Let $p,q \in \mathbb{C}[x,y]$ be two polynomials such that $p\_xp\_yq\_xq\_y \neq 0$
(namely, each partial derivative is non-zero).


Assume that the following four conditions are satisfied:


**(1)** $\frac{\mathbb{C}[x][y]}{\langle p \rangle}$ is a flat $\mathbb{C}[x]$-module.
**(2)** $\frac{\mathbb{C}[y][x]}{\langle p \rangle}$ is a flat $\mathbb{C}[y]$-module.


**(3)** $\frac{\mathbb{C}[x][y]}{\langle q \rangle}$ is a flat $\mathbb{C}[x]$-module.
**(4)** $\frac{\mathbb{C}[y][x]}{\langle q \rangle}$ is a flat $\mathbb{C}[y]$-module.



> 
> **Question:** When $\frac{k[x,y]}{\langle p,q \rangle}$ is a flat $\mathbb{C}[x]$-module or a flat $\mathbb{C}[y]$-module? 
>  Or slightly more generally, when $\langle p,q \rangle= \langle r \rangle$ for some $r \in\mathbb{C}[x,y]$?
> 
> 
> **"Answer":** Perhaps the followig additional condition **(5)** would guarantee 
>  flatness of $\frac{\mathbb{C}[x,y]}{\langle p,q \rangle}$ over $\mathbb{C}[x]$
>  or $\langle p,q \rangle= \langle r \rangle$:
>  **(5)** $\{p\_x,q\_x\}$ are linearly independent over $\mathbb{C}$ and $\{p\_y,q\_y\}$ are linearly independent over $\mathbb{C}$.
> 
> 
> 


**Remarks:**


**(i)** If $(p,q)$ is an automorphic pair so $\mathbb{C}[p,q]=\mathbb{C}[x,y]$, then $x-\lambda, y-\mu \in \langle p,q \rangle$, hence $\langle x-\lambda, y-\mu \rangle \subseteq \langle p,q \rangle$. This impies that $\langle p,q \rangle$ is either non-proper or maximal which equals $\langle x-\lambda,y-\mu \rangle$. If it is maximal, then $\frac{\mathbb{C}[x,y]}{\langle p,q \rangle}=k$ which is not $\mathbb{C}[x]$-flat nor $\mathbb{C}[y]$-flat.


**(ii)** If $\frac{k[x,y]}{\langle p,q \rangle}$ is $\mathbb{C}[x]$-flat, then it can be shown that there exists $r \in k[x,y]$ such that $\langle p,q \rangle =\langle r \rangle$; indeed, apply [Corollary 1.3](https://projecteuclid.org/download/pdf_1/euclid.bams/1183533044) and [Theorem 178(3)](https://books.google.co.il/books?redir_esc=y&id=xmIPAQAAMAAJ&focus=searchwithinvolume&q=invertible%20ideal).


**(iii)** In condition **(5)** both $\{p\_x,q\_x\}$ and $\{p\_y,q\_y\}$ should be $\mathbb{C}$-linearly independent, as can be seen in my second comment below.


**An attempts to find a counterexample:**
$p=x, q=y+x^2$: Condition (1) is not satisfied, since 
$\frac{\mathbb{C}[x][y]}{\langle x \rangle}=\mathbb{C}[y]$ is not a flat $\mathbb{C}[x]$-module, by [this](https://math.stackexchange.com/questions/562243/criterion-for-proving-flatness) criterion (the action of $\mathbb{C}[x]$ on $\mathbb{C}[y]$ is $x r=0$ for every $r \in \mathbb{C}[y]$ and scalars of $\mathbb{C}[x]$ act as usual multiplication in $\mathbb{C}[y]$. $x \otimes\_{\mathbb{C}[x]} 1$ is in the kernel).


Any comments and hints are welcome! Thank you.



Flatness of certain quotient rings
Think about the geometry: let $X = V(p)$ and $Y = V(q)$ in $\mathbf A^2$. The nonvanishing of partial derivatives of $p$ and $q$ is equivalent to the statement that $X$ and $Y$ project surjectively onto each coordinate axis.


Since $\mathbf C[x]$ and $\mathbf C[y]$ are principal ideal domains, flatness is equivalent to torsion-freeness. Thus, a closed subscheme $Z \subseteq \mathbf A^2$ is flat over both coordinate axes if and only if every associated prime of $Z$ maps to the generic points in both coordinate $\mathbf A^1$s. In other words, every irreducible component of $Z$ is a curve, $Z$ has no emdedded points, and $Z$ does not contain a horizontal or vertical line. Then primary decomposition, Krull's principal ideal theorem, and factoriality of $\mathbf C[x,y]$ show that $Z = V(r)$ is principal. The condition that $Z$ contains no horizontal or vertical lines means that no irreducible factor of $r$ is of the form $x - a$ or $y - b$ for $a, b \in \mathbf C$.


Now $V(p,q)$ is the intersection $V(p) \cap V(q)$. By the discussion above, it is flat over both coordinate axes if and only if $(p,q) = (r)$ for a polynomial $r$ that has no irreducible factor of the form $x-a$ or $y-b$. But $r$ is a product of common factors of $p$ and $q$, so since $p$ and $q$ have no factors $x-a$ and $y-b$, the same goes for $r$. In this case $V(r)$ is the union of the components that $V(p)$ and $V(q)$ have in common.


Let $p = rp'$ and $q = rq'$ where $p'$ and $q'$ have no factors in common (but they could contain further powers of factors of $r$). Then $V(p) \cap V(q)$ is flat over both coordinate $\mathbf A^1$s if and only if $(p,q) = (r')$ for some $r'$. We claim that this is only possible if $(r') = (r)$ and $(p',q') = (1)$, i.e. $V(p') \cap V(q') = \varnothing$.


Indeed, suppose $V(p',q') \neq \varnothing$. Since $p'$ and $q'$ have no factors in common, $V(p',q')$ is a finite union of closed points. Then $r \in \mathbf C[x,y]/(p,q)$ is killed by $(p',q')$, hence all primes in $V(p',q')$ occur as associated primes for $(p,q)$, which is impossible when $(p,q)$ is supposed to be principal (see discussion above). Thus we conclude that $(p',q') = (1)$ and hence $(p,q) = (r)(p',q') = (r)$.



> 
> **Conclusion.** If $V(p)$ and $V(q)$ are flat over both coordinate axes, then the same holds for $V(p,q)$ if and only if $(p',q') = (1)$ where $p = rp'$ and $q = rq'$ where $p'$ and $q'$ have no factors in common. Geometrically, this means that $V(p') \cap V(q') = \varnothing$.
> 
> 
> 


**Example.** If $f$ and $g$ are irreducible polynomials not of the form $x-a$ or $y-b$, then $p = f^2g$ and $q = fg^2$ satisfy the hypotheses if and only if $(f,g) = (1)$.


For example, in Moret-Bailly's example $f = x + y$, $g = x - y$, this is not satisfied, because $(x+y,x-y) = (x,y)$ is the origin. We see that also $\mathbf C[x,y]/(f^2g,fg^2)$ is not flat over $\mathbf C[x]$, because the element $fg$ is killed by $x$.


A case where it does work is $f = x+y$ and $g = x + y + 1$.


