Let $A \hookrightarrow B$ be an extension of finitely generated, reduced $k$-algebras, where $k$ is a field of characteristic zero such that $B$ is a free $A$-module of finite rank. Let $A$ be an integral domain and let $\mathfrak{p}$ be a minimal prime ideal of $B$.

Can you give me an example of such a situation such that $A \hookrightarrow B/ \mathfrak{p}$ is not flat?


1 Answer 1


Here's a counterexample: consider the homogeneous primes in $k[x,y,z,w]$

$$P_1 := (xw-yz, x^2z-y^3,w^2y-z^3,xz^2-wy^2) \\ P_2 := (xw-yz,x^2y-z^3,w^2z-y^3,xy^2-wz^2)$$

Then $P_1 \cap P_2 = (xw-yz, y^2z^2-xyzw)$ has height $2$ (as $xw-yz$ is prime). Take a graded Noether normalization $A := k[X,Y] \subseteq B := k[x,y,z,w]/(P_1 \cap P_2)$. Since $A$ is a polynomial ring, if $A'$ is standard graded with $A \subseteq A'$ module-finite, then $A'$ is $A$-flat iff $A'$ is $A$-free iff $A'$ is Cohen-Macaulay. Now $B$ is CM (even a complete intersection), but neither $B/P_1$ nor $B/P_2$ are CM.

The relevant projective algebraic geometry: a curve on the quadric surface $Q = V(xw-yz) \subseteq \mathbb{P}^3$ has a bidegree $(a,b)$, and is ACM iff $|a - b| \le 1$. Combining curves of type $(1,3)$ and $(3,1)$ on $Q$ gives a curve of type $(4,4)$ on $Q$, which is the complete intersection of $Q$ with a quartic hypersurface. The example above is obtained by taking the rational quartic curve, a particular (irreducible) curve of type $(1,3)$ on $Q$, with ideal given by ($4$ of the $6$) $2 \times 2$ minors of

$$\begin{pmatrix} x & z & y^2 & wy \\ y & w & xz & z^2 \end{pmatrix}$$

and the dual curve of type $(3,1)$ obtained by switching $y$ and $z$.


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