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. $\square$

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