On the flatness of a particular ring map on two variable polynomial ring

Question

Is the $\mathbb C$-algebra map $f: \mathbb C[x,y]\to \mathbb C[x,z]$ given by $f(x)=x$ and $f(y)=xz$ a flat map? i.e. is $\mathbb C[x,z]$ a flat $\mathbb C[x,y]$-module via the map $f$ ?

I was trying to use the equational criteria for flatness, but no luck.

Please help

Answer 1

I don't think so. One may rephrase your setting as follows: for $R = k[x,xy] \subset S = k[x,y]$, is $S$ a flat over $R$?

If $S$ were $R$-flat, then the regular sequence $xy,x$ in $R$ will extend to a regular sequence in $S$. However, $x$ is not regular on $S/xyS$.

Answer 2

The map $f$ is flat if and only the dual map $$f^\sharp:\mathbb C^2\to \mathbb C^2:(x,z)\mapsto (x,y)=(x,xz)$$ of algebraic varieties is flat.
However $f^\sharp$ is not flat, so $f$ is not flat.
But why is $f^\sharp$ not flat? It is not flat because a flat map between varieties is open and $f^\sharp$ is not open:
Indeed $f^\sharp(\mathbb C^2)=\mathbb C^*\times \mathbb C\cup \{(0,0)\}$, which is not open in the Zariski topology of $\mathbb C^2$.

Answer 3

The map $f$ is flat if and only the dual map $$f^\sharp:\mathbb C^2\to \mathbb C^2:(x,z)\mapsto (x,y)=(x,xz)$$ of algebraic varieties is flat.
However $f^\sharp$ is not flat, so $f$ is not flat.
But why is $f^\sharp$ not flat? It is not flat because if a morphism $g:X\to Y $ between irreducible varieties is flat, then the fiber of any point $y\in g(X)\subset Y$ must have dimension: $$\operatorname {dim}(g^{-1}(y))=\operatorname {dim}X-\operatorname {dim} Y$$ In our case $\operatorname {dim}X=\operatorname {dim} Y=\operatorname {dim}\mathbb C^2=2$, so that all fibers of $f^\sharp$ should have dimension $0$.
But that condition is violated since the fiber of $y=(0,0)\in \mathbb C^2$ is $$(f^\sharp)^{-1}(y)=\{0\}\times \mathbb C$$ which has dimension $1$.