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Flat and finite type is not open without finite presentation

Statement refuted

False claim. Every flat morphism of finite type is open, and a quotient $R \to R / I$ of finite type is finitely presented.

Counterexample with witness. Assume the Axiom of Choice. Let $k$ be a field, $R = \prod_{n \geq 1} k, I = ⨁_{n \geq 1} k \subseteq R$ the ideal of finite-support sequences. Then the quotient map $Spec (R / I) \to Spec R$ is flat and of finite type, but it is not open, and $R / I$ is not a finitely presented $R$ -algebra.

Facts & Assumptions

Given: The data and hypotheses displayed in the Statement, with the conventions fixed there.

[F1]

A morphism $f : X \to S$ is flat at $x$ when the local ring map makes $O_{X, x}$ flat over $O_{S, f (x)}$ , and flat when this holds at every point (Flat morphism of schemes).

[F2]

An $R$ -module $M$ is flat if and only if the multiplication map $J \otimes_{R} M \to M$ is injective for every finitely generated ideal $J \subseteq R$ (Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests, criteria (1) and (4)).

[F3]

Let $f : X \to S$ be a morphism, $U = Spec B \subseteq X$ and $V = Spec A \subseteq S$ affine opens with $f (U) \subseteq V$ . Then $f$ is flat at every point of $U$ if and only if $B$ is flat over $A$ ; the direction from module flatness to pointwise flatness is choice-free (Affine-local flatness).

[F4]

$A$ is of finite type over $R$ when $A = R [a_{1}, \dots, a_{n}]$ for some $n$ ; at $n = 0$ this is the image of the structure map $R \to A$ , so a quotient $R / I$ with its quotient structure map is of finite type, and a morphism of affine schemes whose ring map is of finite type is (locally) of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras, Locally finite presentation morphisms).

[F5]

A finitely presented $R$ -algebra $R \to A$ has: for every surjection $R [x_{1}, \dots, x_{n}] \to A$ of $R$ -algebras, the kernel is a finitely generated ideal (Stacks, Algebra, Lemma 10.6.3). In particular, if the quotient map $R \to R / I$ (zero generators, no relations besides the kernel) presents a finitely presented $R$ -algebra, then $I$ is finitely generated (Finitely presented modules and finitely presented algebras).

[F6]

Assume the Axiom of Choice. If $C \subseteq Spec R$ is clopen, then there is an idempotent $e \in R$ with $C = V (e) = D (1 - e)$ (A clopen decomposition of the spectrum comes from a nontrivial idempotent).

[F7]

Assume the Axiom of Choice. If $Z = V (J) \subseteq Spec R$ is closed, then the unique radical ideal defining $Z$ is $\sqrt{J}$ (Every Zariski-closed subset has a unique radical defining ideal).

[F8]

A ring map $φ : R \to S$ induces the contraction map on spectra, $q \mapsto φ^{- 1} (q)$ ; for the quotient map $R \to R / I$ the primes of $R / I$ correspond to the primes of $R$ containing $I$ , and the image of $Spec (R / I)$ consists of exactly those primes (The map of affine spectra induced by a ring homomorphism, The underlying space of an affine spectrum).

[F9]

In a field, $x^{n} = 0$ implies $x = 0$ (Prime ideals and maximal ideals in a commutative ring); consequently a sequence $(x_{n}) \in R$ has the same support as its $k$ -th power $(x_{n}^{k})$ , and $R$ is reduced.

Counterexample

technique · direct

1.1algebra

Write elements of $R$ as sequences $x = (x_{n})_{n \geq 1}$ with $x_{n} \in k$ . For $x \in R$ define $e (x) \in R$ by $e (x)_{n} = 1$ if $x_{n} \neq 0$ and $e (x)_{n} = 0$ if $x_{n} = 0$ , and let $y \in R$ have $y_{n} = x_{n}^{- 1}$ when $x_{n} \neq 0$ and $y_{n} = 0$ otherwise. Then $e (x)^{2} = e (x)$ , $x e (x) = x$ and $x y = e (x)$ , so $(x) = (e (x))$ is generated by an idempotent.

1.2givenalgebra

The ideal $I = ⨁_{n \geq 1} k$ is not finitely generated: if $I = (x_{1}, \dots, x_{m})$ , each $x_{j} \in I$ has finite support $S_{j}$ , every $R$ -linear combination of the $x_{j}$ is supported in the finite set $S_{1} \cup \dots \cup S_{m}$ , but for any $n \notin S_{1} \cup \dots \cup S_{m}$ the element with $1$ at $n$ and $0$ elsewhere lies in $I$ and is not supported there.

1.3F4

The map is of finite type: $R / I$ is generated as an $R$ -algebra by the empty set, since its structure map $R \to R / I$ is surjective, so it is of finite type over $R$ by [F4]; on the affine charts this is a finite-type ring map.

1.4F8

The underlying map of $Spec (R / I) \to Spec R$ is the contraction of primes along the surjection $R \to R / I$ , so its image is the set of primes of $R$ containing $I$ , which is exactly the closed set $V (I)$ of [F8]; thus the morphism is open only if $V (I)$ is open.

1.5F9given

$I$ is a radical ideal: if $x \in R$ satisfies $x^{k} \in I$ then the support ${n : x_{n} \neq 0}$ equals the support of $x^{k}$ , which is finite because $x^{k} \in ⨁_{n \geq 1} k$ , so $x \in I$ ; hence $R / I$ is reduced and $I = \sqrt{I}$ .

2.1step 1.1algebra

If $e, f \in R$ are idempotent then $(e + f - e f)^{2} = e + f - e f$ and $(e, f) = (e + f - e f)$ : the generator is a combination of $e, f$ , while $e = e (e + f - e f)$ and $f = f (e + f - e f)$ . Hence by induction on the number of generators every finitely generated ideal of $R$ is principal, generated by an idempotent.

2.2F6F7step 1.5

Suppose $V (I)$ is open. It is closed by definition, hence clopen, so by [F6] there is an idempotent $e \in R$ with $V (I) = V (e)$ ; applying [F7] to the closed set $Z = V (I) = V (e)$ and to the ideals $I$ and $(e)$ gives $\sqrt{I} = I (Z) = \sqrt{(e)}$ , so $I = \sqrt{(e)}$ because $\sqrt{I} = I$ by step 1.5.

2.3F5step 1.2

Finally $R / I$ is not a finitely presented $R$ -algebra: the quotient map $R \to R / I$ is a surjection from a polynomial ring in $0$ variables, so by [F5] finite presentation of $R / I$ would force its kernel $I$ to be finitely generated, contrary to step 1.2.

3.1F2step 2.1

For every finitely generated ideal $J \subseteq R$ we have $J = (e)$ with $e$ idempotent by step 2.1, and then $e R \cap I = e I = I \cdot e R$ : an element of $e R \cap I$ equals $e r$ and lies in $I$ , hence equals $e (e r) = e r^{'}$ with $e r^{'} \in e I$ ; conversely $e I \subseteq e R \cap I$ . Therefore $J \cap I = J I$ , so the kernel $(J \cap I) / J I$ of the multiplication map $J \otimes_{R} R / I \to R / I$ is zero, i.e. the map is injective. Since this holds for every finitely generated ideal, $R / I$ is a flat $R$ -module by [F2].

3.2F9step 2.2

The ideal $(e)$ is radical: if $z \in R$ satisfies $z^{k} = e r$ then coordinatewise $z_{n}^{k} = e_{n} r_{n}$ , and $e_{n} \in {0, 1}$ because $e$ is idempotent; for $e_{n} = 0$ this gives $z_{n} = 0 = e_{n} z_{n}$ by [F9], and for $e_{n} = 1$ it gives $z_{n} = e_{n} z_{n}$ , so $z = e z \in (e)$ . Hence $\sqrt{(e)} = (e)$ , and step 2.2 yields $I = (e)$ , a principal ideal, hence a finitely generated ideal.

4.1F1F3step 3.1

The morphism $Spec (R / I) \to Spec R$ is flat: on the affine charts $U = Spec (R / I)$ , $V = Spec R$ with $f (U) \subseteq V$ the ring $R / I$ is flat over $R$ by step 3.1, and [F3] converts this into flatness at every point.

5.1step 1.2step 4.1step 1.3step 1.4step 3.2

Step 3.2 contradicts step 1.2, so $V (I)$ is not open; by step 1.4 the morphism $Spec (R / I) \to Spec R$ is not open, while by steps 4.1 and 1.3 it is flat and of finite type.

6.1

The Axiom of Choice is used exactly twice: in step 2.2 through [F6] to convert the clopen set $V (I)$ into an idempotent, and through [F7], which produces prime ideals. Steps 1.1 through 5.1 use no choice principle. [F6, F7, step 2.2] $□$

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Sources

The Stacks Project, Morphisms of Schemes, Section 29.26 (flat morphisms) (standard reference, not scraped)
The Stacks Project, Commutative Algebra, Tag 00R2 (Lemma 10.6.3: finite presentation and kernels of surjections) (standard reference, not scraped)
The Stacks Project, Commutative Algebra, Section 10.21 (idempotents and connected components) (standard reference, not scraped)