Proof. The fact that $0\leqslant t_{Z}$ follows from the definition. From (2.3) we have

$$\begin{eqnarray}\dim [I_{Z}]_{j+1}-\binom{j+1}{2}=2j+3-h_{Z}(j+1)\geqslant 2j+3-|Z|,\end{eqnarray}$$

with equality if and only if $h_{Z}(j+1)=|Z|$ . Thus

$$\begin{eqnarray}\displaystyle t_{Z} & = & \displaystyle \min \{j\mid 2j+3-h_{Z}(j+1)>0\}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \min \{j\mid 2j+3-|Z|>0\}\nonumber\\ \displaystyle & = & \displaystyle \min \biggl\{j\mid j\geqslant \frac{|Z|-2}{2}\biggr\}\nonumber\\ \displaystyle & = & \displaystyle \biggl\lfloor\frac{|Z|-1}{2}\biggr\rfloor.\nonumber\end{eqnarray}$$

For (b), suppose first that $t_{Z}=\lfloor (|Z|-1)/2\rfloor$ . Since $2(t_{Z}-1)+3-h_{Z}(t_{Z})\leqslant 0$ , we have

$$\begin{eqnarray}2\biggl\lfloor\frac{|Z|-1}{2}\biggr\rfloor+1-h_{Z}(t_{Z})\leqslant 0.\end{eqnarray}$$

Recalling that $h_{Z}(j)\leqslant |Z|$ for all $j$ , this gives:

1. – if $|Z|$ is odd and $t_{Z}=\lfloor (|Z|-1)/2\rfloor thenh_{Z}(t_{Z})=|Z|$ ;

2. – if $|Z|$ is even and $t_{Z}=\lfloor (|Z|-1)/2\rfloor thenh_{Z}(t_{Z})\geqslant |Z|-1$ .

For the converse, assume that the parity condition holds. In both cases, $h_{Z}(t_{Z}+1)=|Z|$ since $h_{Z}$ is strictly increasing until it reaches the value $|Z|$ . So the only inequality in the calculation in (a) is an equality, and we are done.◻

Proof. To simplify notation, put $t=t_{Z}$ . We use $\unicode[STIX]{x1D6E5}h_{Z}$ to denote the first difference of the Hilbert function of $Z$ ; that is, $\unicode[STIX]{x1D6E5}h_{Z}(j)=h_{Z}(j)-h_{Z}(j-1)$ .

First assume $t=0$ . By Example 2.9, the points of $Z$ are collinear, so (a) holds if and only if $|Z|>1$ , and (b) holds if and only if $|Z|\geqslant 2$ , so (a) and (b) are equivalent, and clearly $0=t\leqslant (|Z|-2)/2$ for $|Z|>1$ . Thus it is now enough to consider the case where $t\geqslant 1$ , that is, where $Z$ is not collinear.

Assume (a) holds. By the definition of $t_{Z}$ , Equation (2.3) and the fact that $h_{Z}$ is strictly increasing until it stabilizes at the value $|Z|$ , this forces

$$\begin{eqnarray}2t+1\leqslant h_{Z}(t)<h_{Z}(t+1)\leqslant 2t+2,\end{eqnarray}$$

and thus $h_{Z}(t+1)=2t+2=1+h_{Z}(t)$ . In particular, $\unicode[STIX]{x1D6E5}h_{Z}(t+1)=1$ . By standard results (see, for example, [Reference Davis, Geramita and MarosciaDGM84, Proposition 3.9]), this implies that the values of $\unicode[STIX]{x1D6E5}h_{Z}$ are as follows (where $s$ is the regularity of $I_{Z}$ ):

(2.4) $$\begin{eqnarray}\begin{array}{@{}ccccccccc@{}}j & : & 0 & 1 & \ldots & t+1 & \ldots & s-1 & s\\ \unicode[STIX]{x1D6E5}h_{Z}(j) & : & 1 & 2 & \ldots & 1 & \ldots & 1 & \,0.\end{array}\end{eqnarray}$$

Thus, $h_{Z}(t+1)=2t+2$ implies

$$\begin{eqnarray}|Z|=2t+2+(s-t-2)=s+t.\end{eqnarray}$$

Using $h_{Z}(t+1)=2t+2\leqslant |Z|$ , we conclude that $s=|Z|-t\geqslant t+2$ .

Now we consider two cases.

Case 1. Assume $Z$ does not lie on a conic, that is, $\unicode[STIX]{x1D6E5}h_{Z}(2)=3$ . Hence

$$\begin{eqnarray}2t+2=h_{Z}(t+1)=\mathop{\sum }_{j=0}^{t+1}\unicode[STIX]{x1D6E5}h_{Z}(j)\end{eqnarray}$$

forces $\unicode[STIX]{x1D6E5}h_{Z}(t)=1$ , and thus $\unicode[STIX]{x1D6E5}h_{Z}(s-2)=\unicode[STIX]{x1D6E5}h_{Z}(s-1)=1>\unicode[STIX]{x1D6E5}h_{Z}(s)$ . By [Reference DavisDav86, (2.3)] (or by applying results of [Reference Bigatti, Geramita and MiglioreBGM94]), it follows that $[I_{Z}]_{s-1}$ has a linear form $\ell$ as a common divisor. Since $I_{Z}$ has a minimal set of homogeneous generators all of whose degrees are at most $s$ , there must be a generator $f$ of degree $s$ and by [Reference CampanellaCam86, Theorem 2.1] there is only one generator of degree $s$ in a minimal set of homogeneous generators. Moreover, since $Z$ is reduced, the curves defined by $f$ and $\ell$ must intersect transversely. Thus the ideal $(\ell ,f)$ defines a subset of $s$ collinear points of $Z$ and clearly $\ell$ vanishes at no point of $Z$ other than these $s$ . Therefore, condition (ii) is satisfied.

Case 2. Assume $Z$ is contained in a conic, defined, say, by a homogeneous form $q$ . Again taking into account $h_{Z}(t+1)=2t+2$ , we get

(2.5) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}h_{Z}(j)=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if }j=0\text{ or }t+1\leqslant j<s,\\ 2\quad & \text{if }1\leqslant j\leqslant t,\\ 0\quad & \text{otherwise.}\end{array}\right.\end{eqnarray}$$

If $t=1$ , then $\unicode[STIX]{x1D6E5}h_{Z}=(1,2,1,\ldots ,1)$ . Thus $Z$ is either four general points (i.e. a complete intersection) or $Z$ consists of three or more collinear points and one point off the line; in both cases it is easy to check that the assertions hold. Now assume that $t>1$ . It follows that $q$ is a common factor for $[I_{Z}]_{j}$ for $j\leqslant t$ , but not for $j=t+1$ , so any minimal set of homogeneous generators for $I_{Z}$ must contain $q$ and a generator $g$ of degree $t+1$ . If $q$ and $g$ are coprime, then $|Z|\leqslant \deg (q)\deg (g)=2t+2$ . Since $|Z|=t+s$ and $s\geqslant t+2$ , this means $s=t+2$ and $|Z|=2t+2$ , so $Z$ is a complete intersection as claimed in (i). Otherwise, $q$ and $g$ have a linear common factor $\ell$ and $I_{Z}$ has another minimal generator $f$ of degree $s$ . As in Case 1 we conclude that the line defined by $\ell$ contains precisely $s$ points of $Z$ , and so condition (ii) is met.

Conversely, assume one of the conditions in (b) is true. Thus, $|Z|-t\geqslant t+2$ (and hence $t\leqslant (|Z|-2)/2$ ), by hypothesis for part (ii) and using the fact that $Z$ is a transverse complete intersection of a conic with a curve of degree $t+1$ for part (i). Again, we consider two cases.

If (i) is true, then $h_{Z}(t)=2t+1<2t+2=|Z|$ , as desired. Moreover, here we have $t=(|Z|-2)/2$ .

Finally, assume (ii) is true, let $Y\subset Z$ be a subset of $|Z|-t$ collinear points and let $U$ be the complement of $Y$ in $Z$ . Then $t=|U|$ and $U$ is reduced, so $U$ imposes independent conditions on forms of degree $t-1$ ; that is, $h_{U}(t-1)=t$ and thus $\dim [I_{U}]_{t-1}=\binom{t+1}{2}-t$ . But the linear form $\ell$ vanishing on $Y$ is, by Bezout’s theorem, a common divisor of $[I_{Z}]_{t}$ , so $\dim [I_{Z}]_{t}=\dim [I_{U}]_{t-1}$ , and we have $h_{Z}(t)=2t+1<2t+2\leqslant |Z|$ .◻

Proof. By definition, we clearly have $t_{Z}\leqslant t_{Z+Q}$ . It remains to show the second inequality. Suppose $t_{Z+Q}\geqslant t_{Z}+2$ . Then the definition gives $h_{Z}(t_{Z}+1)\leqslant 2t_{Z}+2$ and $h_{Z+Q}(t_{Z}+1)\geqslant 2t_{Z}+3$ . Since the Hilbert functions of $Z$ and $Z+Q$ differ at most by $1$ in each degree, we conclude that

$$\begin{eqnarray}h_{Z+Q}(j)=h_{Z}(j)+1\quad \text{whenever }j>t_{Z},\end{eqnarray}$$

and, in particular, $h_{Z}(t_{Z}+1)=2t_{Z}+2$ . Considering degree $t_{Z}+2\leqslant t_{Z+Q}$ , we get $h_{Z+Q}(t_{Z}+2)\geqslant 2t_{Z}+5$ , which implies

$$\begin{eqnarray}|Z|\geqslant h_{Z}(t_{Z}+2)\geqslant 2t_{Z}+4=h_{Z}(t_{Z}+1)+2.\end{eqnarray}$$

It follows that $h_{Z}(t_{Z})<|Z|$ and $t_{Z}\leqslant (|Z|-4)/2$ . Hence, Proposition 2.10 shows that $|Z|-t_{Z}$ of the points in $Z$ are collinear. Denote by $Y$ this subset of $Z$ , and so $|Y|=|Z|-t_{Z}\geqslant t_{Z}+4$ . Now, using that the points in $Y$ are collinear, we obtain

$$\begin{eqnarray}h_{Z+Q}(t_{Z}+2)\leqslant h_{Y}(t_{Z}+2)+|Z+Q-Y|=t_{Z}+3+t_{Z}+1=2t_{Z}+4,\end{eqnarray}$$

contradicting our estimate above that $h_{Z+Q}(t_{Z}+2)\geqslant 2t_{Z}+5$ .◻

Proof. Let $q:Y\rightarrow \mathbb{P}^{2}$ be the blow up of $\mathbb{P}^{2}$ at the point $P$ . Let $H$ be the pullback of a line and $E=q^{-1}(P)$ the exceptional curve coming from $P$ . The proper transforms of the lines through $P$ give the linear system $|H-E|$ , which gives a morphism $p:Y\rightarrow L=\mathbb{P}^{1}$ with fibers the elements of $|H-E|$ , making $Y$ a $\mathbb{P}^{1}$ -bundle over $L$ . (In fact, $Y$ is just the Hirzebruch surface $H_{1}$ .)

Let ${\mathcal{I}}_{Z}$ be the sheaf of ideals of $Z$ on $\mathbb{P}^{2}$ . Since $P$ is general (and hence the $t_{Z,y}$ appearing in [Reference Faenzi and VallèsFV14, Theorem 4.3] is 0), we have the isomorphism $p_{\ast }(q^{\ast }({\mathcal{I}}_{Z}(1)))\cong {\mathcal{O}}_{L}(-a_{Z})\oplus {\mathcal{O}}_{L}(-b_{Z})$ from [Reference Faenzi and VallèsFV14, Theorem 4.3]. Since $P\not \in Z$ , $q$ is an isomorphism on an open set containing $Z$ , so we can regard $Z$ as being on $\mathbb{P}^{2}$ or on $Y$ , hence there is a natural identification of ${\mathcal{I}}_{Z}$ with $q^{\ast }({\mathcal{I}}_{Z})$ . Under this identification we can regard $q^{\ast }({\mathcal{I}}_{Z}(1))$ as being the sheaf ${\mathcal{I}}_{Z}\otimes {\mathcal{O}}_{Y}(H)$ . Thus we have $p_{\ast }({\mathcal{I}}_{Z}\otimes {\mathcal{O}}_{Y}(H))\cong {\mathcal{O}}_{L}(-a_{Z})\oplus {\mathcal{O}}_{L}(-b_{Z})$ . Now tensor through by ${\mathcal{O}}_{L}(j)$ to get

$$\begin{eqnarray}\displaystyle p_{\ast }({\mathcal{I}}_{Z}\otimes {\mathcal{O}}_{Y}((j+1)H-jE)) & \cong & \displaystyle p_{\ast }({\mathcal{I}}_{Z}\otimes {\mathcal{O}}_{Y}(H)\otimes p^{\ast }({\mathcal{O}}_{L}(j)))\cong p_{\ast }({\mathcal{I}}_{Z}\otimes {\mathcal{O}}_{Y}(H))\otimes {\mathcal{O}}_{L}(j)\nonumber\\ \displaystyle & \cong & \displaystyle {\mathcal{O}}_{L}(j-a_{Z})\oplus {\mathcal{O}}_{L}(j-b_{Z}).\nonumber\end{eqnarray}$$

Since $p_{\ast }$ preserves global sections, taking global sections gives

$$\begin{eqnarray}\displaystyle [I_{Z+jP}]_{j+1} & \cong & \displaystyle \unicode[STIX]{x1D6E4}(\mathbb{P}^{2},{\mathcal{I}}_{Z+jP}\otimes {\mathcal{O}}_{\mathbb{P}^{2}}((j+1)H))\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D6E4}(\mathbb{P}^{2},{\mathcal{I}}_{Z}\otimes {\mathcal{I}}_{jP}\otimes {\mathcal{O}}_{\mathbb{P}^{2}}((j+1)H))\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D6E4}(Y,{\mathcal{I}}_{Z}\otimes {\mathcal{O}}_{Y}((j+1)H-jE))\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D6E4}(L,p_{\ast }({\mathcal{I}}_{Z}\otimes {\mathcal{O}}_{Y}((j+1)H-jE)))\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D6E4}(L,{\mathcal{O}}_{L}(j-a_{Z})\oplus {\mathcal{O}}_{L}(j-b_{Z}))\nonumber\\ \displaystyle & \cong & \displaystyle \unicode[STIX]{x1D6E4}(L,{\mathcal{O}}_{L}(j-a_{Z}))\oplus \unicode[STIX]{x1D6E4}(L,{\mathcal{O}}_{L}(j-b_{Z})).\nonumber\end{eqnarray}$$

The result now follows by taking dimensions. ◻

Proof. Part (a) follows immediately from Lemma 3.3, while (b) follows from the definition of $m_{Z}$ . For (c), note that for any $j$ ,

(3.1) $$\begin{eqnarray}\binom{j+3}{2}-\binom{j+1}{2}-|Z|=2j+3-|Z|=(j-a_{Z}+1)+(j-b_{Z}+1),\end{eqnarray}$$

the latter since $a_{Z}+b_{Z}=|Z|-1$ . Because $a_{Z}\leqslant b_{Z}$ , the result follows from the definition of $u_{Z}$ , Lemma 3.3 and $a_{Z}+b_{Z}=|Z|-1$ . Part (d) comes from Lemma 2.7 and the definitions.

For (e), assume that $m_{Z}=a_{Z}<t_{Z}$ . It is enough to prove that $a_{Z}\leqslant j<t_{Z}$ implies $j<u_{Z}$ . But $a_{Z}\leqslant j<t_{Z}$ implies that $Z+jP$ does not impose independent conditions on forms of degree $j+1$ , so $j<u_{Z}$ and we are done.◻

Proof. From Lemma 3.5 we have $a_{Z}\leqslant t_{Z}$ and $u_{Z}=b_{Z}-1$ . Since $a_{Z}<b_{Z}-1$ , applying Lemma 3.3 and (3.1) with $j=a_{Z}$ , we get

$$\begin{eqnarray}1=\dim [I_{Z+a_{Z}P}]_{a_{Z}+1}>\binom{a_{Z}+3}{2}-|Z|-\binom{a_{Z}+1}{2}.\end{eqnarray}$$

Now suppose that $a_{Z}=t_{Z}$ . Since $h_{Z}(t_{Z})=|Z|$ , the points of $Z$ impose independent conditions on curves of degree $t_{Z}=a_{Z}$ and hence also on curves of degree $a_{Z}+1$ , $\binom{a_{Z}+3}{2}-|Z|-\binom{a_{Z}+1}{2}=\dim [I_{Z}]_{a_{Z}+1}-\binom{a_{Z}+1}{2}>0$ (by definition of $t_{Z}$ ). Combining with the previous inequality, we obtain

$$\begin{eqnarray}1>\binom{a_{Z}+3}{2}-|Z|-\binom{a_{Z}+1}{2}>0,\end{eqnarray}$$

which is impossible since the middle expression is an integer. Thus $m_{Z}=a_{Z}<t_{Z}$ .◻

Proof. See Theorem 4.3. ◻

Proof. Assume $Z$ admits an unexpected curve. Then, for a general point $P$ , there is an $m\geqslant a_{Z}=m_{Z}$ such that

$$\begin{eqnarray}\dim [I_{Z+mP}]_{m+1}>\max \biggl\{0,\dim [I_{Z}]_{m+1}-\binom{m+1}{2}\!\biggr\}.\end{eqnarray}$$

By Theorem 3.7 we have $h_{Z}(t_{Z})=|Z|$ . We now claim that $m_{Z}<u_{Z}$ . Indeed, if it were true that $u_{Z}\leqslant m_{Z}\leqslant m$ then

$$\begin{eqnarray}\begin{array}{@{}rclll@{}}\displaystyle \max \biggl\{0,\dim [I_{Z}]_{m+1}-\binom{m+1}{2}\!\biggr\} & {\geqslant} & \displaystyle \binom{m+3}{2}-\binom{m+1}{2}-|Z| & & (by(2.2))\\ & = & \displaystyle \dim [I_{Z+mP}]_{m+1} & & \displaystyle (sincem\geqslant u_{Z})\\ & {>} & \displaystyle \max \biggl\{0,\dim [I_{Z}]_{m+1}-\binom{m+1}{2}\!\biggr\} & & (bychoiceofm).\end{array}\end{eqnarray}$$

Lemma 3.6 now implies $m_{Z}<t_{Z}$ .

Conversely, if $m_{Z}<t_{Z}$ , then

$$\begin{eqnarray}\dim [I_{Z+m_{Z}P}]_{m_{Z}+1}>0=\max \biggl\{0,\dim [I_{Z}]_{m_{Z}+1}-\binom{m_{Z}+1}{2}\!\biggr\},\end{eqnarray}$$

and so $Z$ admits an unexpected curve of degree $m_{Z}+1$ .

For the rest, assume $m_{Z}<t_{Z}$ and $h_{Z}(t_{Z})=|Z|$ . Then $t_{Z}\leqslant u_{Z}$ by Lemma 3.5(e). If $m_{Z}\leqslant j<t_{Z}$ , we have $\dim [I_{Z+jP}]_{j+1}>0=\max \{0,\dim [I_{Z}]_{j+1}-\binom{j+1}{2}\!\}$ , and so there are unexpected curves for each such $j$ .

Now assume that $t_{Z}\leqslant j<u_{Z}=b_{Z}-1$ (Lemma 3.5(c)). Since $h_{Z}(t_{Z})=|Z|$ , we know that $Z$ imposes independent conditions on curves of degree $j$ . Then using Lemma 3.3, we have

$$\begin{eqnarray}\displaystyle \dim [I_{Z+jP}]_{j+1} & = & \displaystyle j-m_{Z}+1\nonumber\\ \displaystyle & {>} & \displaystyle (j+1-a_{Z})+(j-b_{Z}+1)\nonumber\\ \displaystyle & = & \displaystyle \binom{j+3}{2}-|Z|-\binom{j+1}{2}\nonumber\\ \displaystyle & = & \displaystyle \dim [I_{Z}]_{j+1}-\binom{j+1}{2}\nonumber\\ \displaystyle & = & \displaystyle \max \biggl\{0,\dim [I_{Z}]_{j+1}-\binom{j+1}{2}\!\biggr\},\nonumber\end{eqnarray}$$

hence there are unexpected curves for each such  $j$ .

Finally, if $j\geqslant u_{Z}$ , we have

$$\begin{eqnarray}\dim [I_{Z+jP}]_{j+1}=\binom{j+3}{2}-|Z|-\binom{j+1}{2}=\max \biggl\{0,\dim [I_{Z}]_{j+1}-\binom{j+1}{2}\!\biggr\},\end{eqnarray}$$

so there are no unexpected curves of any such degree $j+1$ .◻

Proof. Assume $X$ fails to impose the expected number of conditions on $V=[I_{Z}]_{m+1}$ ; that is, $Z$ has an unexpected curve of degree $m+1$ . Then $a_{Z}\leqslant m<u_{Z}$ by Theorem 3.9 and $h_{Z}(t_{Z})=|Z|$ by Theorem 3.7.

Conversely, if $a_{Z}<u_{Z}$ and $h_{Z}(t_{Z})=|Z|$ , then $a_{Z}<t_{Z}$ by Lemma 3.6, and hence by Theorem 3.9 there are unexpected curves of degree $m+1$ for each $m$ in the range $a_{Z}\leqslant m<u_{Z}$ .◻

Proof. This follows from the exact sequence

$$\begin{eqnarray}\displaystyle 0 & \rightarrow & \displaystyle H^{0}({\mathcal{I}}_{Z+jP}(j+1))\rightarrow H^{0}({\mathcal{O}}_{\mathbb{P}^{2}}(j+1))\rightarrow H^{0}({\mathcal{O}}_{Z+jP}(j+1))\nonumber\\ \displaystyle & \rightarrow & \displaystyle H^{1}({\mathcal{I}}_{Z+jP}(j+1))\rightarrow 0.\Box \nonumber\end{eqnarray}$$

Proof. If $f\in [I_{Z+jP}]_{j}$ then any line joining $P$ to a point of $Z$ is a component of $f$ , since the restriction of $f$ to a line is either zero or has at most $j$ roots up to multiplicity. If $P$ is general, any such line contains no other points of $Z$ . Hence $\dim [I_{Z+jP}]_{j}=0$ if $j<|Z|$ and $\dim [I_{Z+|Z|\cdot P}]_{|Z|}=1$ . In particular, by adding suitable lines through $P$ we obtain the first assertion.

Now, $1=\dim [I_{Z+|Z|\cdot P}]_{|Z|}\geqslant \dim [I_{Z}]_{|Z|}-\binom{|Z|+1}{2}\geqslant \binom{|Z|+2}{2}-|Z|-\binom{|Z|+1}{2}=1$ , hence $Z+|Z|\cdot P$ (and thus $Z$ ) imposes independent conditions on forms of degree $|Z|$ . This means $h^{1}({\mathcal{I}}_{Z}(j))=0$ for $j=|Z|$ (and hence for $j\geqslant |Z|$ ), and it means $h^{1}({\mathcal{I}}_{Z+jP}(j))=0$ for $j=|Z|$ . Replacing $Z$ by $Z+Q$ for any point $Q\notin Z$ , we now get $h^{1}({\mathcal{I}}_{Z+Q+jP}(j))=0$ for $j=|Z+Q|=|Z|+1$ and hence $Z+Q+jP$ imposes independent conditions on forms of degree $j=|Z|+1$ , and therefore $Z+jP$ also imposes independent conditions on forms of degree $j=|Z|+1$ . Continuing in this way, we see that for any $j\geqslant |Z|$ , $Z+jP$ imposes independent conditions on forms of degree $j$ ; hence for such $j$ we have $h^{1}({\mathcal{I}}_{Z+jP}(j))=0$ . Thus $\dim [I_{Z+jP}]_{j}=\binom{j+2}{2}-|Z|-\binom{j+1}{2}=j+1-|Z|$ as asserted. Since $h^{1}({\mathcal{I}}_{Z}(j))=0$ for $j\geqslant |Z|$ , we also have $\dim [I_{Z}]_{j}=\binom{j+2}{2}-|Z|$ , so $\dim [I_{Z+jP}]_{j}$ can in addition be written as $\dim [I_{Z}]_{j}-\binom{j+1}{2}$ .

Since we have already shown that $h^{1}({\mathcal{I}}_{Z+jP}(j))=0$ for $j\geqslant |Z|$ , it remains only to prove that $h^{1}({\mathcal{I}}_{Z}(j-1))=0$ for $j\geqslant |Z|$ . But this is true for any finite set of points, so we are done.◻

Proof. If $t_{Z}=0$ , then the points of $Z$ are collinear, in which case it is not hard to check that the claims hold. So we may assume $t_{Z}>0$ . By Proposition 2.10 we have to consider two cases.

Case 1. Assume $Z$ is defined by an ideal $I_{Z}=(q,g)$ , where $q,g\in R$ are forms of degree $2$ and $t_{Z}+1$ , respectively. We have (from the proof of Proposition 2.10)

$$\begin{eqnarray}h_{Z}(t_{Z})=2t_{Z}+1,\quad h_{Z}(t_{Z}+1)=2t_{Z}+2=|Z|.\end{eqnarray}$$

Then for $j+1<t_{Z}+1$ , we get

$$\begin{eqnarray}[I_{Z}\cap I_{P}^{j}]_{j+1}=[(q)\cap I_{P}^{j}]_{j+1}=q\cdot [I_{P}^{j}]_{j-1}=0,\end{eqnarray}$$

which implies $m_{Z}=t_{Z}$ by Lemma 3.5(d). Since $a_{Z}+b_{Z}=|Z|-1=2t_{Z}+1$ , we also obtain $b_{Z}=u_{Z}+1=t_{Z}+1$ . Then Lemma 3.3 gives $\dim [I_{Z+m_{Z}P}]_{m_{Z}+1}=1$ as desired. And since $|Z|=2t_{Z}+2$ , we have $t_{Z}<(|Z|-1)/2$ .

To show $\dim [I_{Z+jP}]_{j+1}=\dim [I_{Z}]_{j+1}-\binom{j+1}{2}$ for $j\geqslant m_{Z}$ , first note that we have

$$\begin{eqnarray}\displaystyle 1 & = & \displaystyle \dim [I_{Z+m_{Z}P}]_{m_{Z}+1}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \dim [I_{Z}]_{m_{Z}+1}-\binom{m_{Z}+1}{2}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \binom{t_{Z}+3}{2}-(2t_{Z}+2)-\binom{t_{Z}+1}{2}\nonumber\\ \displaystyle & = & \displaystyle 1,\nonumber\end{eqnarray}$$

hence $Z+m_{Z}P$ imposes independent conditions on forms of degree $m_{Z}+1$ . This also means that the points of $Z$ impose independent conditions on $[I_{P}^{m_{Z}}]_{m_{Z}+1}$ . By adding lines through $P$ , it is then clear that the points of $Z$ also impose independent conditions on $[I_{P}^{m_{Z}+k}]_{m_{Z}+1+k}$ , which implies $\dim [I_{Z+jP}]_{j+1}=\dim [I_{Z}]_{j+1}-\binom{j+1}{2}$ for all $j\geqslant m_{Z}$ as desired.

Case 2. Assume a line defined by a linear form $\ell \in R$ contains precisely $|Z|-t_{Z}\geqslant t_{Z}+2$ points of $Z$ (and hence $t_{Z}<(|Z|-1)/2$ ) and let $Y$ be the set of these points. Let $U\subset Z$ be the subset of the other $t_{Z}$ points. Then $I_{Y}=(\ell ,f)$ for some form $f$ , where $\deg f\geqslant t_{Z}+2$ . Thus, for each integer $j$ and any general point $P\in \mathbb{P}^{2}$ , we get

$$\begin{eqnarray}[I_{Z}\cap I_{P}^{j}]_{j+1}=[(\ell ,f)\cap I_{U}\cap I_{P}^{j}]_{j+1}.\end{eqnarray}$$

Since $\deg f\geqslant t_{Z}+2$ , it follows for $j+1\leqslant t_{Z}+1$ that

$$\begin{eqnarray}[I_{Z+jP}]_{j+1}=[I_{Z}\cap I_{P}^{j}]_{j+1}=[(\ell )\cap I_{U}\cap I_{P}^{j}]_{j+1}=\ell \cdot [I_{U}\cap I_{P}^{j}]_{j}\end{eqnarray}$$

because $\ell$ does not vanish at $P$ or at any of the points in $U$ . Since $|U|=t_{Z}$ , Lemma 4.2 gives $\dim [I_{U+jP}]_{j}\leqslant 1$ for $j\leqslant t_{Z}$ , with equality exactly when $j=t_{Z}$ . Thus $m_{Z}=t_{Z}$ and $\dim [I_{Z+m_{Z}P}]_{m_{Z}+1}=1$ .

Now assume $j\geqslant m_{Z}=t_{Z}$ . Using Equation (2.4), we have

$$\begin{eqnarray}h_{Z}(j+1)=\min \{t_{Z}+j+2,|Z|\}=\min \{m_{Z}+j+2,|Z|\}.\end{eqnarray}$$

Hence

$$\begin{eqnarray}\dim [I_{Z}]_{j+1}=\max \biggl\{\!\binom{j+3}{2}-(m_{Z}+j+2),\binom{j+3}{2}-|Z|\biggr\}.\end{eqnarray}$$

Then

$$\begin{eqnarray}\displaystyle \displaystyle \dim [I_{Z}]_{j+1}-\binom{j+1}{2} & = & \displaystyle \displaystyle \max \{2j+3-(m_{Z}+j+2),2j+3-|Z|\}\nonumber\\ \displaystyle & = & \displaystyle \max \{j+1-m_{Z},(j+1-m_{Z})+(j+1-b_{Z})\}\nonumber\\ \displaystyle & = & \displaystyle \dim [I_{Z+jP}]_{j+1},\nonumber\end{eqnarray}$$

the latter thanks to Lemma 3.3. ◻

Proof. Note that the multiplicity of an irreducible curve at a point is at most the degree of the curve and that the multiplicity is equal to the degree if and only if the curve is a line. Since the degree of $C$ at $P$ is precisely one more than its multiplicity at $P$ , it follows that $C$ has a unique irreducible component $C^{\prime }$ whose multiplicity at $P$ is $-1+\deg C^{\prime }$ and that this component has multiplicity $1$ . Thus, $f=f^{\prime }\cdot \ell _{1}\cdots \ell _{k}$ , where $f$ and $f^{\prime }$ define the curves $C$ and $C^{\prime }$ , respectively, $k\geqslant 0$ , and each $\ell _{i}$ is a linear form in $I_{P}$ , so $\deg (C^{\prime })=\deg (C)-k=m_{Z}+1-k$ . The genus formula implies that $C^{\prime }$ is rational and smooth at all points other than $P$ .

Put $Z^{\prime }=Z\cap C^{\prime }$ . Since $P$ is general, each of the $k$ components of $C$ other than $C^{\prime }$ passes through at most one point of $Z$ . The union of these lines must contain $Z^{\prime \prime }=Z-Z^{\prime }$ , and so $|Z^{\prime \prime }|\leqslant k$ .

We have seen that $f^{\prime }$ is in $[I_{Z^{\prime }+(m_{Z}-k)P}]_{m_{Z}-k+1}$ , which implies $m_{Z^{\prime }}\leqslant m_{Z}-k$ . Hence, the estimate $m_{Z^{\prime }}=m_{Z-Z^{\prime \prime }}\geqslant m_{Z}-|Z^{\prime \prime }|$ gives $k\leqslant |Z^{\prime \prime }|$ . Therefore, we obtain $|Z^{\prime \prime }|=k$ , so $m_{Z^{\prime }}=m_{Z}-k=m_{Z}-|Z^{\prime \prime }|$ and $\deg (C^{\prime })=\deg (C)-k=\deg (C)-|Z^{\prime \prime }|=m_{Z}+1-|Z^{\prime \prime }|=m_{Z^{\prime }}+1$ . It also follows that each of the $k$ lines defined by some $\ell _{i}$ passes through one of the $k$ points of $Z^{\prime \prime }$ , and no two lines pass through the same point of $Z^{\prime \prime }$ . Thus, the curve $C$ is reduced.◻

Proof. The existence of a curve $C$ with the desired properties is guaranteed by the definition of the multiplicity index $m_{Z}$ . Suppose $C$ is defined by a form $f$ . Then $f\cdot [I_{(j-m_{Z})P}]_{j-m_{Z}}\subset [I_{Z+jP}]_{j+1}$ , and, comparing dimensions by using Lemma 3.3, we see that

$$\begin{eqnarray}[I_{Z+jP}]_{j+1}=f\cdot [I_{(j-m_{Z})P}]_{j-m_{Z}}\quad \text{if }m_{Z}\leqslant j\leqslant u_{Z}.\end{eqnarray}$$

Since, by definition, $[I_{Z+jP}]_{j+1}=0$ if $j<m_{Z}$ , this proves our claim if $j\leqslant u_{Z}$ . If $j=u_{Z}+1$ , then Lemma 3.3 gives that there is a form $g$ of degree $u_{Z}+2$ such that

$$\begin{eqnarray}[I_{Z+(u_{Z}+1)P}]_{u_{Z}+2}=\{f\cdot [I_{(u_{Z}+1-m_{Z})P}]_{u_{Z}+1-m_{Z}}\}\oplus \langle g\rangle .\end{eqnarray}$$

We will now show that $f$ and $g$ are relatively prime. Using the notation of Lemma 5.1, write $f=f^{\prime }\cdot \ell _{1}\cdots \ell _{k}$ , where $f^{\prime }$ defines the irreducible curve $C^{\prime }$ and each $\ell _{i}$ defines a line though $P$ and one of the $k$ points of $Z^{\prime \prime }=Z-Z^{\prime }$ . Assume first that $f^{\prime }$ divides $g$ . Then the curve defined by $g/f^{\prime }$ has multiplicity $u_{Z}+1-(m_{Z}-k)=\deg g/f^{\prime }$ . Thus, $g/f^{\prime }$ is a product of linear forms of $I_{P}$ that must vanish on $Z^{\prime \prime }$ . Hence, $\ell _{1}\cdots \ell _{k}$ divides $g/f^{\prime }$ , which implies $g\in f\cdot [I_{(u_{Z}+1-m_{Z})P}]_{u_{Z}+1-m_{Z}}$ , in contradiction to the choice of $g$ .

Assume next that $k\geqslant 1$ and that one of the linear forms $\ell _{i}$ divides $g$ . Let $P_{i}\in Z^{\prime \prime }$ be the point of $Z$ on which $\ell _{i}$ vanishes. By Lemma 5.1, we know that $m_{Z^{\prime }}=m_{Z-Z^{\prime \prime }}=m_{Z}-|Z^{\prime \prime }|$ . This gives $m_{Z-P_{i}}=m_{Z}-1$ , and thus $u_{Z-P_{i}}=u_{Z}$ . It follows that $f/\ell _{i}\in [I_{Z-P_{i}+m_{Z-P_{i}}P}]_{m_{Z-P_{i}}+1}$ and $g/\ell _{i}\in [I_{Z-P_{i}+u_{Z-P_{i}}P}]_{u_{Z-P_{i}}+1}$ . Applying the part of the statement we have already shown to $Z-P_{i}$ , we conclude that $g/\ell _{i}\in f/\ell _{i}\cdot [I_{(u_{Z-P_{i}}-m_{Z-P_{i}})P}]_{u_{Z-P_{i}}-m_{Z-P_{i}}}=f/\ell _{i}\cdot [I_{(u_{Z}-m_{Z}+1)P}]_{u_{Z}-m_{Z}+1}$ , which is again in contradiction to the choice of $g$ .

Thus, we have shown that $f$ and $g$ form a regular sequence. We claim that

$$\begin{eqnarray}\{f\cdot [I_{(j-m_{Z})P}]_{j-m_{Z}}\}\cap \{g\cdot [I_{(j-u_{Z}-1)P}]_{j-u_{Z}-1}\}=0\quad \text{if }j\geqslant u_{Z}+1.\end{eqnarray}$$

This is clear if $j\leqslant |Z|$ because the degrees of the syzygies of the ideal $(f,g)$ are at least $m_{Z}+1+u_{Z}+2=|Z|+1$ . Assume now that the claim is false for some $j\geqslant |Z|+1$ . That is, there are forms $h_{1},h_{2}$ of suitable degrees such that $fh_{1}=gh_{2}$ , where $h_{2}$ is a product of $j-u_{Z}-2\geqslant m_{Z}+1$ linear forms in $I_{P}$ . Since $f$ and $g$ are coprime, it follows that $f$ is a product of $m_{Z}+1$ linear forms in $I_{P}$ . By the generality of $P$ , each of these linear forms vanishes on at most one point of $Z$ . However, by definition $f$ vanishes at each point of $Z$ , which implies $m_{Z}+1\geqslant |Z|=m_{Z}+u_{Z}+2$ . This is a contradiction because $u_{Z}\geqslant 0$ (as $|Z|\geqslant 2$ ). Thus, the above claim is shown. It gives that the sum

$$\begin{eqnarray}\{f\cdot [I_{(j-m_{Z})P}]_{j-m_{Z}}\}+\{g\cdot [I_{(j-u_{Z}-1)P}]_{j-u_{Z}-1}\}\subset [I_{Z+(u_{Z}+1)P}]_{j+1}\end{eqnarray}$$

is a direct sum. Since both sides have the same dimension we get equality, as desired. ◻

Proof. By Proposition 5.2, the vector space $[I_{Z+m_{Z}P}]_{m_{Z}+1}$ contains two polynomials, $F$ and $G$ , that form a regular sequence. By Lemma 5.1, any $K$ -linear combination of $F$ and $G$ which is not irreducible has a linear factor. Suppose there are two such linear combinations, for example $aF+G$ and $bF+G$ for distinct scalars $a$ and $b$ , which have a common linear factor $L$ . We have $aF+G=LH_{1}$ and $bF+G=LH_{2}$ for some forms $H_{1}$ and $H_{2}$ , so $(a-b)F=L(H_{1}-H_{2})$ . Then $L$ is a factor of $F$ , so from the equation $aF+G=LH_{1}$ we also have that $L$ is a factor of $G$ , and hence of every curve in the linear system. This contradicts the fact that $F$ and $G$ are a regular sequence.

By Lemma 5.1 there are only finitely many possible linear factors, each of which corresponds to a point of $Z$ , so we can conclude that at most $|Z|$ curves in the pencil defined by $F$ and $G$ are reducible. Since $K$ is infinite, the general element must be irreducible.◻

Proof. This is an immediate consequence of Proposition 5.2 and the fact for $j\geqslant m_{Z}$ that the non-zero forms in $[I_{(j-m_{Z})P}]_{j-m_{Z}}$ are precisely the products of $j-m_{Z}$ linear forms vanishing at $P$ (the $j$ in Proposition 5.2 corresponds to $t-1$ here).◻

Proof. (a) Since $C^{\prime }$ is a component of $C$ and is defined by an element of $[I_{Z^{\prime }+m_{Z^{\prime }}P}]_{m_{Z^{\prime }}+1}$ , we see $\dim [I_{Z^{\prime }+m_{Z^{\prime }}P}]_{m_{Z^{\prime }}+1}=1$ , so $C^{\prime }=C_{P}(Z^{\prime })$ . Moreover, $m_{Z^{\prime }}+t=m_{Z}$ and $m_{Z}+u_{Z}+2=|Z|=|Z^{\prime }|+t=m_{Z^{\prime }}+u_{Z^{\prime }}+2+t$ , so $u_{Z^{\prime }}=u_{Z}$ . Thus we have $m_{Z^{\prime }}+t=m_{Z}\leqslant u_{Z}=u_{Z^{\prime }}$ .

(b) We first show unexpectedness. If $h_{Z}(t_{Z})=|Z|$ then by Theorem 3.10 we obtain that $Z$ has an unexpected curve if $m_{Z}<u_{Z}$ . (We did not explicitly need to use $1\leqslant m_{Z^{\prime }}$ yet.) Thus, it remains to rule out that $h_{Z}(t_{Z})<|Z|$ . Indeed, if $h_{Z}(t_{Z})<|Z|$ , then Theorem  4.3 gives $m_{Z}=t_{Z}$ . By Proposition 2.10, there are two cases.

1. – In case (b)(i) we have that $Z$ is the complete intersection of a conic and a curve of degree $t_{Z}+1$ , so $|Z|=2t_{Z}+2=2m_{Z}+2$ . But then $u_{Z}+1=|Z|-m_{Z}-1=2m_{Z}+2-m_{Z}-1=m_{Z}+1$ , contradicting our hypothesis. (We still did not explicitly need to use $1\leqslant m_{Z^{\prime }}$ .)

2. – In case (b)(ii), there is a line through $|Z|-t_{Z}\geqslant t_{Z}+2$ of the points, that is, through $|Z|-m_{Z}\geqslant m_{Z}+2$ of the points. Thus this line is a component of any curve of degree $m_{Z}+1$ containing $Z+m_{Z}P$ , but does not itself contain $P$ ; that is, this line is $C^{\prime }$ , hence $m_{Z^{\prime }}=0$ , contrary to assumption.

Conversely, $0<m_{Z}$ by Remark 2.3, and $m_{Z}<u_{Z}$ by Theorem 3.9.

(c) We have $m_{Z^{\prime }}<u_{Z^{\prime }}$ by (a) and $C^{\prime }$ is irreducible, so $C^{\prime }$ is unexpected for $Z^{\prime }$ by (b) if $1\leqslant m_{Z^{\prime }}$ . Conversely, if $C^{\prime }$ is unexpected for $Z^{\prime }$ , then $1\leqslant m_{Z^{\prime }}$ by Remark 2.3.◻

Proof. For convenience we set $d=|Z|$ , the number of points. Since $d-m_{Z}-2=u_{Z}$ by Lemma 3.5(c), the range of degrees in which unexpected curves can occur is due to Theorem 3.9. Now assume $Z$ admits an unexpected curve. Then it has one (call it $C$ ) of degree $m_{Z}+1$ , so by Corollary 5.6(b), $m_{Z}<u_{Z}$ and hence $2m_{Z}+2<d$ . However, if there were a subset of $m_{Z}+2$ (or more) points of $Z$ on a line $L$ , let $Z^{\prime }$ be the points of $Z$ on $L$ and let $Z^{\prime \prime }$ be the rest. By Bezout’s theorem, $L$ is a component of $C_{P}(Z)$ not through $P$ , so $L=C_{P}(Z^{\prime })$ by Lemma 5.1. Then clearly $m_{Z^{\prime }}=0$ , and $t_{Z^{\prime }}=0$ by Example 2.9, so again by Lemma 5.1 we obtain $|Z^{\prime \prime }|=m_{Z}$ . Furthermore, $m_{Z}\leqslant t_{Z}\leqslant t_{Z^{\prime }}+|Z^{\prime \prime }|=m_{Z}$ by Corollary 2.11. Thus $m_{Z}=t_{Z}$ so $Z$ cannot admit an unexpected curve, by Theorem 1.1, contrary to hypothesis.

Conversely, assume $2m_{Z}+2<d$ but no subset of $m_{Z}+2$ (or more) of the points is collinear. Then $m_{Z}<u_{Z}$ , hence $\dim [I_{Z+m_{Z}P}]_{m_{Z}+1}=1$ by Proposition 5.2, so we can speak of $C_{P}(Z)$ . If we now show that $m_{Z^{\prime }}\geqslant 1$ , then $C_{P}(Z)$ is unexpected by Corollary 5.6 and we will be done. If $m_{Z^{\prime }}=0$ , then $C_{P}(Z)$ consists of the line $C_{P}(Z^{\prime })$ through $s\leqslant m_{Z}+1$ points of $Z$ , plus $m_{Z}$ additional lines, one each for the remaining $d-s\geqslant d-m_{Z}-1>2m_{Z}+2-m_{Z}-1=m_{Z}+1$ . Thus $m_{Z}\geqslant d-s>m_{Z}+1$ , which contradicts $m_{Z^{\prime }}=0$ .◻

Proof. The assumption on the characteristic guarantees that the derivation bundle ${\mathcal{D}}_{Z}$ is isomorphic to the syzygy bundle of $\operatorname{Jac}(f)$ (cf. Lemma A.1), and thus each form $s_{i}$ has degree $m_{Z}=a_{Z}$ .

For a polynomial $g\in R$ , denote by $\bar{g}$ its restriction to $L$ . It is a polynomial in two variables, and thus a product of linear forms since $K$ is algebraically closed. The fact that $\deg s_{i}=a_{Z}$ implies (by definition of $a_{Z}$ ) that $\bar{\unicode[STIX]{x1D70E}}=(\bar{s}_{0},\bar{s}_{1},\bar{s}_{2})$ is a syzygy of minimal degree of the restriction of $\operatorname{Jac}(f)$ . It follows that the ideal generated by $\bar{s}_{0},\bar{s}_{1}$ and $\bar{s}_{2}$ has codimension $2$ , that is, that these polynomials do not have a common factor. Hence, the rational map

$$\begin{eqnarray}\unicode[STIX]{x1D70E}=(s_{0}:s_{1}:s_{2}):(\mathbb{P}^{2})^{\vee }{\dashrightarrow}(\mathbb{P}^{2})^{\vee }\end{eqnarray}$$

induces a morphism $\bar{\unicode[STIX]{x1D70E}}:L\rightarrow (\mathbb{P}^{2})^{\vee }$ .

For each $i$ , $1\leqslant i\leqslant d$ , let $Q_{i}$ be the point of intersection of $L$ with $L_{i}$ . Since $L$ is general, the $Q_{i}$ are distinct. Note that the line $L_{Q_{i}}$ dual to $Q_{i}$ contains $P$ and $P_{i}$ .