53.9 Plane curves

Let $k$ be a field. A plane curve will be a curve $X$ which is isomorphic to a closed subscheme of $\mathbf{P}^2_ k$. Often the embedding $X \to \mathbf{P}^2_ k$ will be considered given. By Divisors, Example 31.31.2 a curve is determined by the corresponding homogeneous ideal

$I(X) = \mathop{\mathrm{Ker}}\left( k[T_0, T_2, T_2] \longrightarrow \bigoplus \Gamma (X, \mathcal{O}_ X(n)) \right)$

Recall that in this situation we have

$X = \text{Proj}(k[T_0, T_2, T_2]/I)$

as closed subschemes of $\mathbf{P}^2_ k$. For more general information on these constructions we refer the reader to Divisors, Example 31.31.2 and the references therein. It turns out that $I(X) = (F)$ for some homogeneous polynomial $F \in k[T_0, T_1, T_2]$, see Lemma 53.9.1. Since $X$ is irreducible, it follows that $F$ is irreducible, see Lemma 53.9.2. Moreover, looking at the short exact sequence

$0 \to \mathcal{O}_{\mathbf{P}^2_ k}(-d) \xrightarrow {F} \mathcal{O}_{\mathbf{P}^2_ k} \to \mathcal{O}_ X \to 0$

where $d = \deg (F)$ we find that $H^0(X, \mathcal{O}_ X) = k$ and that $X$ has genus $(d - 1)(d - 2)/2$, see proof of Lemma 53.9.3.

To find smooth plane curves it is easiest to write explicit equations. Let $p$ denote the characteristic of $k$. If $p$ does not divide $d$, then we can take

$F = T_0^ d + T_1^ d + T_2^ d$

The corresponding curve $X = V_+(F)$ is called the Fermat curve of degree $d$. It is smooth because on each standard affine piece $D_+(T_ i)$ we obtain a curve isomorphic to the affine curve

$\mathop{\mathrm{Spec}}(k[x, y]/(x^ d + y^ d + 1))$

The ring map $k \to k[x, y]/(x^ d + y^ d + 1)$ is smooth by Algebra, Lemma 10.137.16 as $d x^{d - 1}$ and $d y^{d - 1}$ generate the unit ideal in $k[x, y]/(x^ d + y^ d + 1)$. If $p | d$ but $p \not= 3$ then you can use the equation

$F = T_0^{d - 1}T_1 + T_1^{d - 1}T_2 + T_2^{d - 1}T_0$

Namely, on the affine pieces you get $x + x^{d - 1}y + y^{d - 1}$ with derivatives $1 - x^{d - 2}y$ and $x^{d - 1} - y^{d - 2}$ whose common zero set (of all three) is empty1. We leave it to the reader to make examples in characteristic $3$.

More generally for any field $k$ and any $n$ and $d$ there exists a smooth hypersurface of degree $d$ in $\mathbf{P}^ n_ k$, see for example [Poonen].

Of course, in this way we only find smooth curves whose genus is a triangular number. To get smooth curves of an arbitrary genus one can look for smooth curves lying on $\mathbf{P}^1 \times \mathbf{P}^1$ (insert future reference here).

Lemma 53.9.1. Let $Z \subset \mathbf{P}^2_ k$ be a closed subscheme which is equidimensional of dimension $1$ and has no embedded points (equivalently $Z$ is Cohen-Macaulay). Then the ideal $I(Z) \subset k[T_0, T_1, T_2]$ corresponding to $Z$ is principal.

Proof. This is a special case of Divisors, Lemma 31.31.3 (see also Varieties, Lemma 33.34.4). The parenthetical statement follows from the fact that a $1$ dimensional Noetherian scheme is Cohen-Macaulay if and only if it has no embedded points, see Divisors, Lemma 31.4.4. $\square$

Lemma 53.9.2. Let $Z \subset \mathbf{P}^2_ k$ be as in Lemma 53.9.1 and let $I(Z) = (F)$ for some $F \in k[T_0, T_1, T_2]$. Then $Z$ is a curve if and only if $F$ is irreducible.

Proof. If $F$ is reducible, say $F = F' F''$ then let $Z'$ be the closed subscheme of $\mathbf{P}^2_ k$ defined by $F'$. It is clear that $Z' \subset Z$ and that $Z' \not= Z$. Since $Z'$ has dimension $1$ as well, we conclude that either $Z$ is not reduced, or that $Z$ is not irreducible. Conversely, write $Z = \sum a_ i D_ i$ where $D_ i$ are the irreducible components of $Z$, see Divisors, Lemmas 31.15.8 and 31.15.9. Let $F_ i \in k[T_0, T_1, T_2]$ be the homogeneous polynomial generating the ideal of $D_ i$. Then it is clear that $F$ and $\prod F_ i^{a_ i}$ cut out the same closed subscheme of $\mathbf{P}^2_ k$. Hence $F = \lambda \prod F_ i^{a_ i}$ for some $\lambda \in k^*$ because both generate the ideal of $Z$. Thus we see that if $F$ is irreducible, then $Z$ is a prime divisor, i.e., a curve. $\square$

Lemma 53.9.3. Let $Z \subset \mathbf{P}^2_ k$ be as in Lemma 53.9.1 and let $I(Z) = (F)$ for some $F \in k[T_0, T_1, T_2]$. Then $H^0(Z, \mathcal{O}_ Z) = k$ and the genus of $Z$ is $(d - 1)(d - 2)/2$ where $d = \deg (F)$.

Proof. Let $S = k[T_0, T_1, T_2]$. There is an exact sequence of graded modules

$0 \to S(-d) \xrightarrow {F} S \to S/(F) \to 0$

Denote $i : Z \to \mathbf{P}^2_ k$ the given closed immersion. Applying the exact functor $\widetilde{\ }$ (Constructions, Lemma 27.8.4) we obtain

$0 \to \mathcal{O}_{\mathbf{P}^2_ k}(-d) \to \mathcal{O}_{\mathbf{P}^2_ k} \to i_*\mathcal{O}_ Z \to 0$

because $F$ generates the ideal of $Z$. Note that the cohomology groups of $\mathcal{O}_{\mathbf{P}^2_ k}(-d)$ and $\mathcal{O}_{\mathbf{P}^2_ k}$ are given in Cohomology of Schemes, Lemma 30.8.1. On the other hand, we have $H^ q(Z, \mathcal{O}_ Z) = H^ q(\mathbf{P}^2_ k, i_*\mathcal{O}_ Z)$ by Cohomology of Schemes, Lemma 30.2.4. Applying the long exact cohomology sequence we first obtain that

$k = H^0(\mathbf{P}^2_ k, \mathcal{O}_{\mathbf{P}^2_ k}) \longrightarrow H^0(Z, \mathcal{O}_ Z)$

is an isomorphism and next that the boundary map

$H^1(Z, \mathcal{O}_ Z) \longrightarrow H^2(\mathbf{P}^2_ k, \mathcal{O}_{\mathbf{P}^2_ k}(-d)) \cong k[T_0, T_1, T_2]_{d - 3}$

is an isomorphism. Since it is easy to see that the dimension of this is $(d - 1)(d - 2)/2$ the proof is finished. $\square$

Lemma 53.9.4. Let $Z \subset \mathbf{P}^2_ k$ be as in Lemma 53.9.1 and let $I(Z) = (F)$ for some $F \in k[T_0, T_1, T_2]$. If $Z \to \mathop{\mathrm{Spec}}(k)$ is smooth in at least one point and $k$ is infinite, then there exists a closed point $z \in Z$ contained in the smooth locus such that $\kappa (z)/k$ is finite separable of degree at most $d$.

Proof. Suppose that $z' \in Z$ is a point where $Z \to \mathop{\mathrm{Spec}}(k)$ is smooth. After renumbering the coordinates if necessary we may assume $z'$ is contained in $D_+(T_0)$. Set $f = F(1, x, y) \in k[x, y]$. Then $Z \cap D_+(X_0)$ is isomorphic to the spectrum of $k[x, y]/(f)$. Let $f_ x, f_ y$ be the partial derivatives of $f$ with respect to $x, y$. Since $z'$ is a smooth point of $Z/k$ we see that either $f_ x$ or $f_ y$ is nonzero in $z'$ (see discussion in Algebra, Section 10.137). After renumbering the coordinates we may assume $f_ y$ is not zero at $z'$. Hence there is a nonempty open subscheme $V \subset Z \cap D_{+}(X_0)$ such that the projection

$p : V \longrightarrow \mathop{\mathrm{Spec}}(k[x])$

is étale. Because the degree of $f$ as a polynomial in $y$ is at most $d$, we see that the degrees of the fibres of the projection $p$ are at most $d$ (see discussion in Morphisms, Section 29.57). Moreover, as $p$ is étale the image of $p$ is an open $U \subset \mathop{\mathrm{Spec}}(k[x])$. Finally, since $k$ is infinite, the set of $k$-rational points $U(k)$ of $U$ is infinite, in particular not empty. Pick any $t \in U(k)$ and let $z \in V$ be a point mapping to $t$. Then $z$ works. $\square$

[1] Namely, as $x^{d - 1} = y^{d - 2}$, then $0 = x + x^{d - 1}y + y^{d - 1} = x + 2 x^{d - 1} y$. Since $x \not= 0$ because $1 = x^{d - 2}y$ we get $0 = 1 + 2x^{d - 2}y = 3$ which is absurd unless $3 = 0$.

Comment #1847 by Keenan KIdwell on

In the second block of text following the third displayed equation of OBYA, the word "curve" should be plural.

Comment #4886 by QGravity on

Does the genus of the curve in $\mathbb{P}^1\times \mathbb{P}^1$ fix the equation of the curve uniquely? Could you please add references for the curves in $\mathbb{P}^1\times \mathbb{P}^1$?

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