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Exercise 1.1.(c) in Hartshorne's Deformation Theory:

Over an algebraically closed field $k$, we define a curve in $\mathbb P^2_k$ to be the closed subscheme, defined by a homogeneous polynomial $f(x,y,z)$ of degree $d$ in the coordinate ring $S=k[x,y,z]$.

(c) For any finitely generated $k$-algebra $A$, we define a family of curves of degree $d$ in $\mathbb P^2$ over $A$ to be a closed subscheme $X\subseteq\mathbb P^2_A$, flat over $A$, whose fibers above closed points of $\mathrm{Spec}\,A$ are curves in $\mathbb P^2$. Show that the ideal $I_X\subseteq A[x,y,z]$ is generated by a single homogeneous polynomial $f$ of degree $d$ in $A[x,y,z]$.

My attempts: the condition on fibers above closed points is equivalent to the condition that for any $\mathfrak m\in\mathrm{Specm}\,A$ the ideal $(I_X,\mathfrak m)/\mathfrak m$ is principal. Equivalently, for any $\mathfrak m$ there is $f_{\mathfrak m}\in I_X$ such that $I_X\subset (f_{\mathfrak m},\mathfrak m)$. Also, it is clearly sufficient to prove that $I_X$ contains a homogeneous element of degree $d$.

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    $\begingroup$ Here is MSE copy: Exercise 1.1.(c) in Hartshorne's Deformation Theory $\endgroup$
    – Martin Sleziak
    Commented Feb 9, 2018 at 9:08
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    $\begingroup$ I do not have the book in front of me, but it sounds to me like the formulation above is false. Certainly $I_X$ is generated as an $A[x,y,z]$-module by the degree-$d$ graded component $I_{X,d} := I_X\cap A[x,y,z]_d$. Also the $A$-module $I_{X,d}$ is locally free of rank $1$. However, if the class group of $A$ is nontrivial, I believe that $I_{X,d}$ can be non-free. Thus, although $I_X$ is generated by a single homogeneous polynomial locally on $\text{Spec}\ A$, this need not be true globally. $\endgroup$
    – Jason Starr
    Commented Feb 9, 2018 at 10:05
  • $\begingroup$ @Jason Starr: now I quote the book verbatim. $\endgroup$
    – danneks
    Commented Feb 9, 2018 at 15:25
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    $\begingroup$ That formulation is false. $\endgroup$
    – Jason Starr
    Commented Feb 9, 2018 at 16:54
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    $\begingroup$ Indeed, it is. And the local version is sufficient for the rest of the exercise (which, by the way, aims to show that the complete linear system of $\mathcal{O}_{\mathbb{P}^2}(d)$ is the Hilbert scheme of curves of degree $d$ in $\mathbb{P}^2$ where the universal family is the tautological one). $\endgroup$
    – Ben
    Commented Feb 9, 2018 at 17:14

2 Answers 2

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I agree. Thanks for discovering the error. And by the way there is another error on the same page, line -1, there is a -2 that should be a -4. Robin Hartshorne

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I am just writing my comment as an answer. The stated exercise is true locally on $\text{Spec}\ A$, however it is not true globally. For instance, let $A$ be $k[s,t,u,v]/\langle st-uv,u+v-1\rangle$. Let $J$ denote the ideal in $A$ generated by $s$ and $u$. This has a presentation, $$A^{\oplus 2}\xrightarrow{M} A^{\oplus 2} \xrightarrow{q} J \to 0,$$ where $q(f,g)$ equals $fs-gu$ and where $M$ is the following $2\times 2$ matrix, $$M = \left[\begin{array}{cc}u & t \\ s & v \end{array} \right].$$ Denote by $I_{X,1}$ the image of the $A$-module homomorphism, $$\phi:A^{\oplus 2} \to A[x,y,z]_1,\ \ (f,g) \mapsto (fs-gu)x + (fv-gt)y.$$ The ideal $I_X\subset A[x,y,z]$ generated by $I_{X,1}$ is a radical homogeneous ideal whose corresponding zero scheme, $X\subset \mathbb{P}^2_A$, is flat over $\text{Spec}\ A$ of relative degree $1$. The $A$-module $I_{X,1}$ is locally free of rank $1$, isomorphic to $J$. Yet the ideal $I_X$ is not principal since the $A$-module $I_{X,1}\cong J$ is not free.

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