Flatness of $\mathbb{C}[x_1,\ldots,x_n]$ over $\mathbb{C}[f]$, $f \in \mathbb{C}[x_1,\ldots,x_n]$, $n \geq 1$

Question

Let $f \in \mathbb{C}[x_1,\ldots,x_n]$, $n \geq 1$. Call $f$ 'good' if $\mathbb{C}[x_1,\ldots,x_n]$ is flat over $\mathbb{C}[f]$.

Is it true that every $f \in \mathbb{C}[x_1,\ldots,x_n]$ is good?

• $f \in \mathbb{C}$ is good, since $\mathbb{C}[x_1,\ldots,x_n]$ is free (hence flat) over $\mathbb{C}$.
• $n=1$: $\mathbb{C}[x]$ is a free $\mathbb{C}[x^2]$-module (of rank two), hence flat, so $\mathbb{C}[x]$ is flat over $\mathbb{C}[x^2]$. Similarly, if $f \in \mathbb{C}[x]$ is of degree $d$, then $\mathbb{C}[x]$ is free (hence flat) over $\mathbb{C}[f]$ of rank $d$. Therefore, every $f$ is good.
• $n=2$: If $f \in \mathbb{C}[x]$ then $\mathbb{C}[x,y]$ is free (of infinite rank) over $\mathbb{C}[f]$ and if $f \in \mathbb{C}[y]$ then $\mathbb{C}[x,y]$ is free (of infinite rank) over $\mathbb{C}[f]$. But what about a general $f \in \mathbb{C}[x,y]$? It seems to me that the extension is still free: If $\deg_x(f)=d_1 < \infty$ and $\deg_y(f)=d_2 < \infty$, then $\mathbb{C}[x,y]$ is free of rank $d_1d_2$. If $\deg_x(f) = -\infty$ or $\deg_y(f) = -\infty$, then $\mathbb{C}[x,y]$ is free of infinite rank.
• $n \geq 3$, similarly to the $n=2$ case, every $f$ is good.

Summarizing, it seems that $\mathbb{C}[x_1,\ldots,x_n]$ is free (hence flat) over $\mathbb{C}[f]$, for every $f \in \mathbb{C}[x_1,\ldots,x_n]$.

Please, am I right or wrong? Thank you very much.

Answer 1

Here is an algebraic-geometric way to prove this, though it might be considered cracking a nut with a sledgehammer. For non-constant $f$, the inclusion $\mathbb C[f] \subset \mathbb C[x_1, \dotsc, x_n]$ induces a morphism of schemes $$f: \mathbb A^n \to \mathbb A^1, \quad (p_1, \dotsc, p_n) \mapsto f(p_1, \dotsc, p_n).$$ Hartshorne's Algebraic Geometry, Chapter III.9 contains the following

Proposition 9.7. Let $f: X \to Y$ be a morphism of schemes with $Y$ integral and regular of dimension $1$. Then $f$ is flat if and only if every associated point $x \in X$ maps to the generic point of $Y$. In particular, if $X$ is reduced, this say that every irreducible component of $X$ dominates $Y$.

As $\mathbb A^n$ is reduced with a single irreducible component, we see that the morphism $f$ is flat because $f$ is non-constant.