# Is this particular module flat?

Let $A=k[x^2,xy,y^2]\hookrightarrow B=k[x,y]$, where $k$ is a field. Is $B$ flat over $A$?

I am guessing the answer is no. My first thought is, since $B$ is integral over $A$, so it's finitely generated as an $A$-module, but I don't know how to go any further. However, I have the following theorem in hand but don't know how to apply properly in this situation.

Theorem. If $A$ is a local ring and $M$ a finitely generated flat $A$-module, then $M$ is free.

• Commented Jan 22, 2017 at 21:55

The answer is negative since $A\subset B$ flat and $B$ regular implies $A$ regular; see Bruns and Herzog, Theorem 2.2.12. But in this case $A\simeq k[a,b,c]/(ac-b^2)$, so $A$ is not regular.
Edit. A simpler approach: let $I=(x^2,xy)$ and $A/I\to A/I$ be the multiplication by $y^2$. Since $A/I\simeq k[y^2]$ this is injective, but on $A/I\otimes_AB\to A/I\otimes_AB$ it is not (why?).
• Q1) Thm. 2.2.12 refers to local rings. How can you see immediately that this transfers to $^*$local rings? Q2) Why is $A$ not regular? Q3) Where is the flaw in my argument? Commented Dec 16, 2014 at 2:03
• @Manos We can localize at the prime ideal $(x^2,xy,y^2)A$. The embedding dimension is $3$ but the dimension is $2$. Commented Dec 16, 2014 at 2:06