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.

## 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

Recall that in this situation we have

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

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

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

The ring map $k \to k[x, y]/(x^ d + y^ d + 1)$ is smooth by Algebra, Lemma 10.136.15 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

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 empty^{1}. 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).

**Proof.**
This is a special case of Divisors, Lemma 31.31.3 (see also Varieties, Lemma 33.33.4). 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

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

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

is an isomorphism and next that the boundary map

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 10.136). 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

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 29.54). 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. $\square$

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