Normalisation of Quotient of Polynomial Ring

Question

My question refers to a former thread of mine: Normalization of a Scheme Example

The setting is the following: We have the ring $A=\mathbb {C} [X,Y]/(XY)$ and consider an arbitrary prime ideal $p$ of $A$. I want to verify that for each such ideal $p$ the ring $$B=(\mathbb {C} [X,Y]/(Y)\times \mathbb {C} [X,Y]/(Y)))_{\mathcal{p}} \cong (\mathbb {C} [X,Y]/(X))_{\mathcal{p}} \times (\mathbb {C} [X,Y]/(Y)))_{\mathcal{p}}$$

is the normalisation of $\mathbb {C} [X,Y]/(XY)_p$. Therefore $(\mathbb {C} [X,Y]/(X))_{\mathcal{p}} \times (\mathbb {C} [X,Y]/(Y)))_{\mathcal{p}}$ is integral closure of $A_p$ with respect to $Frac(A_p)$.

Honestly, I have no idea, since I can't apply Chinese remainder theorem to $A$ and therefore get the desired direct product structure.

Firstly, I would like to consider the more easy case $P \neq (X,Y)$. Then we have already $A_p=Frac(A_p)$, the problem is how to bring $B$ into relation to $A_p$. Is here Chinese reminder applicable?

A further problem is to compare $Frac(A_p)$ and $Frac((\mathbb {C} [X,Y]/(X))_{\mathcal{p}}) \times Frac((\mathbb {C} [X,Y]/(Y))_{\mathcal{p}})$.

Remark: In this thread I'm keen interested how to show this on ring level, not geometrically as often just used hand weaving arguments for normalisation of schemes/curves.

Answer 1

I don't think passing to localizations simplifies anything in this case, so let's just directly show that $\Bbb C[X,Y]/(X) \times \Bbb C[X,Y]/(Y)$ is the integral closure of $\Bbb C[X,Y]/(XY)$ in its total ring of fractions.

Let $R$ be a commutative ring and let $p$ and $q$ be distinct prime elements in $R$ such that $\bigcap_{n \geq 1} (p^n) = \bigcap_{n \geq 1} (p^n) = 0$ (e.g. $R$ is a UFD or a Noetherian domain.) In general, we have an injective homomorphism $f:R/((p) \cap (q)) \to R/(p) \times R/(q)$ constructed as in the Chinese remainder theorem. (Note that $(pq) \neq (p) \cap (q)$ in general and we can't say anything about surjectivity.)

$R/((p)\cap (q))$ and $R/(p)\times R/(q)$ have the same total ring of fractions.

Let $S$ be the set of non zero divisors in $R/((p) \cap (q))$. By the injectivity of $f$, $f(S)$ is contained in the set of non-zero divisors of $R/(p) \times R/(q)$. This implies that we get an induced (injective) homomorphism $\tilde{f}:\operatorname{Frac}(R/((p)\cap (q))) \to \operatorname{Frac}(R/(p) \times R/(q))$

Consider the element $p+q$ in $R/(pq)$. $f(p+q)=(q,p) \in R/(p) \times R/(q)$ which is not a zero-divisor since each component is non-zero (and $R/(p)$ and $R/(q)$ are integral domains), so by injectivity of $f$, $p+q$ is not a zero-divisor in $R/((p) \cap (q))$. Thus we can consider

$\tilde{f}(\frac{q}{p+q})=\frac{(q,0)}{(q,p)}=\frac{(1,0)}{(1,1)}$, since $(q,0)(1,1)-(q,p)(1,0)=0$. By symmetry, we have $\tilde{f}(\frac{p}{p+q})=\frac{(0,1)}{(1,1)}$

Now let $\frac{(r_1,r_2)}{(s_1,s_2)}$ be an element in $\operatorname{Frac}(R/(p) \times R/(q))$. We get $\frac{(r_1,r_2)}{(s_1,s_2)}=(\frac{(1,0)}{(1,1)}+\frac{(0,1)}{(1,1)})\frac{(r_1,r_2)}{(s_1,s_2)}=\frac{(r_1,0)}{(s_1,s_2)}+\frac{(0,r_2)}{(s_1,s_2)}$. So by symmetry and the previous computations, we may assume that $r_2=0$

$\frac{(r_1,0)}{(s_1,s_2)}=\frac{(r_1,0)}{(s_1,1)}$, since $(r_1,0)(s_1,1)-(r_1,0)(s_1,s_2)$ so assume $s_2=1$.

Since $\frac{(r_1,0)}{(s_1,1)}=\frac{(r_1,r_1)}{(1,1)} \frac{(1,0)}{(s_1,1)}=\tilde{f}(r_1) \frac{(1,0)}{(s_1,1)}$, we may also assume that $r_1=1$.

Lift $s_1$ to an element $s$ in $R$. Since we assumed that $\bigcap_{n \geq 1} (q^n)$, we can write $s=q^ns'$ for some $n \in \Bbb N_0$ and some $s' \in R$ such that $s' \not \in (q)$. $s'$ is mapped to a non-zero element in both $R/(p)$ and $R/(q)$, so using the injectivity of $R/((p) \cap (q)) \to R/(p) \times R/(q)$, we get that $s'$ is not a zero-divisor in $R/((p) \cap (q))$.

Thus we can consider $\tilde{f}(\frac{1}{s'})=\frac{(1,1)}{(s',s')}$.

We use a variant of a previous argument and get that $q^{n+1}+p$ is not a zero divisor in $R/((p) \cap (q))$ and obtain $\tilde{f}(\frac{q}{q^{n+1}+p})=\frac{(q,0)}{(q^{n+1},p)}=\frac{(1,0)}{(q^n,1)}$

Thus the image will also contain the product $\frac{(1,0)}{(q^n,1)} \frac{(1,1)}{(s',s')}=\frac{(1,0)}{(s'q^n,1)}=\frac{(1,0)}{(s_1,1)}$. By the previous reductions, this is sufficient for the surjectivity of $\tilde{f}$. $\square$

For any ring $R$, the diagonal $R \to R \times R$ is an integral homomorphism.

Clearly $R \times R$ is generated by $(1,0)$ and $(0,1)$ as an $R$-module. It's well-known that finite homomorphisms are integral. $\square$

For two integral homomorphisms $R \to S$ and $R \to T$, the induced homomorphism $R \to S \times T$ is integral.

We can factorize over the diagonal $R \to R \times R \to S \times T$.
$R \times R \to S \times T$ is integral as we can clearly check this for each coordinate separately. $R \to R \times R$ is integral by the previous lemma. Now apply transitivity. $\square$

Note that in your situation, the first result applies to $\Bbb C[X,Y]$ with $p=(X)$ and $q=(Y)$ and we have $(X) \cap (Y) = (XY)$ (e.g. by unique factorization.) Also the maps $\Bbb C[X,Y]/(XY) \to \Bbb C[X,Y]/(X)$ and $\Bbb C[X,Y]/(XY) \to \Bbb C[X,Y]/(Y)$ are clearly integral since they are surjective.