Examples of flatness and nonflatness

Let \(x\in X\), put \(y=f(x)\), and write \[ A=\OO_{Y,y},\qquad B=\OO_{X,x}. \] Because \(f\) is finite, the local homomorphism \(A\to B\) is finite. Both rings are regular local rings because \(X\) and \(Y\) are nonsingular.

The finite surjective morphism is dominant, so \(A\to B\) is an injective integral extension of domains. The regular local ring \(A\) is normal. Going-up for integral extensions and going-down over the integrally closed domain \(A\) show that chains of primes ending at the maximal ideal of \(B\) correspond in length to chains ending at the maximal ideal of \(A\). Hence \[ \dim A=\dim B=d. \] Choose a regular system of parameters \[ t_1,\ldots,t_d \] for \(A\). Since \(B\) is finite over \(A\), the ideal \[ (t_1,\ldots,t_d)B=\mathfrak m_AB \] is \(\mathfrak m_B\)-primary, so the images of the \(t_i\) form a system of parameters of \(B\). The regular local ring \(B\) is Cohen–Macaulay, hence every system of parameters is a regular sequence. Thus \[ t_1,\ldots,t_d \] is a \(B\)-regular sequence.

The Koszul complex on the \(t_i\) is a free resolution of the residue field \(k(y)\) over the regular local ring \(A\). Tensoring it with \(B\) remains exact in positive degrees because the \(t_i\) form a \(B\)-regular sequence. Therefore \[ \operatorname{Tor}_1^A(k(y),B)=0. \]

Choose a minimal set of generators of the finite \(A\)-module \(B\), giving an exact sequence \[ 0\longrightarrow K\longrightarrow A^m\longrightarrow B\longrightarrow0. \] The ring \(A\) is noetherian, so \(K\) is finite. Tensoring with \(k(y)\), minimality makes \[ k(y)^m\longrightarrow B/\mathfrak m_AB \] an isomorphism, while the Tor vanishing shows that \[ K/\mathfrak m_AK=0. \] Nakayama’s lemma gives \(K=0\). Hence \(B\cong A^m\) is free, in particular flat, over \(A\).

This holds for every \(x\in X\), so \(f\) is flat. This proves part (a).

Let \[ I_1=(z,w),\qquad I_2=(x+z,y+w) \] in \(k[x,y,z,w]\). The two components are the planes \[ V(I_1)\cong\AA^2_{x,y}, \qquad V(I_2)\cong\AA^2_{x,y}, \] and their intersection is the origin. For a union of two closed affine subschemes, the coordinate ring is the fibre product of their coordinate rings over the coordinate ring of the intersection. Thus \[ k[x,y,z,w]/(I_1\cap I_2) \cong A\times_k A, \] where both maps \(A\to k\) evaluate at \((0,0)\). The morphism to \(Y=\Spec A\) acts diagonally on this fibre product.

The isomorphism is \[ A\times_k A\longrightarrow A\oplus\mathfrak m, \qquad (u,v)\longmapsto(v,u-v). \] The condition \(u(0)=v(0)\) says exactly that \(u-v\in\mathfrak m\).

If \(R\) were flat over \(A\), its direct summand \(\mathfrak m\) would be flat. Localize at the maximal ideal \(\mathfrak m\). The finite module \(\mathfrak mA_{\mathfrak m}\) would then be finite flat over the local ring \(A_{\mathfrak m}\), hence free. It has rank one, since \[ \mathfrak m\otimes_A k(x,y)\cong k(x,y), \] so it would have to be free of rank one. But \[ \dim_k\mathfrak m/\mathfrak m^2=2, \] so Nakayama’s lemma says that \(\mathfrak mA_{\mathfrak m}\) requires two generators and cannot be free of rank one. This contradiction proves nonflatness in part (b).

The relations \(z^2=zw=w^2=0\) give \[ N^2=0, \] so \(N\) is nilpotent and is contained in the nilradical. The quotient is \[ R/N\cong A, \] which is reduced. Hence the nilradical is exactly \(N\), and \[ R_{\mathrm{red}}=R/N\cong A. \] Therefore \(X_{\mathrm{red}}\cong\Spec A=Y\).

The elements \(z,w\) generate \(N\), and the only \(A\)-linear relation imposed by the defining ideal is \[ xz-yw=0. \] Thus \[ N\cong A^2/A(x,-y), \] where the standard basis maps to \(z,w\).

Define \[ A^2\longrightarrow(x,y), \qquad (a,b)\longmapsto ay+bx. \] This is surjective, and \((x,-y)\) lies in its kernel. Conversely, if \[ ay+bx=0, \] coprimality of \(x\) and \(y\) in the UFD \(A\) gives \[ a=xc,\qquad b=-yc \] for some \(c\in A\). Hence the kernel is exactly \(A(x,-y)\), and \[ N\cong(x,y). \] In particular \(N\) is torsion-free as an \(A\)-module.

As an \(A\)-module, \[ R\cong A\oplus N \cong A\oplus(x,y). \] Both summands are torsion-free over the domain \(A\), so \(R\) is torsion-free as an \(A\)-module. Let \(\mathfrak q=\operatorname{Ann}_R(r)\) be an associated prime of \(R\), with \(0\ne r\in R\). Then \[ \mathfrak q\cap A=\operatorname{Ann}_A(r)=0, \] because \(R\) is torsion-free over \(A\).

The finite extension \(A\to R\) induces the homeomorphism \[ \Spec R\xrightarrow{\sim}\Spec A \] because the nilpotent ideal \(N\) is the kernel of \(R\to A\). Thus the only prime of \(R\) contracting to \((0)\) is the minimal prime \[ N=(z,w). \] It follows that \[ \operatorname{Ass}_R(R)=\{N\}. \] Hence every associated point is minimal, so \(X\) has no embedded points.

Step 5 gives \[ R\cong A\oplus(x,y) \] as an \(A\)-module. If \(R\) were flat, the direct summand \((x,y)\) would be flat. Step 3 already proves that \((x,y)\) is not flat over \(A\). Therefore \(R\) is not flat over \(A\), and the morphism \(X\to Y\) is not flat. This proves part (c).

Step 1 proves part (a), steps <1>2–<1>3 prove part (b), and steps <1>4–<1>7 prove part (c).