Flatness and Faithful Flatness

Let $\{M_i{\}}_{i\in I}$ be flat and put $M={⨁}_{i\in I}{M}_i$. For any exact sequence $A\to B\to C$, [L2] gives $(A{\otimes }_{R}M)\to (B{\otimes }_{R}M)\to (C{\otimes }_{R}M)$ as the direct sum over $i$ of the exact sequences obtained by tensoring with $M_i$. Therefore the displayed sequence is exact, so $M$ is flat by [L1].

Suppose $F\cong N\oplus N^{\mathrm{\prime }}$ is flat. For any exact sequence $A\to B\to C$, tensoring with $F$ gives $(A{\otimes }_{R}N)\oplus (A{\otimes }_R{N}^{\mathrm{\prime }})\to (B{\otimes }_{R}N)\oplus (B{\otimes }_R{N}^{\mathrm{\prime }})\to (C{\otimes }_{R}N)\oplus (C{\otimes }_R{N}^{\mathrm{\prime }})\mathrm{.}$ If an element of $B{\otimes }_{R}N$ maps to zero in $C{\otimes }_{R}N$, then the same element viewed in the direct sum lies in the image of $A{\otimes }_{R}F$ because $F$ is flat. Projecting back to the first summand shows exactness for tensoring with $N$. Thus $N$ is flat.

Steps 1.1 and 1.2 prove the two claims.

For an exact sequence $A\to B\to C$ of $R$-modules, [L2] gives an exact sequence $S^{-1}A\to S^{-1}B\to S^{-1}C\mathrm{.}$ By [L3] this is $(S^{-1}R{\otimes }_{R}A)\to (S^{-1}R{\otimes }_{R}B)\to (S^{-1}R{\otimes }_{R}C),$ so $S^{-1}R$ is flat over $R$ by [L1].

Let $M$ be an $S^{-1}R$-module. If $M$ is flat over $S^{-1}R$, then for any exact sequence over $R$ we first tensor with $S^{-1}R$ as in step 1.1 and then tensor over $S^{-1}R$ with $M$; the result is exact, so $M$ is flat over $R$. Conversely, if $M$ is flat over $R$, then every exact sequence of $S^{-1}R$-modules is in particular exact over $R$, and tensoring it with $M$ over $R$ agrees with tensoring over $S^{-1}R$ because the scalars already act through the localization. Hence $M$ is flat over $S^{-1}R$.

If $N$ is flat over $R$, then $S^{-1}N\cong S^{-1}R{\otimes }_{R}N$ by [L3]. Applying step 2.1 to the $S^{-1}R$-module $S^{-1}N$ proves it is flat over $S^{-1}R$.

Step 2.1 applies in particular to localization at a prime ideal by [L4].

Therefore all three claims hold.

If $M$ is flat, then every localization $M_{\mathfrak{p}}$ is flat over $R_{\mathfrak{p}}$ by [L1]. Therefore 1 implies 2, and 2 implies 3 trivially.

Assume 3. By [L2] it is enough to prove that for every finitely generated ideal $I\subseteq R$, the map $I{\otimes }_{R}M\to M$ is injective. Let $K$ be its kernel. If $K\mathrm{\ne }0$, then some maximal ideal $\mathfrak{m}$ contains $\mathrm{Ann}(k)$ for a nonzero $k\in K$, so $K_{\mathfrak{m}}\mathrm{\ne }0$.

But localization commutes with tensor products, so localizing the map of algebra at $\mathfrak{m}$ gives $I_{\mathfrak{m}}{\otimes }_{R_{\mathfrak{m}}}{M}_{\mathfrak{m}}\to M_{\mathfrak{m}}\mathrm{.}$ By assumption 3 and criterion [L2], this map is injective. Hence its kernel $K_{\mathfrak{m}}$ is zero, contradicting algebra. Therefore $K=0$.

So the ideal-injection criterion [L2] holds globally, and therefore $M$ is flat. Thus 3 implies 1.

The three conditions are equivalent.

Assume $M$ is flat, and let $I=(a_1,\dots ,a_n)\subseteq R$. The surjection $\pi :R^n\to I$ sending $e_i$ to $a_i$ has kernel $K$. Tensoring the exact sequence $K⟶R^n\stackrel{\pi }{\to }I⟶0$ with $M$ remains exact by [L1].

Conversely, assume the stated relation-lifting property. To prove flatness it is enough by [L3] to show that for every finitely generated ideal $I=(a_1,\dots ,a_n)$, the multiplication map $I{\otimes }_{R}M\to M$ is injective. Take an element ${\sum }_{i=1}^n{a}_i\otimes x_i$ in its kernel, so ${\sum }_i{a}_i{x}_i=0$. By the relation-lifting property, $x_i={\sum }_j{b}_{ij}{y}_j$ and ${\sum }_i{a}_i{b}_{ij}=0$ for every $j$. Then ${\sum }_i{a}_i\otimes x_i={\sum }_j\left({\sum }_i{a}_i{b}_{ij}\right)\otimes y_j=0$ in $I{\otimes }_{R}M$. Hence the map is injective.

The ideal-injection criterion [L3] now shows that $M$ is flat.

The relation ${\sum }_{i=1}^n{a}_i{x}_i=0$ says exactly that the tensor $\eta ={\sum }_i{a}_i\otimes x_i\in I{\otimes }_{R}M$ maps to $0$ under the multiplication map $I{\otimes }_{R}M\to M$. By [L3], that map is injective, so $\eta =0$. Therefore the tensor $\tilde{\eta }:={\sum }_i{e}_i\otimes x_i\in R^n{\otimes }_{R}M$ lies in the image of $K{\otimes }_{R}M\to R^n{\otimes }_{R}M$ from step 1.1.

Write a preimage of $\tilde{\eta }$ as a finite sum ${\sum }_{j=1}^m{k}_j\otimes y_j$ with $k_j\in K$ and $y_j\in M$. Under the identification $R^n{\otimes }_{R}M\cong M^n$ from [L2], if $k_j=(b_{1j},\dots ,b_{nj})$, then comparing coordinates gives $x_i={\sum }_{j=1}^m{b}_{ij}{y}_j\text{for every }i\mathrm{.}$ Since each $k_j$ lies in $\ker \pi$, one also has ${\sum }_{i=1}^n{a}_i{b}_{ij}=0\text{for every }j\mathrm{.}$ This is the required decomposition.

Therefore the equational criterion is equivalent to flatness.

Assume $R\mathrm{/}I$ is flat. Apply [L1] to the ideal $I$ and the module $R\mathrm{/}I$. The multiplication map $I{\otimes }_{R}R\mathrm{/}I\to R\mathrm{/}I$ is injective, but it is also the zero map because every $a\in I$ acts trivially on $R\mathrm{/}I$. Hence $I{\otimes }_{R}R\mathrm{/}I=0.$ By [L2], this tensor product is $I\mathrm{/}{I}^2$, so $I=I^2$.

If $I=Re$ with $e^2=e$, then $R\cong Re\oplus R(1-e),$ and the quotient $R\mathrm{/}Re$ identifies with the direct summand $R(1-e)$. Since $R$ is free, hence flat, [L3] shows that $R(1-e)$ is flat. Thus $R\mathrm{/}I$ is flat.

Now assume $I$ is finitely generated and $R\mathrm{/}I$ is flat. Step 1.1 gives $I=I^2$, so [L4] applied to the finite module $I$ gives $e\in I$ with $(1-e)I=0$. Thus every $x\in I$ satisfies $x=ex$, whence $I\subseteq Re$; the reverse inclusion follows from $e\in I$. Moreover $(1-e)e=0$, so $e^2=e$. Therefore $I=Re$ is generated by an idempotent.

Step 1.1 proves clause 1, step 1.2 proves clause 2 and the reverse implication in clause 3, and step 2.1 proves the forward implication in clause 3.

If $M$ is faithfully flat and $N\mathrm{\ne }0$, then the map $0\to N$ is nonzero. If $N{\otimes }_{R}M$ were zero, tensoring would turn the nonzero map $0\to N$ into the zero map, contradicting exactness reflection in [L1]. Thus 1 implies 2.

Condition 2 implies 3 by taking $N=\kappa (\mathfrak{p})$, and 3 implies 4 by restricting to maximal primes.

Assume 4. Let $N\mathrm{\ne }0$ and choose $0\mathrm{\ne }x\in N$. The cyclic submodule $Rx\cong R\mathrm{/}\mathrm{Ann}(x)$ injects into $N$. Choose a maximal ideal $\mathfrak{m}$ containing $\mathrm{Ann}(x)$. Then there is a surjection $R\mathrm{/}\mathrm{Ann}(x)\twoheadrightarrow R\mathrm{/}\mathfrak{m}\mathrm{.}$ Tensoring with $M$ preserves the injection by [L2], and the target tensor is nonzero by 4. Hence $N{\otimes }_{R}M\mathrm{\ne }0.$ So 4 implies 2.

Assume 2 and let $N_1\to N_2\to N_3$ be a complex whose tensor with $M$ is exact. Since $M$ is flat, it is enough to prove exactness at $N_2$. If $x\in \ker (N_2\to N_3)$ is not in the image of $N_1\to N_2$, then it defines a nonzero element of the quotient $Q:=\ker (N_2\to N_3)\mathrm{/}\mathrm{im}(N_1\to N_2)\mathrm{.}$ But tensoring with $M$ kills $Q$, because the tensor complex is exact. This contradicts 2. Hence $Q=0$, and the original complex is exact. Therefore $M$ is faithfully flat.

The four conditions are equivalent.

By [L2], the flatness of $f$ is the flatness of the $R$-module $S$. For a proper ideal $I$, the quotient $R\mathrm{/}I$ is nonzero, and $(R\mathrm{/}I){\otimes }_{R}S\cong S\mathrm{/}IS\mathrm{.}$ Therefore [L1] says that $f$ is faithfully flat exactly when $S\mathrm{/}IS\mathrm{\ne }0$ for every proper ideal $I$, which is equivalent to $IS\mathrm{\ne }S$. So 1 and 2 are equivalent.

A prime of $S$ lying over a maximal ideal $\mathfrak{m}$ exists exactly when the fibre ring $S{\otimes }_R\kappa (\mathfrak{m})$ is nonzero. Since $R\mathrm{/}\mathfrak{m}{\otimes }_{R}S\cong S\mathrm{/}\mathfrak{m}S$ and $\mathfrak{m}$ is maximal, this is precisely the maximal-ideal fibre condition in [L1]. Thus 1 and 3 are equivalent.

Assume 1 and let $\mathfrak{p}\subset R$ be prime. By the prime-fibre form of [L1], the ring $T:=S{\otimes }_R\kappa (\mathfrak{p})$ is nonzero. Choose a prime ideal $\mathfrak{n}\subset T$, and let $\mathfrak{q}$ be its preimage in $S$. Because the composite $R\to S\to T$ factors through $\kappa (\mathfrak{p})$, every element of $\mathfrak{p}$ maps to $0$ in $T$, so $\mathfrak{p}\subseteq \mathfrak{q}\cap R$. If $r\in R\setminus \mathfrak{p}$, then its image in the field $\kappa (\mathfrak{p})$ is a unit, hence its image in $T$ is a unit and cannot lie in the prime $\mathfrak{n}$. Therefore $r\mathrm{\notin }\mathfrak{q}$, so $\mathfrak{q}\cap R=\mathfrak{p}$. Thus a prime of $S$ lies over $\mathfrak{p}$, and 1 implies 4.

Condition 4 obviously implies 3. Since step 1.2 gives 3 if and only if 1, condition 4 is equivalent to 1 as well.

Therefore 1, 2, 3, and 4 are equivalent.

Let $K=\ker f$. Since $S$ is flat over $R$ by [L1], tensoring the exact sequence $0\to K\to R\stackrel{f}{\to }S$ with $S$ remains exact: $0\to K{\otimes }_{R}S\to R{\otimes }_{R}S\to S{\otimes }_{R}S\mathrm{.}$

Under the canonical identification $R{\otimes }_{R}S\cong S$, the middle map in step 1.1 is $s\mapsto 1\otimes s$. Multiplication $\mu :S{\otimes }_{R}S\to S$, $\mu (s\otimes t)=st$, is a left inverse to this map, so the middle map is injective. Therefore $K{\otimes }_{R}S=0$.

Because $S$ is faithfully flat over $R$ by [L1], [L2] applied to $N=K$ shows that $K=0$.

Hence $f$ is injective.

If $N$ is flat over $R$, then $N{\otimes }_{R}S$ is flat over $S$ by [L1].

Conversely, assume $N{\otimes }_{R}S$ is flat over $S$. Let $A\to B\to C$ be an exact sequence of $R$-modules. Tensoring with $N$ and then with $S$ gives $((A{\otimes }_{R}N){\otimes }_{R}S)\to ((B{\otimes }_{R}N){\otimes }_{R}S)\to ((C{\otimes }_{R}N){\otimes }_{R}S)\mathrm{.}$ Associativity of tensor product identifies this with $A{\otimes }_R(N{\otimes }_{R}S)\to B{\otimes }_R(N{\otimes }_{R}S)\to C{\otimes }_R(N{\otimes }_{R}S),$ which is exact because $N{\otimes }_{R}S$ is flat over $S$ and [L2] transports that flatness back along $R\to S$.

Since $R\to S$ is faithfully flat, [L3] reflects exactness. Therefore the sequence $A{\otimes }_{R}N\to B{\otimes }_{R}N\to C{\otimes }_{R}N$ was already exact, so $N$ is flat over $R$.

Hence flatness descends and ascends along faithfully flat base change.

Choose generators $u_1,\dots ,u_r$ of $M{\otimes }_{R}S$. Write each $u_i$ as a finite sum of simple tensors and collect the finitely many elements of $M$ occurring there, say $m_1,\dots ,m_n$. Let $N\subseteq M$ be the submodule they generate.

By construction the images of the $m_j\otimes 1$ generate $M{\otimes }_{R}S$, so $(M\mathrm{/}N){\otimes }_{R}S=0.$ If $M\mathrm{/}N\mathrm{\ne }0$, then [L1] applied to the faithfully flat $R$-module $S$ from [L2] would force $(M\mathrm{/}N){\otimes }_{R}S\mathrm{\ne }0$, contradiction. Therefore $M\mathrm{/}N=0$.

Thus $M=N$ is finitely generated.

Let $\mathfrak{m}$ be the maximal ideal of $R$. We first prove that the multiplication map $\mathfrak{m}{\otimes }_{R}M\to M$ is injective. Take an element $z={\sum }_{i=1}^n{f}_i\otimes x_i$ in its kernel, so $f_i\in \mathfrak{m}$ and ${\sum }_i{f}_i{x}_i=0$. Reducing modulo $I$, the module $M\mathrm{/}IM$ is flat over $R\mathrm{/}I$, so [L1] applied over $R\mathrm{/}I$ yields elements ${\bar{y}}_1,\dots ,{\bar{y}}_t\in M\mathrm{/}IM$ and coefficients ${\bar{a}}_{ij}\in R\mathrm{/}I$ with ${\bar{x}}_i={\sum }_{j=1}^t{\bar{a}}_{ij}{\bar{y}}_j\text{and}{\sum }_{i=1}^n({\bar{f}}_i){\bar{a}}_{ij}=0\text{ for every }j\mathrm{.}$ Choose lifts $y_j\in M$ and $a_{ij}\in R$. Then $x_i-{\sum }_j{a}_{ij}{y}_j\in IM$ and ${\sum }_i{f}_i{a}_{ij}\in I$ for every $j$. Writing each $x_i-{\sum }_j{a}_{ij}{y}_j$ as a finite sum of terms $bm$ with $b\in I$, one sees that $z$ is the image in $\mathfrak{m}{\otimes }_{R}M$ of an element of $I{\otimes }_{R}M$ whose product in $M$ is also $0$. Hypothesis 2 makes that element zero, hence $z=0$.

Let $k=R\mathrm{/}\mathfrak{m}$. Choose elements $x_1,\dots ,x_r\in M$ whose images form a $k$-basis of $M\mathrm{/}\mathfrak{m}M$. By [L2], these elements generate $M$, so they define a surjection $\pi :F:=R^r\twoheadrightarrow M\mathrm{.}$ Let $K:=\ker (\pi )$. Because $R$ is Noetherian, [L3] makes $K$ finitely generated.

Tensoring the exact sequence $K\to F\to M\to 0$ with the ideal $\mathfrak{m}$ gives an exact sequence $\mathfrak{m}{\otimes }_{R}K⟶\mathfrak{m}{\otimes }_{R}F⟶\mathfrak{m}{\otimes }_{R}M⟶0$ by [L4]. Step 1.1 identifies $\mathfrak{m}{\otimes }_{R}M$ with its image $\mathfrak{m}M\subseteq M$, and [L4] identifies $\mathfrak{m}{\otimes }_{R}F$ with $\mathfrak{m}F$. Under these identifications, the kernel of $\mathfrak{m}F\to \mathfrak{m}M$ is exactly $K\cap \mathfrak{m}F$, while the image of $\mathfrak{m}{\otimes }_{R}K$ is $\mathfrak{m}K$. Therefore $K\cap \mathfrak{m}F=\mathfrak{m}K\mathrm{.}$

The induced map $F\mathrm{/}\mathfrak{m}F\to M\mathrm{/}\mathfrak{m}M$ sends the standard basis of $k^r$ to the chosen basis from step 1.2, so it is an isomorphism. Its kernel is $(K+\mathfrak{m}F)\mathrm{/}\mathfrak{m}F\cong K\mathrm{/}(K\cap \mathfrak{m}F)=K\mathrm{/}\mathfrak{m}K$ by step 2.1. Hence $K\mathrm{/}\mathfrak{m}K=0$, so $\mathfrak{m}K=K$. Now [L5] gives $K=0$. Therefore $\pi$ is an isomorphism, $M\cong R^r$ is free, and in particular $M$ is flat over $R$.

Thus $M$ is flat over $R$.

The ring $R\mathrm{/}\mathfrak{m}$ is a field. Since $M\mathrm{/}\mathfrak{m}M$ is flat over the $S\mathrm{/}\mathfrak{m}S$-algebra in hypothesis 1, it is in particular a vector space over $R\mathrm{/}\mathfrak{m}$, hence flat over $R\mathrm{/}\mathfrak{m}$.

Hypothesis 2 is exactly the injectivity condition required in [L1] for the ideal $I=\mathfrak{m}$. Therefore [L1] applies and yields that $M$ is flat over $R$.

This is the claimed closed-fibre criterion.

Choose elements $x_1,\dots ,x_r\in M$ whose residue classes form a basis of the vector space $M\mathrm{/}\mathfrak{m}M$. By [L1], they generate $M$. Thus there is a surjection $\phi :R^r\twoheadrightarrow M$ sending the $i$th standard basis vector to $x_i$.

Let $K=\ker \phi$. Because $R$ is Noetherian and $R^r$ is finite, [L2] makes $K$ finitely generated. Tensoring $0\to K\to R^r\to M\to 0$ with the residue field $k=R\mathrm{/}\mathfrak{m}$ remains exact by [L3], so $0\to K\mathrm{/}\mathfrak{m}K\to k^r\to M\mathrm{/}\mathfrak{m}M\to 0.$ The last map is an isomorphism by the choice of the $x_i$, hence $K\mathrm{/}\mathfrak{m}K=0$.

Applying [L1] to the finite module $K$ and the empty list of generators gives $K=0$. Thus $\phi$ is an isomorphism and $M\cong R^r$ is free.

Therefore every finite flat module over a Noetherian local ring is free.

By [L1], choose a finite presentation $R^m\stackrel{D}{\to }{R}^n\stackrel{\pi }{\to }M\to 0$. Put $x_i=\pi (e_i)$, and write the columns of $D$ as relation rows $u^{(1)},\dots ,u^{(m)}\in R^n$, so $\ker \pi$ is generated by those rows. Start with $x=Ay$, where $A$ is the identity matrix and $y$ is the column of the $x_i$.

Process the $m$ relation rows in order. Suppose $x=Ay$ and the first $h-1$ rows satisfy $u^{(j)}A=0$. For the next row $u^{(h)}$, the equality $u^{(h)}x=0$ says $(u^{(h)}A)y=0$. Applying [L2] to this one finite relation gives a finite column $z$ of elements of $M$ and a matrix $B$ with $y=Bz$ and $(u^{(h)}A)B=0$. Replace $(A,y)$ by $(AB,z)$. The new matrix satisfies $u^{(h)}AB=0$, and each earlier relation remains zero because $u^{(j)}AB=(u^{(j)}A)B=0$. After the finite last stage, write $x=Cz$ with $u^{(j)}C=0$ for every $j$. This finite induction makes only finitely many choices.

Let $z$ have $t$ entries, and define $s_0:R^n\to R^t$ by sending $e_i$ to the $i$th row of $C$. Every generator $u^{(j)}$ of $\ker \pi$ is killed by $s_0$ because $u^{(j)}C=0$. Thus $s_0$ descends to $s:M\to R^t$. Define $q:R^t\to M$ by sending its basis vectors to the entries of $z$. The identity $x=Cz$ says $qs(x_i)=x_i$ for every generator $x_i$, hence $qs=1_M$.

Let $E\twoheadrightarrow N$ be any epimorphism and $g:M\to N$ any map. For each of the finitely many basis vectors of $R^t$, choose a preimage in $E$ of its image under $gq$, and extend linearly to a lift $h:R^t\to E$. Then $hs$ lifts $g$ because $qs=1_M$. By [L3], $M$ is projective. It is finite by hypothesis, so it is finite projective.

Localize at ${\mathfrak{p}}_2$ and then at ${\mathfrak{q}}_2$. We obtain a flat local homomorphism $R_{{\mathfrak{p}}_2}\to S_{{\mathfrak{q}}_2}$ by [L1].

Because this localized map is local and flat, [L2] makes it faithfully flat. Applying the spectral characterization of faithful flatness to the prime ideal ${\mathfrak{p}}_1{R}_{{\mathfrak{p}}_2}$ of the source produces a prime ${\mathfrak{q}}_1{S}_{{\mathfrak{q}}_2}\subseteq S_{{\mathfrak{q}}_2}$ lying over it. Contracting back to $S$ gives a prime ${\mathfrak{q}}_1\subseteq {\mathfrak{q}}_2$ with ${\mathfrak{q}}_1\cap R={\mathfrak{p}}_1\mathrm{.}$

Therefore flat ring maps satisfy going down.

As an $R$-module, $R[x]={⨁}_{n\ge 0}R{x}^n,$ so it is free on the monomial basis. By [L1], it is flat over $R$.

If $I⊊R$, then every polynomial in $IR[x]$ has all coefficients in $I$, so $1\mathrm{\notin }IR[x]$. Thus $IR[x]$ is proper. By [L2], the map $R\to R[x]$ is faithfully flat.

Therefore polynomial algebras give basic faithfully flat examples.

By [L1], $\mathbb{Z}[1\mathrm{/}2]$ is flat over $\mathbb{Z}$.

The proper ideal $(2)\subset \mathbb{Z}$ becomes the unit ideal after localization, since $2$ is invertible in $\mathbb{Z}[1\mathrm{/}2]$. Thus $(2)\mathbb{Z}[1\mathrm{/}2]=\mathbb{Z}[1\mathrm{/}2]$. By [L2], the map is not faithfully flat.

So a proper localization can be flat without being faithfully flat.

Since $\mathbb{Q}$ is the localization of $\mathbb{Z}$ at the nonzero integers, [L1] gives that $\mathbb{Q}$ is flat over $\mathbb{Z}$.

Assume the Axiom of Choice. If $\mathbb{Q}$ were projective, [L3] would make it a direct summand of a free abelian group. Every direct summand of a free abelian group is reduced, while $\mathbb{Q}$ is nonzero and divisible. Hence $\mathbb{Q}$ is not projective.

Thus a fraction field can be flat without being projective.

The element $e=(1,0)\in A\times B$ satisfies $e^2=e$, and $I=Re$.