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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.

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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?).

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  • $\begingroup$ 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? $\endgroup$
    – Manos
    Commented Dec 16, 2014 at 2:03
  • $\begingroup$ @Manos We can localize at the prime ideal $(x^2,xy,y^2)A$. The embedding dimension is $3$ but the dimension is $2$. $\endgroup$
    – Youngsu
    Commented Dec 16, 2014 at 2:06
  • $\begingroup$ Q1) Localizing both rings at their irrelevant maximal ideals preserves the flatness. Q2) When specialize a regular (local) ring by an element in the square of the maximal ideal you get a non-regular ring. $\endgroup$
    – user26857
    Commented Dec 16, 2014 at 9:18

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