This is exercise IV.6.2 in Hartshorne, which asks us to find an example of a nonsingular rational degree $5$ curve in $\mathbb{P}^3$ which is not contained in a quadric.

I'm somewhat at a loss for how to do this. Similar problems in this section relate these kind of questions to intrinsic properties of the curve (like having certain linear systems, etc.), but this certainly cannot work here, since all rational nonsingular curves are isomorphic.

Can anyone offer me any hints as to how I can find such a curve? Is it just a matter of trying out different parametrizations?

I also thought there was maybe some kind of moduli dimension argument here, but that also doesn't seem to work, since the moduli of rational nonsingular curves is just a point.


PS: In a solutions document I saw someone claim that $(s: t) \mapsto (s^5: s^4t: s^4t + as^3t^2: t^3)$ which could very well be right, but I don't really understand how one can find such curves.

  • $\begingroup$ @coudy Sorry, it was supposed to be a quadric, that it's not contained in. Every degree 5 rational curve is contained in a cubic $\endgroup$
    – Daniel
    Commented Mar 8, 2023 at 23:43
  • $\begingroup$ +1 for this nice answer $\endgroup$
    – TShiong
    Commented Mar 9, 2023 at 18:50

2 Answers 2


You're right, there's some amount of leg work to track down an actual example of a curve here. Let's go through why this result is possible and how to make the exploration as easy as we can.

The way to check whether a nonsingular rational curve $X$ of degree $d$ in $\Bbb P^3$ must be contained in a surface of degree $e$ is twisting the exact sequence $0\to \mathcal{I}_X\to\mathcal{O}_{\Bbb P^3}\to\mathcal{O}_X\to 0$ and then taking global sections: twisting by $e$ gives you $$0\to\mathcal{I}_X(e)\to\mathcal{O}_{\Bbb P^3}(e)\to\mathcal{O}_X(e)\cong\mathcal{O}_{\Bbb P^1}(de)\to 0.$$ The global sections of $\mathcal{O}_{\Bbb P^1}(de)$ are of dimension $de+1$, while the global sections of $\mathcal{O}_{\Bbb P^3}(e)$ are of dimension $\binom{3+e}{3}$. When $d=5$ and $e=3$, we have a map from a vector space of dimension $20$ to a vector space of dimension $16$, so $\Gamma(\Bbb P^3,\mathcal{I}_X(3))\neq 0$ and $X$ lies on a cubic surface; when $d=5$ and $e=3$, we have a map from a vector space of dimension $10$ to a vector space of dimension $11$, so we could have $\Gamma(\Bbb P^3,\mathcal{I}_X(3))=0$ (i.e. $X$ is not on a quadric surface), but we're not locked in to this outcome without doing more computations further on in the long exact sequence of cohomology.

To explain how to set up the guessing process to make it as easy as possible, let's start with what we know: the map $\Bbb P^1\to\Bbb P^3$ is

  1. a closed immersion
  2. by $\mathcal{O}_{\Bbb P^1}(5)$.

The fact the morphism is given by $\mathcal{O}(5)$ means it's given by $[a:b:c:d]$ for $a,b,c,d$ quintic polynomials in the coordinates $[s:t]$ on $\Bbb P^1$, and they can't have any common vanishing locus. The easiest way to get that to happen is if $a=s^5$ and $d=t^5$. The fact that this is a closed immersion means that we probably want one of $b$ or $c$ to be $s^4t$ or $st^4$ so that we get that our map is a closed immersion on $D(s)$ or $D(t)$, respectively, and then put a little more work in to showing that our map is a closed immersion on the other patch. This leaves us with three of the entries of our map picked and one to play with:

$$ [s^5: s^4t: ? : t^5]$$

Since we can subtract off any multiples of $s^5$, $s^4t$, or $t^5$ from the $?$ by means of a linear automorphism of $\Bbb P^3$, we might as well assume that $?$ is of the form $ps^3t^2+qs^2t^3+rst^4$. We'll want $r\neq 0$ to get something of valuation $1$ in the local ring at $s=0$ in order to get a closed immersion there, so set $r=1$. Now we need to do two things: find some $p,q$ so that the only quadric vanishing on our curve is $0$, and double check that with those $p,q$ we get a closed immersion on $D(t)\subset \Bbb P^1$. I'll spoiler-tag those calculations so you get a fair shot at doing them yourself. (Personally, I'd say these are potentially a little tedious depending on your appetite for this sort of thing, but they should be doable, and there's some fun in trying to set everything up to give you the best shot at the least work.)

Finding $p,q$:

Now let's write down the form of a quadric in 4 variables and plug in. Writing our quadric as $$c_{00}x_0^2+c_{01}x_0x_1+c_{02}x_0x_2+c_{03}x_0x_3+c_{11}x_1^2+c_{12}x_1x_2+c_{13}x_1x_3+c_{22}x_2^2+c_{23}x_2x_3+c_{33}x_3^2$$ and substituting our equations, we get $$c_{00}s^{10} + c_{01} s^9t + c_{02} (ps^8t^2 + qs^7t^3+s^6t^4) + c_{03} s^5t^5 +$$ $$+ c_{11}s^8t^2 + c_{12} (ps^7t^3 + q s^6t^4 + s^5t^5) + c_{13} s^4t^6 + $$ $$+ c_{22}(p^2s^6t^4+2pqs^5t^5+q^2s^4t^6+ps^4t^6+2qs^3t^7+s^2t^8) + c_{23}(ps^3t^7+qs^2t^8+st^9) + $$ $$+c_{33}t^{10}.$$ If this is to vanish, we can quickly see that $c_{00}=c_{01}=c_{23}=c_{33}=0$. Plugging these in, we get $$ (c_{02}p +c_{11})s^8t^2 + (c_{02}q+c_{12}p)s^7t^3 + (c_{02}+c_{12}q+c_{22}p^2)s^6t^4 + (c_{03}+c_{12}+c_{22}2pq)s^5t^5 + (c_{13}+c_{22}q^2+c_{22}p)s^4t^6 + (c_{22}2q+c_{23}p)s^3t^7 + c_{22}s^2t^8. $$ So $c_{11}=-c_{02}p$ and $c_{22}=0$, giving $$ (c_{02}q+c_{12}p)s^7t^3 + (c_{02}+c_{12}q)s^6t^4 + (c_{03}+c_{12})s^5t^5 + c_{13}s^4t^6 + c_{23}ps^3t^7,$$ which leads to $c_{13}=c_{23}=0$ and $c_{03}=-c_{12}$. With these determinations, we're at $$ (c_{02}q+c_{12}p)s^7t^3 + (c_{02}+c_{12}q)s^6t^4.$$ If $q^2-p\neq 0$, then $c_{02}=c_{12}=0$, which forces all of our coefficients to be zero.

Checking we get a closed immersion:

It's relatively quick to see that our curve is covered by $D(x_0)$ and $D(x_3)$, and the coordinate algebra of our curve intersected with $D(x_0)$ is $k[\frac{t}{s}]$, so we have a closed immersion there. The challenge is going to be testing whether we have a closed immersion on $D(x_3)$ - that is, expressing $x=\frac{s}{t}$ as a polynomial in $x^5$, $x^4$, and $px^3+qx^2+x$. You can do this by hand, and for some choices of $p,q$ it's not that bad, but a quick proof is to note that you can express any $x^n$ for $n\geq 12$ as a product of $x^4$ and $x^5$, while you can always "fix" $x^n$ at the cost of some terms of higher exponent by subtracting off $(x+qx^2+px^3)^n$. So we really do have a closed immersion here.

  • $\begingroup$ Thanks for the help, and for the spoiler tags! This clears this process up a lot. $\endgroup$
    – Daniel
    Commented Mar 9, 2023 at 14:41
  • 1
    $\begingroup$ It is much easier to take a smooth cubic surface, identify it with the blowup of a plane in 6 points, and consider the strict transform of a smooth conic passing exactly through one of the blown points. This will be a smooth rational quintic curve, and it is evident it is not contained in a quadric. $\endgroup$
    – Sasha
    Commented Mar 9, 2023 at 14:57
  • 1
    $\begingroup$ @Sasha This is an interesting idea, and indeed sounds less arithmetic-heavy. Unfortunately, I have not read about surfaces yet so I haven't seen the results which you're invoking. I also don't quite see why this strict transform would not be contained in a quadric. $\endgroup$
    – Daniel
    Commented Mar 9, 2023 at 15:20
  • 1
    $\begingroup$ @Sasha that's definitely quicker, but these tools haven't been introduced yet at the point in the book where this exercise is mentioned. My goal here with this answer was to try and show a way that one could solve the exercise using the technology available and a little experimentation. I'm sure if you expanded on your comment in an answer it would be well-received - I've learned a lot from you here and on MO over the years. $\endgroup$
    – KReiser
    Commented Mar 9, 2023 at 15:48
  • 1
    $\begingroup$ @KReiser: OK, I expanded my comment, as you suggested. $\endgroup$
    – Sasha
    Commented Mar 9, 2023 at 17:28

Let $p_1,\dots,p_6 \in \mathbb{P}^2$ be points in general position, and let $$ X = \mathrm{Bl}_{p_1,\dots,p_6}(\mathbb{P}^2) $$ be their blowup. Then the linear system $$ H = 3\ell - e_1 - \dots - e_6 $$ (where $\ell$ is the line class and $e_i$ are the exceptional divisors of the blowup) embeds $X$ into $\mathbb{P}^3$ as a smooth cubic surface. Let $$ C \subset X $$ be the strict transform of a smooth conic on $\mathbb{P}^2$ passing through $p_1$, but not through any of other $p_i$, so that $C$ belongs to the linear system $2\ell - e_1$. Then $C$ is a smooth rational curve, and since $$ H \cdot C = 3 \cdot 2 - 1 = 5, $$ its image in $\mathbb{P}^3$ has degree 5. So, we only need to check that $C$ does not lie on a quadric.

Indeed, if $C$ lies on a quadric $Q \subset \mathbb{P}^3$ then $Q \cap X$ is a curve of degree $2 \cdot 3 = 6$, and since it contains the quintic curve $C$, we must have $$ Q \cap X = C \cup L, $$ where $L$ is a line. Since $Q$ is linearly equivalent to $2H$, we obtain a linear equivalence on $X$: $$ 6\ell - 2e_1 - \dots - 2e_6 = 2\ell - e_1 + L, $$ which implies that $$ L = 4\ell - e_1 - 2e_2 - \dots -2e_6. $$ But it is well known that any line on $X$ belongs to one of the 27 classes $$ e_i,\qquad \ell - e_i - e_j,\qquad 2\ell + e_i - e_1 - \dots - e_6, $$ and since the class of $L$ is distinct from either of these, this is a contradiction.


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