This relates to problem 3.14(b) of Hartshorne, wherein I am required to show that the projection of the twisted cubic curve (viewed as a subset of $\mathbb P^3$) from the point $(0,0,1,0)$ onto the hyperplane $\mathbb P^2$ defined by the equation $z=0$ is a cuspidal cubic curve. I have been able to show that the said projection is the zero set of the homogenous polynomial $f(x,y,z):= y^3-x^2z \in k[x,y,z]$ so it is a cubic curve. From some searching, I learnt that a cuspidal curve is one where all singularities are cusps and by using the characterization of singular points in projective curves defined by a single equation as points where all partial derivatives vanish, I have been able to show that the only singularity of $f$ occurs at the point $(0, 0, 1) \in \mathbb P^2$. Now I am not sure how to show that this point is a cusp, because I require a rigorous definition of "cusp" in this context which is easy to verify.

Most references I came across give a definition to the effect that a "cusp" is a point where two "branches" have a common semi-tangent, and the notion of "branch" is either handwaved or explained only for Euclidean spaces. The only rigorous definition I have found after searching of a cusp $P$ on a variety $Y$ involves completions of $\mathcal O_{P, Y}$. Now I am not familiar with the idea of completion in this context and would really appreciate a definition which is rigorous and can be used in the aforementioned context.

Edit: Here (https://arxiv.org/pdf/1511.02691.pdf) is a reference, where "cusp" is defined on pg. 6, but it uses the notion of a "branch", which has not been defined. I would really like to know a simple rigorous definition of that terminology.

  • $\begingroup$ Oh that's fun, I never noticed that before: Hartshorne asks you to do this before defining cusps. Cusps are defined in a rather roundabout manner in the exercises of Chapter I Section 5. Basically, a curve has a cusp at the origin if its Taylor expansion looks like $y^2 - x^3 + (\text{higher-order terms})$, which is precisely your curve in the chart $z=1$. $\endgroup$ Nov 20, 2020 at 17:36
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    $\begingroup$ You've shown that the projection is $V(y^3-x^2z)$, which is the standard cuspidal cubic. I would say that this is enough in the context of this exercise. (Hartshorne has gaps in his text, this is one of them, congrats on finding it. Yes, it's frustrating, but coming up with a good definition of "cuspidal" is something that should wait until you have some more tools on your belt - you should at least make it to chapter I section 5 first.) $\endgroup$
    – KReiser
    Nov 20, 2020 at 19:10
  • $\begingroup$ In the reference you linked, by branch the author really means tangent line. So a cusp is a singular point of the curve at which there is only one (repeated) tangent line. For instance, at $(0,0)$ on $y^2 = x^3$, the tangent line is $y = 0$ with multiplicity $2$. $\endgroup$ Nov 21, 2020 at 6:01

1 Answer 1


The plane curve $f(x,y)=x^2(x+1)-y^2=0\subset \Bbb{C}^2$ doesn't have a cusp at $(0,0)$ because we have a parametrization $(x,x\sqrt{1+x}),(x,-x\sqrt{1+x})$ of two distinct "analytic curves" around $(0,0)$.

In contrary $x^2-y^3=0$ has a cusp because $(y\sqrt{y},y)$ is not an analytic curve around $(0,0)$.

Over $\Bbb{C}$ we have the ring of all functions that are analytic in some neighborhood of $x_0=0$, but this doesn't generalize well to arbitrary fields.

We may try to replace it by the ring of formal power series in $x$, which is $k[[x]]=\varprojlim k[x]/(x^n)$, or in $y$, which is $k[[y]]=\varprojlim k[y]/(y^n)$, but it appears that the most insightful ring is $\varprojlim k[x,y]/(f(x,y))/(x,y)^n=k[[x,y]]/(f(x,y))$.

If $char(k)\ne 2$ then $\sqrt{1+x}=\sum_{k\ge 0} {1/2\choose k}x^k$ is in this ring and $(y-x\sqrt{1+x}),(y+x\sqrt{1+x})$ are two distinct prime ideals. $k[[x,y]]/(f(x,y))$ is not an integral domain as $(y-x\sqrt{1+x})(y+x\sqrt{1+x})=0$. In contrary to $k[[x,y]]/(x^2-y^3)$ which is an integral domain.

For an arbitrary variety this generalizes to $\varprojlim O_P / \mathfrak{p}^n$ where $O_P$ is the ring of rational functions regular at $P$ (or the localization thereof for a non-closed point) and $\mathfrak{p}$ is the prime ideal of those vanishing at $P$ (its unique maximal ideal).

For a curve, when a parametrization $(x,h(x))$ exists with $h $ a formal power series then in fact $h$ is algebraic over $k[x]$ so that it can be seen as a rational function on a ramified covering of our curve.

  • $\begingroup$ Thanks for your answer. Could you let me know what an analytic curve is? (I am relatively new to algebraic geometry). Also could you please look at the above edit? $\endgroup$
    – asrxiiviii
    Nov 20, 2020 at 17:09
  • $\begingroup$ The analytic curve is $\{ (x,x\sqrt{1+x})\in \Bbb{C}^2,|x|<1/2\}$. $\qquad x\sqrt{1+x}$ is analytic at $0$ whereas $y\sqrt{y}$ is not. $\endgroup$
    – reuns
    Nov 20, 2020 at 17:11
  • $\begingroup$ Is it reasonable to define a cusp in an arbitrary curve (e.g. a curve in $\mathbb P^n$) as a point s.t. the completion of its local ring is $\ne k$ and an integral domain? (Also - do you know any references about singularities of curves in spaces of $\dim > 2$?) $\endgroup$ Mar 13 at 14:48

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