A certain Noetherian domain of dimension $1$ is normal away from a prime ideal

Question

Disclaimer. I know that I have posted the same exercise a few days ago, but this time I am asking if it is reasonable that a particular argument (that has not been brought up in the answers to my previous post) does not work, because I am quite convinced that this exercise should be solved, in my course, with such kind of argument, and the fact that it almost work, but not completely, does not sit right with me. Let me know if you consider it a duplicate and I will remove it.

Preliminary result.

Let $k$ be a ring and $k\to R$ an algebra that is a finite $k$-module. For any prime ideal $\mathfrak p\subset k$, then $R\otimes _k(k_\mathfrak p/\mathfrak pk_\mathfrak p)$ is a finite $k_\mathfrak p/\mathfrak pk_\mathfrak p$-vector space, so $R\otimes _k(k_\mathfrak p/\mathfrak pk_\mathfrak p)$ is Artinian.

So $R\otimes _k(k_\mathfrak p/\mathfrak pk_\mathfrak p)\cong R_1\times\dots\times R_n$, where the ring $R_i$ is local for all $1\le i\le n$, with maximal ideal $\mathfrak P_i$. Let $\mathfrak p_i$ be the contraction of $\mathfrak P_i$ via $R\to R\otimes _k(k_\mathfrak p/\mathfrak pk_\mathfrak p)\cong R_1\times\dots\times R_n\to R_i$. The prime ideals $\mathfrak p_1,\dots,\mathfrak p_n$ are precisely the prime ideals of $R$ which contract to $\mathfrak p$ and, for $1\le i\le n$: $$R_{\mathfrak p_i}/\mathfrak pR_{\mathfrak p_i}\cong (R/\mathfrak p)_{\mathfrak p_i}\cong (R\otimes_k(k_\mathfrak p/\mathfrak pk_\mathfrak p))_{\mathfrak p_i}\cong R_i.$$

Hence we conclude that $\mathfrak pR_{\mathfrak p_i}$ is the maximal ideal of $R_{\mathfrak p_i}$ if and only if $R_i$ is a field.

Question.

Let now $k$ be an algebraically closed field and $R:=k[X,Y]/((X^2+Y^2)^2-2a^2(X^2-Y^2))$. Denote by $x$, $y$ the images in $R$ of the variables $X$, $Y$ in $k[X,Y]$. I am proving that, for a maximal ideal $\mathfrak m\subset R$ with $\mathfrak m\neq (x,y)$, the localization $R_\mathfrak m$ is a normal domain; $R$ is a Noetherian domain of dimension $1$, so it is sufficient that, for a maximal ideal $\mathfrak m\subset R$ with $\mathfrak m\neq (x,y)$, the maximal ideal $\mathfrak mR_\mathfrak m$ is principal.

The algebra $k[X]\to R$ is a finite $k[X]$-module. For a maximal ideal $\mathfrak m:=(x-\lambda)\subset k[X]$, with $\lambda \in k$, one has $R/\mathfrak m R\cong R_1\times\dots\times R_n$, applying the preliminary result. Let $\mathfrak p_1,\dots,\mathfrak p_n\subset R$ be the corresponding prime ideals, i.e. the prime ideals of $R$ contracting to $\mathfrak m$.

If $R_i$ is a field, then $\mathfrak mR_{\mathfrak p_i}=\mathfrak p_iR_{\mathfrak p_i}$, i.e. the maximal ideal of $R_{\mathfrak p_i}$ is principal, generated by (the image of) $x-\lambda$; in this case one concludes that $R_{\mathfrak p_i}$ is normal.

If $R_i$ is not a field, it means that $\mathfrak p_iR_{\mathfrak p_i}$ is not generated by (the image of) $x-\lambda$; however it still might be a principal ideal I think, so $R_{\mathfrak p_i}$ might be normal.

E.g. take $\mathfrak m_0:=(X)\subset k[X]$. Note $R/\mathfrak m_0 R\cong k[Y]/(Y^2(Y^2+2a^2))\cong k[Y]/(Y^2)\times k\times k$. The maximal ideal of $R$ corresponding to $k[Y]/(Y^2)$ is $(x,y)$, so I cannot conclude that $xR_{(x,y)}=(x,y)R_{(x,y)}$ and that $R_{(x,y)}$ is normal, because $k[Y]/(Y^2)$ is not a field. Instead the two other maximal ideals in the fiber of $\mathfrak m_0$, i.e. $(x,y+\beta a)$ and $(x,y-\beta a)$ with $\beta \in k$ such that $\beta^2=-2$, correspond to a copy of $k$, so they must become principal in the respective localizations, that are then normal.

My issue is with the maximal ideals $\mathfrak m:=(X-c)\subset k[X]$, with $c\neq 0$. In that case $R/\mathfrak m_c R\cong k[Y]/(Y^4+2(a^2+c^2)Y^2-2a^2c^2)$; surely $Y^4+2(a^2+c^2)Y^2-2a^2c^2$ has four roots in $k[Y]$ because $k$ is algebraically closed, but I need to know that there are not multiple roots in order to conclude that $R/\mathfrak m_c R\cong k\times k\times k\times k$ and that the localizations at the maximal ideals in the fiber of $\mathfrak m_c$ are actually normal. However take $c$ such that $a^2+c^2=\beta ac$, with $\beta\in k$ such that $\beta^2=-2$: if $k$ is algebraically closed such a $c$ must exist, but then $Y^4+2(a^2+c^2)Y^2-2a^2c^2=(Y^2+\beta ac)^2$, hence there are two double roots in $k[Y]$, i.e. two maximal ideals $\mathfrak p_1,\mathfrak p_2$ in the fiber over $\mathfrak m_c$. This seem to contradict the fact that $R$ is normal at any maximal ideal except for $(x,y)$; maybe it is not a contradiction, because the ideals $\mathfrak p_1R_{\mathfrak p_1}$, $\mathfrak p_2R_{\mathfrak p_2}$ may still be principal, just not generated by (the image in the respective localizations of) $x-c$. However I believe that this exercise has to be solved using this kind of argument, because I found in the notes of my course among exercises of this kind; so it is good that it fails for $(x,y)$, but it is strange that it does not work for all the other maximal ideals in $R$.

Answer 1

A: Let $f:=y−x^2$ and $R:=k[x,y]/(f)$ and let $u:k[y]→R$ be the canonical map. Let $I:=(y−a)\subseteq k[y]$ be a maximal ideal with $a∈k$ any element. If $a≠0$ you get the map $k[y]/(y−a)=k→k⊕k$. If $a=0$ you get the map $k→k[x]/(x^2)$. The curve $C:=V(f)$ is a parabola hence it is regular, hence any local ring of $R$ is normal. Hence to me it seems your method cannot be used. – hm2020

Note I: The local ring $R_{\mathfrak{m}}$ is the ring of germs of functions defined in a neighborhood of $x:=\mathfrak{m}$. When you look at the map $k[x]/\mathfrak{m} \rightarrow R/\mathfrak{m}R$ you study the fiber $\pi^{-1}(x) \subseteq C$, of the map $\pi: C \rightarrow \mathbb{A}^1_k$, and its local rings.

Note II: The coming argument gives an explicit and elementary proof of the following: Let $\mathcal{O}_{C,p}$ be the local ring of a curve $C$ with $p$ a non singular point. It follows the maximal ideal $\mathfrak{m}_p$ is a principal ideal.

There is for any polynomial $f(x,y) \in k[x,y]$ and any $k$-rational point $p:=\mathfrak{m}:=(x-\alpha, y-\beta)$ with $f(\alpha, \beta)=0$ a relation

$$(*)h(x,y)(x-\alpha) + g(x,y)(y-\beta)=0$$

in $A:=k[x,y]/(f)$. The polynomials $h(x,y), g(x,y)$ have the following property:

$$h(\alpha, \beta)=f_x(p), g(\alpha, \beta)=f_y(p).$$

Using the relation $(*)$ you may prove the following: If $f_x(p) \neq 0$ it follows $\mathfrak{m}A_{\mathfrak{m}}=(y-\beta)$ and if $f_y(p)\neq 0$ it follows $\mathfrak{m}A_{\mathfrak{m}}=(x-\alpha)$. Hence if $p$ is a regular point it follows $\mathfrak{m}A_{\mathfrak{m}}$ is a principal ideal. The polynomial $f(x,y)$ is arbitrary.

$aX + bY$ is an element of $M^2$ if and only if the line $aX + bY = 0$ is tangent to $W$ at $(0, 0)$

Answer 2

You're largely on the right track here. Getting a point in the fiber over $X-c$ of the form $\operatorname{Spec} k[Y]/(Y^2)$ means that the maximal ideal of the local ring of that point of your curve is not generated by $x-c$, but it does not necessarily mean the maximal ideal of the local ring of that point is not principal. After all, it could be generated by $y-d$ - since the maximal ideal is $(x-c,y-d)$, if it is singly generated, we must have $(x-c)=(x-c,y-d)$ or $(y-d)=(x-c,y-d)$ (this is not an exclusive or - both could be true at the same time).

One thing we can do with the argument you lay out is run it in both the $x$-direction and the $y$-direction at the same time. This gives us something: in order for the maximal ideal $(x-c,y-d)\subset R$ to be neither generated by $x-c$ nor generated by $y-d$, we must have that $(c^2+d^2)^2-2a^2(c^2-d^2)=0$, $c$ is a double root of $(X^2+d^2)^2-2a^2(X^2-d^2)$, and $d$ is a double root of $(c^2+Y^2)^2-2a^2(c^2-Y^2)$.

Since $t$ is a double root of a polynomial $p$ iff $t$ is a root of $p$ and its formal derivative $p'$, and the formal derivative $p'$ is not zero, we observe that $c,d$ must solve the equations $(c^2+d^2)^2-2a^2(c^2-d^2)$, $4c(-a^2+d^2+c^2)$, and $4d(a^2 + c^2 + d^2)$, and we must also have $4\neq 0$. Assuming we are in characteristic not two, the only way for this to happen is for $c=d=0$:

• if $c=0$ and $d\neq 0$, then we must have $a^2=-d^2$ by the third equation, giving that the first equation evaluates to $-a^4$,
• if $c\neq 0$ and $d=0$, then we must have $c^2=a^2$ by the second equation, giving that the first equation evaluates to $-a^4$,
• if $c\neq0$ and $d\neq 0$, we must have $-a^2+c^2+d^2=a^2+c^2+d^2=0$, implying $a=0$.

This proves what you're after, once you add the necessary hypothesis of being in characteristic not two. (The characteristic hypothesis is necessary, because no maximal ideal of $k[X,Y]/((X^2+Y^2)^2-2a^2(X^2-Y^2))\cong k[X,Y]/((X^2+Y^2)^2)$ is singly generated.)