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Given the ring $$A = \frac{K[x,y]}{y^2-x^2(x+1)}$$ I know that its normalization is $K[t]$, where $$x\mapsto t^2-1\qquad y\mapsto t^3-t$$ I have to show that the normalization map is not flat.

I know that the problem is with the maximal ideal $m=(x,y)$, since it's the counterimage of two maximal ideals in $K[t]$, that are $(t-1)$ and $(t+1)$. I think I've to show that $$A_m\to K[t]_m$$ doesn't make $K[t]_m$ a free module, but I don't know how.

Some background: this is a counterexample to the lemma

If $X$ is a reduced noetherian scheme, and $Y$ a regular irreducible scheme of dimension 1, then $f:X\to Y$ is flat iff every irreducible component of $X$ dominates $Y$

in the case $Y$ is not regular

Exodd
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1 Answers1

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A much more general statement is true: let $A$ is an integral domain and let $B$ is the integral closure in $\textrm{Frac}(A)$. If $B$ is finitely-generated as an $A$-module, then it is flat if and only if $A$ is integrally closed (i.e. $B = A$). A nice proof of this is given in user26857's answer here.

This result obviously implies that the integral closure $K[t]$ of the nodal cubic $\frac{K[x,y]}{(y^2 - x^2(x+1))}$ is not flat. Though there are certainly easier ways of showing this, we can glean more from normalization maps in general from this strategy.

The commutative algebra statement can be globalized and reinterpreted in the language of schemes. We can say that, if $X$ is a noetherian integral scheme, then the normalization map $f \colon \tilde{X} \to X$ is never flat (unless, of course, $f$ is an isomorphism). This is Example 4.3.5 of Liu.

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msteve
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