_{j\in J}\) of \(M\), an index set \(I\), and elements \(a_{ij}\in A\) such that

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

Proof

First, if such \(x_j'\) and \(a_{ij}\) 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 immediate.

For the forward direction, consider first the case where \(N\) is free and \(\{y_j\}_{j\in J}\) is a basis of \(N\). Then, using the isomorphism

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

(§Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Theorem 6 (\(\otimes\dashv\Hom\))), the element \(\sum_j x_j\otimes y_j\) corresponds to \((x_j)_{j\in J}\); hence it is zero if and only if every \(x_j\) is zero.

For an arbitrary module \(N\), choose a free \(A\)-module \(F\) with a surjection \(\varepsilon: F \rightarrow N\) such that the basis \(\{f_j\}_{j\in J}\) of \(F\) maps to \(\{y_j\}_{j\in J}\) via \(\varepsilon\). ([Multilinear Algebra] §Basis, ⁋Definition 1) This yields the short exact sequence

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

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

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

These combine to give a free presentation of \(N\):

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

Since \(M\otimes_A-\) is right exact, we obtain 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 hypothesis \(\sum_{j\in J} x_j\otimes f_j\) lies in the kernel of \(\id_M\otimes\varepsilon\). Hence, by exactness at \(M\otimes_AF\), there exist suitable \(x_i'\in M\) and \(z_i\in G\) such that

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

Using the basis \(\{f_j\}_{j\in J}\) of \(F\), 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 preceding equation yields

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

whence

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

By the free module case proved above, we conclude that \(x_j=\sum_{i\in I} a_{ij}x_i'\). Now considering the image of \(\eta(z_i)\) under \(\varepsilon\), we have

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

which gives 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 \(a_i\in A\) satisfy \(0=\sum_{i\in I} a_ix_i\), there exist an index set \(J\), elements \(x_j'\in M\), and elements \(b_{ij}\in A\) such that

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

Proof

By Proposition 1, \(M\) is flat if and only if the multiplication map

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

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

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

lies in the kernel of the above multiplication map, then this element must be zero; now applying Lemma 4 to determine when this element vanishes yields the desired result.

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

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

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

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

   

   such that \(K\subseteq \ker v\).

3. Condition 2 remains valid if the hypothesis is weakened by replacing the monogeneous submodule of \(\ker u\) with a finitely generated submodule.

Proof

The equivalence of the first and second conditions follows immediately from Corollary 5. Hence it suffices to assume the second condition and prove the third: choose \(v_1:F \rightarrow G\) killing the monogenous submodule generated by \(x_1\) among the generators \(x_1,\ldots, x_n\) of \(K\), then repeat the same process for the remaining generators \(v_1(x_2),\ldots, v_1(x_n)\) of \(v_1(K)\).

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

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

where \(F\) is a finitely generated free \(A\)-module and \(K\) is also finitely generated. If \(M\) is flat, then by the above corollary \(\im v\subseteq G\) is mapped isomorphically onto \(M\) via \(G \rightarrow M\). Hence \(G \rightarrow M\) splits, and we see that \(M\) is a direct summand of \(G\). That is, a finitely presented flat module is the same as a finitely presented projective module. (§Projective, Injective, and Flat Modules, ⁋Proposition 4)

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

Proof

First, let us reduce the given statement to a simpler form. Since localization preserves \(\otimes\), replacing \(E\) by \(S^{-1}E\) still yields a flat \(A\)-algebra. Moreover, if \(E/\x E\) is a domain, then \(S^{-1}E/\x S^{-1}E\) is also a domain, so we may assume from the outset that \(E=S^{-1}E\). In this situation, the choice of \(S\) as in the hypothesis is equivalent to assuming that every element of \(E\) of the form \(1-\x f\) is a unit.

Now, under these conditions, let \(\mathfrak{a}\) and \(\mathfrak{b}\) be ideals of \(E\) satisfying \(\mathfrak{a}\mathfrak{b}=0\); we must show that \(\mathfrak{a}=0\) or \(\mathfrak{b}=0\). Since \(\mathfrak{a}\mathfrak{b}=0\), we may enlarge \(\mathfrak{a}\) and \(\mathfrak{b}\) if necessary so that they are mutual 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 in \(E\) and \(\mathfrak{a}(\mathfrak{b}:(\x))\x=0\), so \((\mathfrak{b}:(\x))\) annihilates \(\mathfrak{a}\). That is, \((\mathfrak{b}:(\x))\subseteq \mathfrak{b}\), and therefore \(\mathfrak{b}=\x\mathfrak{b}\). The desired result now follows from §Integral Extensions, ⁋Lemma 7.

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References

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

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