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Determine the local ring $\mathcal{O}$ at $(0,0,0)$ of the curve consisting of the three coordinates axes in $\mathbb{A}^3$. Then, determine the local ring at $(0,0)$ of the curve $xy(x - y)= 0$ and prove that this second curve is not isomorphic to the first one.

My attempt By definition of the local ring, $\mathcal{O}$ for the first curve is the set of all rational functions $f/g$ such that $g$ does not vanish on three axes. As for the second curve, the local ring is the set of all rational functions $f/g$ such that $g$ does not vanish at $x=0$ or $y=0$ or $x=y$. Is this what the exercise is really asking or a more specific description of the ring is required here? I can’t seem to think of any, though. About the isomorphism between the two curves, I thought of using the fact that the tangent space is a local invariant: by showing that the two tangent spaces are different, I would have proved that the two curves cannot be isomorphic. Could that be the right path? Any hints?

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  • $\begingroup$ Your definition of the local ring is wrong and should be corrected, but your strategy about tangent spaces is the right approach. You should review the definition of the local ring at a point in your text, and your goal should be to write it down using some basic ring-theoretic tools: quotients, localizations, etc. $\endgroup$
    – KReiser
    Commented Aug 21, 2020 at 19:18
  • $\begingroup$ Yes, sorry, I just edited the definition. Should I write the local ring as a quotient now? $\endgroup$
    – cip
    Commented Aug 21, 2020 at 19:38
  • $\begingroup$ Also, knowing that the tangent space is the union of all the tangent lines at $x$, I divided the curve into its irreducible components and for each one of them I computed the tangent line. For the first curve I got that the tangent space is the curve itself and for the second curve the tangent space is the union of $x=0$ and $y=0$. Is that right? $\endgroup$
    – cip
    Commented Aug 21, 2020 at 20:12
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    $\begingroup$ One can write the local ring using a localization and a quotient. Have you done much with local rings of points on varieties before (have you seen examples and the like)? And no, the tangent space is a vector space while nothing you've written down is. You should review the definition of the Zariski tangent space and how it's obtained from the local ring of a point. $\endgroup$
    – KReiser
    Commented Aug 21, 2020 at 20:20
  • $\begingroup$ Well $xyz=0$ is a hypersurface in $\mathbb{A}^3$. I think you meant $xy,yz,zx=0$ and the point of the problem is to compare planar vs spatial triple points? $\endgroup$ Commented Aug 22, 2020 at 2:22

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Local ring at a point, by definition, is the localization of the coordinate ring at the maximal ideal corresponding to the point. So for example origin in $xy,yz,zx=0$ in $\mathbb{A}^3$, $$\mathcal{O}_{X,0}=\big(\mathbb{C}[x,y,z]/(xy,yz,zx)\big)_{(x,y,z)}$$ Then using the fact that localization commutes with quotients, and that the localization for $\mathbb{C}[x,y,z]$ at $(x,y,z)$ is simply the subring $$\{f/g\,|\,f,g\in\mathbb{C}[x,y,z],\,g(0,0,0)\not=0\}\subset\mathbb{C}(x,y,z),$$ we get the description $$\mathcal{O}_{X,0}=\{f/g\,|\,f,g\in\mathbb{C}[x,y,z]\text{ that kills }xy,yz,zx,\,g(0,0,0)\not=0\}.$$ Similarly $$\mathcal{O}_{Y,0}=\{f/g\,|\,f,g\in\mathbb{C}[x,y]\text{ that kills }xy(x-y),\,g(0,0,0)\not=0\}.$$ To show these two rings are not isomorphic is usually hard. You always need to look for some kind of invariants that are different on two sides. Tangent space is a very good one, as usually the dimension (as a vector space) can differ.

However there is a much much more important one here, and I will try to explain it intuitively:

Let's think about the spatial triple points first. How can I get a function on the whole thing from the functions on each piece? Well because the three lines sit in $\mathbb{A}^3$, there are enough functions (as a global function) to agree with each component. As long as the three small functions agree at origin, I should be able to put them back together to get a global function.

How about the planar triple points? This is different: you cannot hope that any three functions on the three pieces glue to a global function, because there aren't that many! The three functions might not be compatible because of the restraints of $\mathbb{A}^2$. In particular, this is an example of what is called an elliptic singularity, and you need extra condition besides agreeing at origin for them to glue.

The proper math here is the $\delta$-invariant of a singularity, which is defined to be the dimension of $\pi_*(\mathcal{O}_{\tilde{C}})/\mathcal{O}_C$ where $\pi:\widetilde{C}\to C$ is the normalization. This will be different in the two cases here.

If you are using tangent space, then the two definitions you give are the same. The $(\mathcal{m}/\mathcal{m}^2)^\vee$ is usually called the Zariski tangent space. In particular, in the localization of $X$, the unique maximal ideal is $(x,y,z)$, and $m/m^2=(x,y,z)/(x^2,xy,xz,y^2,yz,z^2)$ which is 3-dimensional as a $k$-vector space. On the other hand in $Y$, $m/m^2=(x,y)/(x^2,y^2,xy)$ is 2-dimensional. So yeah they are indeed different.

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  • $\begingroup$ Thanks Matt. I have a few questions: how did you find that the maximal Ideal is $(x,y,z)$? And how did you compute the dimension of the quotient? $\endgroup$
    – cip
    Commented Aug 22, 2020 at 11:14
  • $\begingroup$ It corresponds to all equations being zero, i.e. $x=0,y=0,z=0$ and thus the origin; localization has a unique maximal ideal $pAp$, which is still the one generated by $p=(x,y,z)$. The dimension is easy: the basis is just the single variables $x,y,z$ and $x,y$. $\endgroup$ Commented Aug 22, 2020 at 14:08
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    $\begingroup$ For example, in $k[x]$, $(x)/(x^n)$ has a $n-1$ dimensional vector space structure with a basis given by $x^i$ for $i=1,\cdots,n-1$. The dimension of $(x,y,z)$ is infinite; it’s only finite when you quotient by a sufficiently large subset. When you quotient by all possible exponent 2 monomials, then a basis is given by all the exponent 1 stuff, thus 3 dimensional. $\endgroup$ Commented Aug 25, 2020 at 8:00
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    $\begingroup$ As for why you need to find the maximal ideal, it’s because you are looking the tangent space at that point. Taking localization or not doesn’t change the tangent space structure at origin here. Alternatively using your geometric definition you see planar triple points has 3 tangent directions in $\mathbb{C}^2$, but then as a vector space it must be all of $\mathbb{C}^2$; while for spatial one it is full 3 dimensional. $\endgroup$ Commented Aug 25, 2020 at 8:05
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    $\begingroup$ It's the partial Jacobians: here it would be $\begin{pmatrix} z&0&x\\y&x&0\\0&z&y\end{pmatrix}\cdot v=0$, and you can see it's $0v=0$ so $v$ is full 3-dimensional. $\endgroup$ Commented Aug 26, 2020 at 3:25

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