Flat Smooth and Etale Morphisms — Examples

Flatness. The monomials $T_1^{a_1}\cdots T_n^{a_n}$ form a basis of $P$ as an $A$-module, so $P$ is free, hence flat over $A$ by [F1]. The morphism ${\mathbf{A}}_A^n\to \mathrm{Spec}A$ is the affine morphism $\mathrm{Spec}P\to \mathrm{Spec}A$ by [F6], so it is flat at every point by the affine-local criterion [F1].

Finite presentation and the standard smooth chart. The $A$-algebra $P=A[T_1,\dots ,T_n]$ is a polynomial algebra, hence finitely presented, so the structure morphism is locally of finite presentation by [F5]. The empty equation list exhibits $P$ as $(A[T_1,\dots ,T_n]\mathrm{/}(\mathrm{\varnothing }){)}_1$; by [F2] this is a standard smooth presentation of relative dimension $n-0=n$ with the empty Jacobian having invertible $0\times 0$ minor, in the convention of [F2] in which the case $r=0$ is the localisation of a polynomial ring.

Smoothness of relative dimension n. The chart of step 1.2 is an affine chart of the morphism with an invertible Jacobian minor and with $r=0$, $m=n$; since the morphism is locally of finite presentation by step 1.2, the smoothness direction of the Jacobian criterion [F3] applies at every point and exhibits relative dimension $n$. By [F4] the morphism is smooth with relative dimension $n$ at every point, i.e. pure relative dimension $n$; its fibres are ${\mathbf{A}}^n$ over the residue fields.

The case $n=0$ and accounting. For $n=0$ the polynomial algebra is $P=A$, the morphism is the identity of $\mathrm{Spec}A$, and the same argument gives smoothness of relative dimension $0$, i.e. étaleness, by [F4]. The Axiom of Choice [F7] is assumed in the Statement and is used exactly through the Jacobian criterion [F3] in step 2.1; steps 1.1 and 1.2 are choice-free. [F3, F4, F6, F7, step 1.1] $\mathrm{\square }$

The chart is standard 'etale. Take $P=T^2-a\in A[T]$, which is monic of degree $2$ with formal derivative $P^{\mathrm{\prime }}=2T$, and take $g=2T$ in the presentation $B=(A[T]\mathrm{/}(P){)}_{2T}$. In $B$ the image of $P^{\mathrm{\prime }}$ is the inverted element $2T$, hence a unit, so $B$ is standard 'etale over $A$ by [F1]; the presentation is a quotient of $A[T]$ by the principal ideal $(P)$ followed by a localisation, so $B$ is a finitely presented $A$-algebra, and therefore $\mathrm{Spec}B\to \mathrm{Spec}A$ is 'etale by [F1]. By [F1] again, $A[T]\mathrm{/}(P)$ is free over $A$ with basis $1,T$ before the localisation. This proves claim 1.

The degree-two locus. Let $\mathfrak{p}\in \mathrm{Spec}A$ with $2a\mathrm{\notin }\mathfrak{p}$. Then $2\mathrm{\notin }\mathfrak{p}$ and $a\mathrm{\notin }\mathfrak{p}$, so $2$ and $a$ are units of $A_{\mathfrak{p}}$; in the ring $A_{\mathfrak{p}}[T]\mathrm{/}(T^2-a)$ the relation $T\cdot (T{a}^{-1})=1$ makes $T$ a unit, hence $2T$ is a unit, and localising at a unit does not change the ring by [F2]. Therefore $B_{\mathfrak{p}}\cong A_{\mathfrak{p}}[T]\mathrm{/}(T^2-a)$, which is free of rank $2$ over $A_{\mathfrak{p}}$ with basis $1,T$ by [F1]. For the fibre, [F3] applied over $A_{\mathfrak{p}}$ gives $B{\otimes }_A\kappa (\mathfrak{p})\cong B_{\mathfrak{p}}\mathrm{/}\mathfrak{p}{A}_{\mathfrak{p}}{B}_{\mathfrak{p}}\cong \kappa (\mathfrak{p})[T]\mathrm{/}(T^2-\bar{a})$, where $\bar{a}\mathrm{\ne }0$ and $2\mathrm{\ne }0$ in the field $\kappa (\mathfrak{p})$; in this ring $T$ is a unit (with inverse $T{\bar{a}}^{-1}$), so the image of the derivative $2T$ of the monic polynomial $T^2-\bar{a}$ is a unit and [F1] makes $\kappa (\mathfrak{p})[T]\mathrm{/}(T^2-\bar{a})$ a standard 'etale, hence finite 'etale, $\kappa (\mathfrak{p})$-algebra of rank $2$. This proves claim 2.

The characteristic two boundary. If $2=0$ in $A$, then $2T=0$, so the localisation of $A[T]\mathrm{/}(T^2-a)$ at the powers of $0$ is the zero ring $B=0$ by [F2]; its spectrum is empty, and the structure morphism from the empty scheme to $\mathrm{Spec}A$ has no point at which 'etaleness could fail, so it is 'etale vacuously. In particular, for a field $k$ of characteristic two the chart $\mathrm{Spec}(k[T]\mathrm{/}(T^2-a){)}_{2T}$ is empty for every $a$.

The chart is supported over $D(2a)$. In $B$, the element $2T$ is a unit, so $4a=(2T{)}^2$ is a unit. Since $4a=2(2a)$, the factor $2a$ is a unit in $B$ as well. Hence $\mathrm{Spec}B\to \mathrm{Spec}A$ factors through $D(2a)$. On $D(2a)$, both $2$ and $a$ are units; the relation $T\cdot (T{a}^{-1})=1$ makes $T$ a unit, so localising at $2T$ changes nothing. Thus the restricted algebra is $A_{2a}[T]\mathrm{/}(T^2-a)$, finite free of rank $2$ and 'etale by the derivative calculation of step 1.2. This proves that the rank-two cover occurs exactly over $D(2a)$.

Conclusion and choice accounting. Claims 1, 2 and 3 follow from step 1.1, step 1.2, step 1.3 and step 2.1. The standard 'etale presentation, the free basis from monic division, the unit computations in a localisation and the fibre computation used above are all choice-free, and no Axiom of Choice is assumed or used in this example.

We first identify the total ring. Let $\phi :k[t,x,y]\to k[x,y]$ be the $k$-algebra map with $\phi (t)=h:=y^2-x^2(x+1)$, $\phi (x)=x$, $\phi (y)=y$. Since $F=y^2-x^2(x+1)-t=h-t$ has unit leading coefficient $-1$ and degree one as a polynomial in $t$ over $k[x,y]$, division of any $p\in k[t,x,y]$ by $F$ gives $p=q\cdot F+r$ with $r\in k[x,y]$ and $\phi (p)=r$; hence $\ker \phi =(F)$ and $\phi$ induces an isomorphism $B\cong k[x,y]$ under which the class of $t$ is $h$. For any nonzero $p(t)\in k[t]$ of degree $d$ with leading coefficient $c_d$, the polynomial $p(h)$ has degree $2d$ in $y$ and leading coefficient $c_d$, so $p(h)\mathrm{\ne }0$. Thus $B$ is a domain and $k[t]\to B$ is injective; also $h\in (x,y{)}^2$.

The special fibre. Under the isomorphism of step 1.1 the ideal $tB$ corresponds to $hk[x,y]$, so by [F9] the fibre over the prime $(t)$ of $k[t]$ is $B{\otimes }_{k[t]}k\cong B\mathrm{/}tB\cong k[x,y]\mathrm{/}(h)$. The prime $\mathfrak{q}=(t,x,y)\subseteq B$ corresponds to the maximal ideal $(x,y)\subseteq k[x,y]$ (as $h\in (x,y)$), and its image in the fibre is $\mathfrak{m}=(x,y)A$, where $A:=k[x,y]\mathrm{/}(h)$.

Flatness. The ring $k[t]$ is a principal ideal domain by [F3]. If $p\in k[t]$ is nonzero and $b\in B$ satisfies $p\cdot b=0$, then under the identification $B=k[x,y]$ of step 1.1 this reads $p(h)b=0$ in the domain $k[x,y]$, so $b=0$ because $p(h)\mathrm{\ne }0$ (step 1.1); hence $B$ is torsion-free over $k[t]$. By [F4] $B$ is flat over $k[t]$, so $f$ is flat at every point by the affine-local criterion [F2], which rests on the germwise definition of flatness [F1].

Finite presentation. The $k[t]$-algebra $k[t,x,y]$ is a polynomial algebra, hence finitely presented by [F10], and $B$ is its quotient by the ideal generated by the single element $F$, hence $B$ is a finitely presented $k[t]$-algebra; therefore $f$ is locally of finite presentation by [F10].

We show that $h$ is irreducible in $k[x,y]$, so that $(h)$ is a prime of $k[x,y]$ contained in $(x,y)$. Suppose $h=pq$ with $p,q\in k[x,y]$ nonunits; view $p,q$ as polynomials in $y$ over $k[x]$. Since the coefficient of $y^2$ in $h$ is $1$, the $y$-degrees of $p$ and $q$ add to two and their leading coefficients multiply to $1$, hence are units of $k[x]$, i.e. nonzero constants. If one factor had $y$-degree zero it would be a nonunit of $k[x]$ contributing that nonunit to the leading coefficient of the other factor, impossible; so after absorbing constants $h=(y-\alpha )(y-\beta )$ with $\alpha ,\beta \in k[x]$. Then $\alpha +\beta =0$ and $\alpha \beta =-x^2(x+1)$, so ${\beta }^2=x^2(x+1)$; but ${\beta }^2$ has even degree in $x$ while $x^2(x+1)$ has degree three, a contradiction. Hence $h$ is irreducible, so $(h)$ is a prime of $k[x,y]$ by [F13], and it lies in $(x,y)$ because $h\in (x,y{)}^2$. Its degree-two initial form at the origin is $y^2-x^2=(y-x)(y+x)$; the factors are distinct because $\mathrm{char}k\mathrm{\ne }2$, so the origin is an ordinary node by [F14].

Cotangent and dimension of the special fibre at the origin. Put $S:=k[x,y{]}_{(x,y)}$ and $A_{\mathfrak{m}}=S\mathrm{/}(h)S$ (step 2.1). The ambient local ring $S$ has cotangent space with $k$-basis the classes of $x,y$: every element of $S$ is $f\mathrm{/}g$ with $g(0)\mathrm{\ne }0$, so it is congruent modulo $(x,y)S$ to the constant term $f(0)\mathrm{/}g(0)$, and congruent modulo $(x,y{)}^{2}S$ to $f(0)\mathrm{/}g(0)$ plus the linear part of $f\mathrm{/}g$. Hence $\mathrm{edim}S=2$ and by [F5], $\dim S\le 2$; the chain $(0)⊊(x)S⊊(x,y)S$ shows $\dim S\ge 2$, so $\dim S=2$.

In $A=k[x,y]\mathrm{/}(h)$ we have $h=y^2-x^2(x+1)\in {\mathfrak{m}}^2$; hence $(h)+(x,y{)}^2=(x,y{)}^2$ and $A\mathrm{/}(x,y{)}^2\cong k[x,y]\mathrm{/}(x,y{)}^2$, so $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ has $k$-basis the classes of $x,y$ and $\mathrm{edim}{A}_{\mathfrak{m}}=2$ by [F6].

Dimension of the fibre at the origin. The chain $(0)⊊\mathfrak{m}{A}_{\mathfrak{m}}$ shows $\dim A_{\mathfrak{m}}\ge 1$, since $A_{\mathfrak{m}}$ is a domain by step 2.4 and $x$ is a nonzero element of its maximal ideal. If $\dim A_{\mathfrak{m}}\ge 2$, then, since $A_{\mathfrak{m}}$ is a local domain by step 2.4, there is a strict chain $(0)⊊{\bar{P}}_0⊊{\bar{P}}_1$ in $A_{\mathfrak{m}}$. Lifting to $k[x,y]$ gives primes $(h)⊊P_0⊊P_1\subseteq (x,y)$. Their localizations yield the strict chain $(0)⊊(h)S⊊(P_0)S⊊(P_1)S$ in $S=k[x,y{]}_{(x,y)}$, of length three, contradicting $\dim S=2$ from step 3.1. Hence $\dim A_{\mathfrak{m}}=1$.

Non-regularity and failure of smoothness. By steps 3.2 and 4.1 the local ring $A_{\mathfrak{m}}$ is a nonzero Noetherian local ring with $\dim A_{\mathfrak{m}}=1\mathrm{\ne }2=\mathrm{edim}{A}_{\mathfrak{m}}$, so it is not regular by [F6]. If the fibre were geometrically regular at the image of $\mathfrak{q}$, then by the case $K=\kappa (\mathfrak{p})=k$ of [F7] the localisation $A_{\mathfrak{m}}$ would be regular; it is not, so the fibre is not geometrically regular at the image of $\mathfrak{q}$. By [F8] the morphism $f$ is not smooth at $\mathfrak{q}$, even though by step 2.2 it is flat at $\mathfrak{q}$ and by step 2.3 locally of finite presentation there. Taking $K=k$ in the quantifier of [F7] is legitimate because $k$ is a field extension of $\kappa ((t))=k$; no further choice is made. The Axiom of Choice [F12] is assumed in the Statement and is used exactly through the bound [F5] in step 3.1. [F5, F7, F8, F12, step 2.2, step 3.2, step 4.1] $\mathrm{\square }$

Smoothness of relative dimension one. Take $A=k$ and $n=1$ in [F1]: the algebra $k[T]$ is a free $k$-module (the monomials form a basis) and the structure morphism $f$ is flat, locally of finite presentation and smooth of relative dimension $1$ at every point. This is claim 1.

'Etaleness fails by relative dimension. By [F2] 'etaleness of $f$ at a point $x$ requires relative dimension $0$ at $x$; but by step 1.1 the smooth germ $f$ has relative dimension $1$ at $x$, and by [F2] this integer is well-defined, so $1\mathrm{\ne }0$ and $f$ is not 'etale at $x$. As $x\in {\mathbf{A}}_k^1$ was arbitrary, $f$ is 'etale at no point.

The differentials are locally free of rank one. By [F3] (AC) applied to the smooth morphism $f$, the sheaf ${\mathrm{\Omega }}_{{\mathbf{A}}_k^1\mathrm{/}k}$ is locally free of finite rank near every point and its rank at a point $x$ equals the relative dimension, namely $1$; in particular ${\mathrm{\Omega }}_{{\mathbf{A}}_k^1\mathrm{/}k,x}\mathrm{\ne }0$ for every $x\in {\mathbf{A}}_k^1$. By [F4] (AC), applied at $x$, $f$ is 'etale at $x$ if and only if it is flat at $x$ and unramified at $x$, and unramifiedness at $x$ would force ${\mathrm{\Omega }}_{{\mathbf{A}}_k^1\mathrm{/}k,x}=0$; since $f$ is flat by step 1.1 and the stalk of the differentials does not vanish, $f$ fails to be unramified and hence to be 'etale at $x$. This corroborates claim 2 and proves claim 3.

Choice accounting. The Axiom of Choice [F5] is assumed in the Statement and used exactly through the smoothness of the affine space in [F1] in step 1.1 and the rank computation [F3] with the flat-unramified criterion [F4] in step 2.2; the relative-dimension argument of step 2.1 is choice-free. [F1, F3, F4, F5] $\mathrm{\square }$

Setup and finite presentation. Let $A=k[t]$, $P=A$, $I=(t)\subseteq A$ and $B=A\mathrm{/}I=k$, so that the quotient map $A\to B$ induces the closed immersion $f$ of the Statement. Since $I=(t)$ is generated by one element, $B$ is a finitely presented $A$-algebra by [F6], and the induced morphism is of finite presentation, in particular locally of finite type.

Unramifiedness. By [F1] one has ${\mathrm{\Omega }}_{A\mathrm{/}A}=0$; applying the conormal sequence [F2] to $A\to P=A$ and $I=(t)$ gives the exact sequence $(t)\mathrm{/}(t{)}^2\to k{\otimes }_A{\mathrm{\Omega }}_{A\mathrm{/}A}\to {\mathrm{\Omega }}_{k\mathrm{/}A}\to 0$ in which the middle term vanishes, so ${\mathrm{\Omega }}_{k\mathrm{/}A}=0$ as the cokernel of a map from a zero module. By [F3] the vanishing of the differentials makes the algebra map $A\to k$ formally unramified, and with finite type from step 1.1 it is unramified; on the scheme level $f$ is unramified. This gives claim 1.

Non-flatness. The ideal $(t)\subseteq A$ is a free $A$-module of rank one via $a\mapsto at$, so $(t){\otimes }_{A}k\cong A{\otimes }_{A}k\cong k$ by [F4] applied with $M=A$ and $I=(t)$, which is nonzero because $k$ is a field. The inclusion $\iota \text{\hspace{0.17em}⁣}:(t)\hookrightarrow A$ is injective, and $\iota {\otimes }_A{\mathrm{i}\mathrm{d}}_k$ sends $t\otimes 1$ to $t\otimes 1=1\otimes t=1\otimes 0=0$ in $A{\otimes }_{A}k$, hence is the zero map $k\to k$, which is not injective. By [F4] tensoring with the $A$-module $k$ therefore does not preserve injections, so $k$ is not flat over $A$, and the affine morphism $f\text{\hspace{0.17em}⁣}:\mathrm{Spec}k\to \mathrm{Spec}A$ is not flat. This gives claim 2.

Not 'etale. The morphism $f$ is locally of finite presentation by step 1.1 and unramified by step 2.1, but not flat by step 2.2. By [F5] (AC), 'etaleness at a point would require flatness at that point; hence $f$ is not 'etale at its single point, and in particular not 'etale. So a finite presentation, unramified morphism need not be 'etale, which is claim 3: the flatness clause is indispensable.

Choice accounting. The Axiom of Choice [F7] is assumed in the Statement and used exactly through the flat-plus-unramified criterion [F5] in step 3.1; the conormal computation, the tensor computation and the unramifiedness argument of steps 1.1, 2.1 and 2.2 are choice-free. [F5, F7, step 3.1] $\mathrm{\square }$

Write elements of $R$ as sequences $x=(x_n{)}_{n\ge 1}$ with $x_n\in k$. For $x\in R$ define $e(x)\in R$ by $e(x{)}_n=1$ if $x_n\mathrm{\ne }0$ and $e(x{)}_n=0$ if $x_n=0$, and let $y\in R$ have $y_n=x_n^{-1}$ when $x_n\mathrm{\ne }0$ and $y_n=0$ otherwise. Then $e(x{)}^2=e(x)$, $xe(x)=x$ and $xy=e(x)$, so $(x)=(e(x))$ is generated by an idempotent.

The ideal $I={⨁}_{n\ge 1}k$ 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 $S_1\cup \cdots \cup S_m$, but for any $n\mathrm{\notin }{S}_1\cup \cdots \cup S_m$ the element with $1$ at $n$ and $0$ elsewhere lies in $I$ and is not supported there.

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

The underlying map of $\mathrm{Spec}(R\mathrm{/}I)\to \mathrm{Spec}R$ is the contraction of primes along the surjection $R\to R\mathrm{/}I$, 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.

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

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