Determining the local ring of a curve 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?
 A: 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.
