Fitting ideals of a diagonal two-by-two presentation

Setup: the map $\phi$ is $A$-linear with $\phi (e_1)=x{e}_1$, $\phi (e_2)=y{e}_2$ on the standard basis of $A^2$ [F15], so $A^2\stackrel{\phi }{\to }{A}^2\stackrel{\pi }{\to }M⟶0$ is a presentation of $M=\mathrm{coker}\phi$ with $n=2$ generators and two relations; in particular $M$ is generated by $\pi (e_1),\pi (e_2)$, hence is finitely generated, and $\mathcal{F}=\tilde{M}$ on $X=\mathrm{Spec}A$ is quasi-coherent of finite type [F11, F12]. The Axiom of Choice enters only through the associated-sheaf construction of [F11].

The Fitting ideals: apply [F1] to the presentation of step 1.1 with $R=A$, $N=M$, $n=2$ and matrix $\mathrm{diag}(x,y)$. The only $2\times 2$ minor is the determinant $\det \mathrm{diag}(x,y)=xy$, so $I_2(\phi )=(xy)$; the $1\times 1$ minors are the entries $x,y,0,0$, so $I_1(\phi )=(x,y)$; and $I_0(\phi )=A$ by the convention for $r\le 0$. With ${\mathrm{Fitt}}_k(M)=I_{2-k}(\phi )$ this gives ${\mathrm{Fitt}}_0(M)=(xy)$, ${\mathrm{Fitt}}_1(M)=(x,y)$ and ${\mathrm{Fitt}}_2(M)=A$, and hence on $X=\mathrm{Spec}A$ ${\mathrm{Fitt}}_0(\mathcal{F})=(xy{)}^{\sim },{\mathrm{Fitt}}_1(\mathcal{F})=(x,y{)}^{\sim },{\mathrm{Fitt}}_2(\mathcal{F})={\mathcal{O}}_X\mathrm{.}$ The same convention gives $I_r(\phi )=A$ for every $r\le 0$, so ${\mathrm{Fitt}}_k(\mathcal{F})={\mathcal{O}}_X$ for every $k\ge 2$, and ${\mathrm{Fitt}}_{-1}(\mathcal{F})=0$; the inclusion $(xy)\subseteq (x,y)\subseteq A$ is the nestedness ${\mathrm{Fitt}}_0\subseteq {\mathrm{Fitt}}_1\subseteq {\mathrm{Fitt}}_2$ of [F1].

The fibre at a prime: fix $\mathfrak{p}\in X$. Localising the presentation of step 1.1 at the multiplicative set $A\setminus \mathfrak{p}$ gives the exact sequence $A_{\mathfrak{p}}^2\stackrel{{\phi }_{\mathfrak{p}}}{\to }{A}_{\mathfrak{p}}^2⟶M_{\mathfrak{p}}⟶0,$ whose matrix is again $\mathrm{diag}(x,y)$ over $A_{\mathfrak{p}}$ [F6]. The residue field $\kappa (\mathfrak{p})=A_{\mathfrak{p}}\mathrm{/}\mathfrak{p}{A}_{\mathfrak{p}}$ is an $A_{\mathfrak{p}}$-algebra [F5], and tensoring the sequence over $A_{\mathfrak{p}}$ is right exact [F7], so $\kappa (\mathfrak{p}{)}^2\stackrel{\mathrm{diag}(\bar{x},\bar{y})}{\to }\kappa (\mathfrak{p}{)}^2⟶M_{\mathfrak{p}}{\otimes }_{A_{\mathfrak{p}}}\kappa (\mathfrak{p})⟶0$ is exact, where $\bar{x},\bar{y}$ are the images of $x,y$ in $\kappa (\mathfrak{p})$. Since $\mathcal{F}=\tilde{M}$, its stalk at $\mathfrak{p}$ is $M_{\mathfrak{p}}$ [F4], so by [F3] $\mathcal{F}(\mathfrak{p})\cong M_{\mathfrak{p}}{\otimes }_{A_{\mathfrak{p}}}\kappa (\mathfrak{p})\cong \mathrm{coker}(\mathrm{diag}(\bar{x},\bar{y}):\kappa (\mathfrak{p}{)}^2\to \kappa (\mathfrak{p}{)}^2)\mathrm{.}$

The dimension formula: over the field $\kappa (\mathfrak{p})$ the matrix $\mathrm{diag}(\bar{x},\bar{y})$ has rank $[\bar{x}\mathrm{\ne }0]+[\bar{y}\mathrm{\ne }0]$, because its image is spanned by $\bar{x}{e}_1$ and $\bar{y}{e}_2$, each of which is a nonzero multiple of a basis vector or is zero. Applying rank-nullity [F8] to this map and to the quotient map onto its cokernel gives ${\dim }_{\kappa (\mathfrak{p})}\mathcal{F}(\mathfrak{p})=2-\mathrm{rank}\mathrm{diag}(\bar{x},\bar{y})=[\bar{x}=0]+[\bar{y}=0]\mathrm{.}$ By [F5] one has $\bar{x}=0$ exactly when $x\in \mathfrak{p}$, and likewise for $y$, so ${\dim }_{\kappa (\mathfrak{p})}\mathcal{F}(\mathfrak{p})=[x\in \mathfrak{p}]+[y\in \mathfrak{p}]:$ the dimension is $2$ when both $x,y\in \mathfrak{p}$, it is $1$ when exactly one of them lies in $\mathfrak{p}$, and it is $0$ when neither does.

The rank strata: the homomorphism $\mathrm{ev}:A\to k$, $f\mapsto f(0,0)$, extending ${\mathrm{id}}_k$ and sending $x$ and $y$ to $0$ exists by [F9]; it is surjective because it is the identity on constants, and its kernel is $(x,y)$: a polynomial with zero constant term is a sum of monomials $x^i{y}^j$ with $(i,j)\mathrm{\ne }(0,0)$, each divisible by $x$ or by $y$, while conversely $x,y\in \ker \mathrm{ev}$. Hence $A\mathrm{/}(x,y)\cong k$ is a field and ${\mathfrak{m}}_0=(x,y)$ is a maximal ideal [F10]; a prime contains both $x$ and $y$ exactly when it contains ${\mathfrak{m}}_0$, hence exactly when it equals ${\mathfrak{m}}_0$. By step 3.1 and [F13] the fibre-dimension strata are therefore: $\{{\mathfrak{m}}_0\}$, where the dimension is $2$ (the origin of $X$); the points of $(V(x)\setminus V(y))\cup (V(y)\setminus V(x))=(V(x)\cup V(y))\setminus V(x,y),$ where the dimension is $1$ (the two punctured coordinate axes); and the complement of $V(x)\cup V(y)=V(xy)$, where the dimension is $0$. The generic point $(0)$ contains neither $x$ nor $y$ by [F14], so it lies in the last stratum.

The rank-locus equalities: the stalk of $(xy{)}^{\sim }$ at $\mathfrak{p}$ is $(xy{)}_{\mathfrak{p}}$ [F4], and $(xy{)}_{\mathfrak{p}}\subseteq \mathfrak{p}{A}_{\mathfrak{p}}$ holds exactly when $xy\in \mathfrak{p}$, by [F5] applied to $a=xy$; hence $V({\mathrm{Fitt}}_0(\mathcal{F}))=V(xy)$. Similarly $V({\mathrm{Fitt}}_1(\mathcal{F}))=V(x,y)$, and $V({\mathrm{Fitt}}_2(\mathcal{F}))=V(A)=\mathrm{\varnothing }$. By [F13] and step 4.1 these are exactly the loci $\{{\dim }_{\kappa (\mathfrak{p})}\mathcal{F}(\mathfrak{p})>0\}=V(xy)$, $\{\dim >1\}=V(x,y)$ and $\{\dim >2\}=\mathrm{\varnothing }$, in agreement with the rank-locus theorem [F2] applied to the finite type module $\mathcal{F}$; the equalities display the strata of step 4.1 as the vanishing sets of the three Fitting ideals computed in step 2.1.

Conclusion and choice accounting: suppose $\mathcal{F}$ were finite locally free of some rank $r\ge 0$. By [F2] one would have ${\mathrm{Fitt}}_{r-1}(\mathcal{F})=0$ and ${\mathrm{Fitt}}_r(\mathcal{F})={\mathcal{O}}_X$. For $r=0$ this would give $(xy{)}^{\sim }={\mathcal{O}}_X$, impossible because $(xy)\subseteq (x,y)⊊A$ and $(x,y)$ is proper [F10]; for $r=1$ it would give $(xy{)}^{\sim }=0$, impossible because $\mathrm{\Gamma }(X,(xy{)}^{\sim })=(xy)\mathrm{\ne }0$ by [F11] and $xy\mathrm{\ne }0$ [F14]; and for $r\ge 2$ it would give ${\mathrm{Fitt}}_{r-1}(\mathcal{F})=0$, impossible because ${\mathrm{Fitt}}_{r-1}\supseteq {\mathrm{Fitt}}_1=(x,y{)}^{\sim }$ by the nestedness of [F1] while $(x,y)\mathrm{\ne }0$, as $x\in (x,y)$ with $x\mathrm{\ne }0$ [F14]. Even allowing varying local ranks cannot help: at ${\mathfrak{m}}_0$ the module $M_{{\mathfrak{m}}_0}$ has nonzero fibre, but is annihilated by $xy$, which remains nonzero in the domain $A_{{\mathfrak{m}}_0}$. A nonzero free module over that ring has zero annihilator. Thus the stalk at the origin is not free of any rank, and $\mathcal{F}$ is of finite type but not finite locally free. All computations used the given presentation, the canonical localisations and the canonical comparisons of [F4] and [F5]; no selection of presentations, points or covers is made, so the only use of the Axiom of Choice is the inherited one recorded in [F11].