1. Introduction
Let
$\mathbb P^n$
be the n-dimensional projective space over K, a perfect field, and let
$\{P_1, \ldots , P_s\}$
be a set of s distinct points in
$\mathbb P^n$
. Denote by
$K[n]=K[X_0, \ldots , X_n]$
the homogeneous coordinate ring of
$\mathbb P^n$
and let
$\wp _1, \ldots , \wp _s$
be the defining homogeneous prime ideals of points
$P_1, \ldots , P_s$
in
$K[n]$
. Let
$m_1, \ldots , m_s$
be positive integers and let
${X=\{(P_1, m_1), \ldots , (P_s, m_s)\}}$
be the zero-scheme defined by the ideal
$I=\bigcap _{i=1}^s \wp _i^{m_i}$
. We may view X as a set of fat points in
$\mathbb P^n$
(we write
$X\subset \mathbb P^n$
). Then, I is the defining ideal of X in
$\mathbb P^n$
.
The ring
$A=K[n]/I$
is the homogeneous coordinate ring of X. It is graded,
${A=\oplus _{t\ge 0} A_t}$
. For each t, the tth graded part
$A_t$
is a finite dimensional K-vector space. It is well known that the ring A has multiplicity
$$ \begin{align*}\delta(A) =\underset{i=1}{\overset{s}{\sum}} \binom{m_i+n-1}{n},\end{align*} $$
and the function
$$ \begin{align*}H_X(t) =\dim_K A_t\end{align*} $$
is called the Hilbert function of X (or of A). This function strictly increases until it reaches the multiplicity
$\delta (A)$
and stays constant thereafter. The number
$$ \begin{align*}r(X)=\min\{t \mid H_X(t)=\delta(A)\}\end{align*} $$
is called the regularity index of X (or of A). We also denote
$r(X)$
by
$r(A)$
. It is difficult to estimate
$r(X)$
(see [Reference Catalisano, Trung and Valla2, Reference Davis, Geramita and Geramita3, Reference Thien and Sinh10]), so instead, one tries to find an upper bound for it (see [Reference Benedetti, Fatabbi and Lorenzini1–Reference Fatabbi and Lorenzini5, Reference Nagel and Trok7–Reference Thien9]).
Put
$$ \begin{align*}\sigma(X)=r(X)+1;\end{align*} $$
also denote
$\sigma (X)$
by
$\sigma (I)$
(see [Reference Franceschini and Lorenzini6, page 171]).
Assume that
$X=\{(P_1, m_1), \ldots , (P_s, m_s) \} \subset \mathbb P^n$
is a set of fat points whose support is contained in a linear subspace of dimension r, say
$\beta $
. We may assume that
$\beta $
is defined by
$X_{r+1}=\cdots =X_n=0$
and
$P_i=(a_{i0}, \ldots , a_{ir}, 0, \ldots , 0)$
,
$i=1, \ldots , s$
. Put
$P_{i}^{\mathbb P^r}=(a_{i0}, \ldots , a_{ir})$
,
$i=1, \ldots , s$
. Then,
$X^{\mathbb P^r}=\{(P_1^{\mathbb P^r}, m_1), \ldots , (P_s^{\mathbb P^r}, m_s)\}$
is a set of fat points in
$\mathbb P^r$
and
$I^{0, \mathbb P^r}=\wp ^{m_1}_1 \cap \cdots \cap \wp ^{m_s}_s \cap K[X_0, \ldots , X_r]$
is the defining ideal of
$X^{\mathbb P^r} \subset \mathbb P^r$
(see Corollary 2.4). Franceschini and Lorenzini showed
$\sigma (I) \ge \sigma (I^{0, \mathbb P^r})$
(see [Reference Franceschini and Lorenzini6, Theorem 4.1]). They conjectured that
$$ \begin{align*}\sigma(I) = \sigma(I^{0, \mathbb P^r}),\end{align*} $$
but they have found neither a proof nor a counterexample (see [Reference Franceschini and Lorenzini6, page 180, line 1]). In this paper, we prove the above conjecture (see Theorem 3.3).
2. Preliminaries
For
$m> n$
, let
$\varphi _{n,m}: \mathbb P^n \to \mathbb P^m$
be the map defined for any
${(a_0,\ldots,a_n)\in\mathbb{P}^n}$
by
${\varphi _{n,m}(a_0, \ldots , a_n)=(a_0, \ldots , a_n, 0, \ldots , 0)}$
.
Given a set of fat points
$X=\{(P_1, m_1), \ldots , (P_s, m_s)\}$
in
$\mathbb P^n$
, put
$$ \begin{align*}\varphi_{n,m}(X)=\{(\varphi_{n,m}(P_1), m_1), \ldots, (\varphi_{n,m}(P_s), m_s)\},\end{align*} $$
which is a set of fat points in
$\mathbb P^m$
. Denote by
$I_X$
the defining ideal of
$X \subset \mathbb P^n$
and denote by
$I_{\varphi _{n,m}(X)}$
the defining ideal of
$\varphi _{n,m}(X) \subset \mathbb P^m$
.
From now on, denote by
$\wp _i$
the defining prime ideal of
$P_i\in \mathbb P^n$
and denote by
$\mathfrak q_i$
the defining prime ideal of
$\varphi (P_i) \in \mathbb P^m$
. Then,
$I_X=\bigcap _{i=1}^s \wp _i^{m_i}$
is the defining ideal of X and
$I_{\varphi _{n,m}(X)}=\bigcap _{i=1}^s \mathfrak q_i^{m_i}$
is the defining ideal of
$\varphi _{n,m}(X)$
.
From [Reference Catalisano, Trung and Valla2, Lemma 3], we get the following result.
Lemma 2.1 [Reference Thien9, Lemma 2.2].
Let
$P_1,\ldots , P_s$
be distinct points in
$\mathbb P^n$
and let
$m_1,\ldots , m_s$
be positive integers. Put
$J= \wp ^{m_1}_1\cap \cdots \cap \wp _{s-1}^{m_{s-1}}$
. If
$\wp _s = (X_1,\ldots , X_n)$
, then
$$ \begin{align*}r(K[n]/(J+\wp_s^{m_s})) \le b\end{align*} $$
if and only if
$X_0^{b-i}M \in J+\wp _s^{i+1}$
for every monomial M of degree i in
$X_1,\ldots , X_n$
,
$i = 0,\ldots , m_s-1$
.
We shall need the following observations.
Lemma 2.2. Let
$\varphi _{n,n+1}: \mathbb P^n \to \mathbb P^{n+1}$
be the map defined by
$\varphi _{n,n+1}(a_0, \ldots , a_n)=(a_0, \ldots , a_n, 0)$
and let
$X=\{(P_1, m_1), \ldots , (P_s, m_s)\}$
be a set of fat points in
$\mathbb P^n$
. Let a be a nonnegative integer.
-
(i) If
$f\in \mathfrak q_j^a$
, then
$f(X_0, \ldots , X_n, 0)\in \wp _j^a$
. -
(ii) If
$f\in I_{\varphi _{n,n+1}(X)}$
, then
$f(X_0, \ldots , X_n, 0) \in I_X$
.
Proof. (i) For each
$P_j=(a_{j0}, \ldots , a_{jn})\in \mathbb P^n$
, there are linear forms
$l_{j1}, \ldots , l_{jn} \in K[n]$
such that
$\wp _j=(l_{j1}, \ldots , l_{jn})\subset K[n]$
. Then,
$\varphi _{n,n+1}(P_j)=(a_{j0}, \ldots , a_{jn}, 0)\in \mathbb P^{n+1}$
and
$\mathfrak q_j=(l_{j1}, \ldots , l_{jn}, X_{n+1})\subset K[n+1]$
.
We argue by induction on a. If
$f\in \mathfrak q_j=(l_{j1}, \ldots , l_{jn}, X_{n+1})$
, then
$$ \begin{align*}f=\sum_{i=1}^nl_{ji}f_i+X_{n+1}f_{n+1}\end{align*} $$
with
$f_i\in K[n+1]$
,
$i=1, \ldots , n+1$
. We have
$$ \begin{align*} f(X_0, \ldots, X_n, 0)=\sum_{i=1}^nl_{ji}(X_0, \ldots, X_n)f_i(X_0, \ldots, X_n, 0) \in \wp_j.\end{align*} $$
Thus, item (i) is true with
$a=1$
. We assume that item (i) is true for
$a-1$
. If
$f\in \mathfrak q_j^a$
, then f can be written
$$ \begin{align*}f=\sum_{i=1}^d g_ih_i\end{align*} $$
with some
$d\in \mathbb N$
,
$g_i\in \mathfrak q_j$
and
$h_i\in \mathfrak q_j^{a-1}$
,
$i=1, \ldots , d$
. By the previous result,
$g_i(X_0, \ldots , X_n, 0) \in \wp _j$
. Moreover, by the inductive assumption,
$h_i(X_0, \ldots , X_n, 0)\in \wp _j^{a-1}$
. Therefore,
$$ \begin{align*}f(X_0, \ldots, X_n, 0)=\sum_{i=1}^d g_i(X_0, \ldots, X_n, 0)h_i(X_0, \ldots, X_n, 0) \in \wp_j^a.\end{align*} $$
(ii) If
$f\in I_{\varphi _{n,n+1}(X)}$
, then
$f\in \mathfrak q_j^{m_j}$
,
$j=1, \ldots , s$
. From item (i),
$f(X_0, \ldots , X_n, 0) \in \wp _j^{m_j}$
,
$j=1, \ldots , s$
. This implies
$$ \begin{align*} f(X_0, \ldots, X_n, 0) \in \bigcap_{j=1}^s \wp_j^{m_j}=I_X.\\[-49pt] \end{align*} $$
Corollary 2.3. Let
$\varphi _{n,n+1}: \mathbb P^n \to \mathbb P^{n+1}$
be the map defined by
$\varphi _{n,n+1}(a_0, \ldots , a_n)=(a_0, \ldots , a_n, 0)$
and let
$X=\{(P_1, m_1), \ldots , (P_s, m_s)\}$
be a set of fat points in
$\mathbb P^n$
. Let a be a nonnegative integer. Then:
-
(i)
$\wp _j^a = \mathfrak q_j^a \cap K[n]$
; -
(ii)
$I_X = I_{\varphi _{n,n+1}(X)}\cap K[n]$
.
Proof. (i) It is clear that
$\wp _j^{a} \subset \mathfrak q_j^{a} \cap K[n].$
For
$g\in \mathfrak q_j^a \cap K[n]$
, since
$g\in \mathfrak q_j^a$
,
$$ \begin{align*}g=\sum_{i=1}^d h_ik_i,\end{align*} $$
where
$d \in \mathbb N$
,
$h_i\in \mathfrak q_j$
and
$k_i\in \mathfrak q_j^{a-1}$
,
$i=1, \ldots , d$
. By Lemma 2.2(i), it follows that
$h_i(X_0, \ldots , X_n, 0)\in \wp _j$
and
$k_i(X_0, \ldots , X_n, 0)\in \wp _j^{a-1}$
,
$i=1, \ldots , d$
. Since
$g\in \mathfrak q_j^a \cap K[n]$
, we have
$g=g(X_0, \ldots , X_n, 0)$
. Therefore,
$$ \begin{align*} g=g(X_0, \ldots, X_n, 0)\ =\sum_{i=1}^d h_i(X_0, \ldots, X_n, 0)k_i(X_0, \ldots, X_n, 0) \in \wp_j^a. \end{align*} $$
(ii) For
$j=1, \ldots , s$
, by using item (i),
$\wp _j^{m_j}=\mathfrak q_j^{m_j}\cap K[n].$
This implies
$$ \begin{align*} I_X=\bigcap_{j=1}^s \wp_j^{m_j}=\bigcap_{j=1}^s\mathfrak q_j^{m_j}\bigcap K[n]=I_{\varphi_{n,n+1}(X)}\bigcap K[n].\\[-49pt] \end{align*} $$
Corollary 2.4. Let
$X=\{(P_1, m_1), \ldots , (P_s, m_s) \} \subset \mathbb P^n$
be a set of fat points whose support is contained in a linear proper subspace of dimension r. Assume that
$P_i=(a_{i0}, \ldots , a_{ir}, 0, \ldots , 0)$
,
$i=1, \ldots , s$
. Put
$P_{i}^{\mathbb P^r}=(a_{i0}, \ldots , a_{ir})$
,
$i=1, \ldots , s$
, and put
$X^{\mathbb P^r}=\{(P_1^{\mathbb P^r}, m_1), \ldots , (P_s^{\mathbb P^r}, m_s)\}$
. Then,
$$ \begin{align*}\bigcap_{i=1}^s \wp_i^{m_i} \bigcap K[r]\end{align*} $$
is the defining ideal
$I_{X^{\mathbb P^r}}$
of the set of fat points
$X^{\mathbb P^r} \subset \mathbb P^r$
.
Proof. Consider the map
$ \varphi _{r,n}: \mathbb P^r \to \mathbb P^{n}$
defined by
$$ \begin{align*}\varphi_{r,n}(a_0, \ldots, a_r)= (a_0, \ldots, a_r, 0, \ldots, 0).\end{align*} $$
We have
$$ \begin{align*}\varphi_{r,n}(X^{\mathbb P^r})=X,\quad I_{\varphi_{r,n}(X^{\mathbb P^r})}=I_X=\wp^{m_1}_1 \cap \cdots \cap \wp^{m_s}_s.\end{align*} $$
We prove the corollary by induction on
$n-r$
. Consider
$\varphi _{r, r+1}: \mathbb P^r \to \mathbb P^{r+1}$
and
${X^{\mathbb P^r} \subset \mathbb P^r}$
. From Corollary 2.3(ii),
$$ \begin{align*}I_{X^{\mathbb P^r}} = I_{\varphi_{r,r+1}(X^{\mathbb P^r})}\bigcap K[r].\end{align*} $$
Consider
$\varphi _{r+1, n}: \mathbb P^{r+1} \to \mathbb P^{n}$
and
$\varphi _{r,r+1}(X^{\mathbb P^r}) \subset \mathbb P^{r+1}$
. By the inductive assumption,
$$ \begin{align*}I_{\varphi_{r,r+1}(X^{\mathbb P^r})} = I_{\varphi_{r+1,n}(\varphi_{r,r+1}(X^{\mathbb P^r}))}\bigcap K[r+1]. \end{align*} $$
We have
$\varphi _{r,n} = \varphi _{r+1,n} \circ \varphi _{r,r+1}$
. So
$$ \begin{align*} I_{X^{\mathbb P^r}}= I_{\varphi_{r,r+1}(X^{\mathbb P^r})}\bigcap K[r] &=(I_{\varphi_{r+1,n}(\varphi_{r,r+1}(X^{\mathbb P^r}))}\bigcap K[r+1])\bigcap K[r]\\ &=I_{\varphi_{r,n}(X^{\mathbb P^r})}\bigcap K[r] =\bigcap_{i=1}^s \wp_i^{m_i} \bigcap K[r].\\[-49pt] \end{align*} $$
3. Proof of the conjecture of Franceschini and Lorenzini
Let
$\varphi _{n,m}: \mathbb P^n \to \mathbb P^m$
be the map defined by
$\varphi _{n,m}(a_0, \ldots , a_n)=(a_0, \ldots , a_n, 0, \ldots , 0)$
and let
$X=\{(P_1, m_1), \ldots , (P_s, m_s)\}$
be a set of fat points in
$\mathbb P^n$
. In this section, we prove that
$$ \begin{align*}r(K[n]/I_X) = r(K[m]/I_{\varphi_{n,m}(X)})\end{align*} $$
(Theorem 3.3). From this result, we prove the conjecture of Franceschini and Lorenzini which is an extension of [Reference Franceschini and Lorenzini6, Theorem 4.1]. To prove Theorem 3.3, we need the following results.
Proposition 3.1. Let
$X=\{(P_1, m_1), \ldots , (P_s, m_s)\}$
be a set of fat points in
$\mathbb P^n$
and let
$\varphi _{n,n+1}: \mathbb P^n \to \mathbb P^{n+1}$
be the map defined by
$\varphi _{n,n+1}(a_0, \ldots , a_n)=(a_0, \ldots , a_n, 0)$
. Put
$Y=\{(P_1, m_1), \ldots , (P_{s-1}, m_{s-1})\}$
. Then,
$$ \begin{align*}r(K[n]/(I_Y+\wp_s^{m_s})) = r(K[n+1]/(I_{\varphi_{n,n+1}(Y)}+\mathfrak q_s^{m_s})).\end{align*} $$
Proof. We have
$I_Y{\kern-1pt}={\kern-1pt}\bigcap _{i=1}^{s-1}\wp _i^{m_i}$
and
$I_{\varphi _{n,n+1}(Y)}{\kern-1pt}={\kern-1pt}\bigcap _{i=1}^{s-1} \mathfrak q_i^{m_i}$
. Put
$P_s{\kern-1pt}={\kern-1pt}(1, 0, \ldots , 0)$
. Then,
$\wp _s{\kern-1pt}={\kern-1pt}(X_1, \ldots , X_n) \subset K[n]$
,
$\varphi _{n,n+1}(P_{s}){\kern-1pt}={\kern-1pt}(1, 0, \ldots , 0)\in \mathbb P^{n+1}$
and
$\mathfrak q_{s}{\kern-1pt}={\kern-1pt}(X_1, \ldots , X_n, X_{n+1})\subset K[n+1]$
.
Step 1:
$r(K[n]/(I_Y+\wp _s^{m_s})) \le r(K[n+1]/(I_{\varphi _{n,n+1}(Y)}+\mathfrak q_s^{m_s}))$
. We will show that if
$r(K[n]/(I_Y+\wp _s^{m_s}))> t,$
then
$r(K[n+1]/(I_{\varphi _{n,n+1}(Y)}+\mathfrak q_s^{m_s}))> t.$
In fact, if
$$ \begin{align*}r(K[n]/(I_Y+\wp_s^{m_s}))> t,\end{align*} $$
then by Lemma 2.1, there is a monomial
$M=X_1^{c_1} \cdots X_{n}^{c_{n}}$
,
$c_1+\cdots +c_{n}=i$
,
$i\in \{0, \ldots , m_s-1\}$
, such that
$$ \begin{align} X_0^{t-i}M \notin I_Y+\wp_s^{i+1}. \end{align} $$
If
$X_0^{t-i}M \in I_{\varphi _{n,n+1}(Y)}+\mathfrak q_s^{i+1}$
, then there are
$f\in I_{\varphi _{n,n+1}(Y)}$
and
$g\in \mathfrak q_s^{i+1}$
such that
$$ \begin{align*}X_0^{t-i}M = f + g.\end{align*} $$
Since
$f\in I_{\varphi _{n,n+1}(Y)}$
, we have
$f(X_0, \ldots , X_n, 0)\in I_Y$
by Lemma 2.2(ii). Since
$g\in \mathfrak q_s^{i+1}$
, we have
$g(X_0, \ldots , X_n, 0)\in \wp _s^{i+1}$
by Lemma 2.2(i). Since
$X_0^{t-i}M\in K[n]$
,
$$ \begin{align*}X_0^{t-i}M = f(X_0, \ldots, X_n, 0)+ g(X_0, \ldots, X_n, 0)\in I_Y + \wp_s^{i+1}, \end{align*} $$
which is in contradiction with (3.1). So,
$X_0^{t-i}M \notin I_{\varphi _{n,n+1}(Y)}+\mathfrak q_s^{i+1}.$
By Lemma 2.1,
$$ \begin{align*}r(K[n+1]/(I_{\varphi_{n,n+1}(Y)}+\mathfrak q_s^{m_s}))> t.\end{align*} $$
Step 2:
$r(K[n]/(I_Y+\wp _s^{m_s})) \ge r(K[n+1]/(I_{\varphi _{n,n+1}(Y)}+\mathfrak q_s^{m_s}))$
. If
$$ \begin{align*}r(K[n+1]/(I_{\varphi_{n,n+1}(Y)}+\mathfrak q_s^{m_s}))> t,\end{align*} $$
then by Lemma 2.1, there is a monomial
$M=X_1^{c_1} \cdots X_{n+1}^{c_{n+1}}$
,
$c_1+\cdots +c_{n+1}=i$
,
$i\in \{0, \ldots , m_s-1\}$
, such that
$$ \begin{align} X_0^{t-i}M \notin I_{\varphi_{n,n+1}(Y)}+\mathfrak q_{s}^{i+1}. \end{align} $$
From Corollary 2.3(i),
$\wp _{s}^{i+1} = \mathfrak q_s^{i+1}\cap K[n]$
. By Corollary 2.3(ii),
$I_Y= I_{\varphi _{n,n+1}(Y)} \cap K[n]$
. Therefore,
$$ \begin{align} I_Y+\wp_{s}^{i+1} \subset I_{\varphi_{n,n+1}(Y)}+\mathfrak q_{s}^{i+1}. \end{align} $$
Consider the monomial
$N=X_1^{c_1} \cdots X_{n}^{c_{n}}$
. If
$X_0^{t-i}N \in I_Y+\wp _{s}^{i+1}$
, then
$$ \begin{align} X_{n+1}^{c_{n+1}}X_0^{t-i}N\in I_Y+\wp_{s}^{i+1}. \end{align} $$
From (3.3) and (3.4),
$X_0^{t-i}M=X_{n+1}^{c_{n+1}} ,X_0^{t-i}N\in I_{\varphi _{n,n+1}(Y)}+\mathfrak q_s^{i+1},$
which is in contradiction with (3.2). So,
$X_0^{t-i}N \notin I_Y+\wp _{s}^{i+1}.$
Then, by Lemma 2.1,
$$ \begin{align*}r(K[n]/(I_Y+\wp_{s}^{m_s}))> t.\end{align*} $$
The proof of Proposition 3.1 is now complete.
Proposition 3.2. Let
$X=\{(P_1, m_1), \ldots , (P_s, m_s)\}$
be a set of fat points in
$\mathbb P^n$
, where
$s\ge 2$
and
$m_1 \ge m_2\ge \cdots \ge m_s$
. Let
$\varphi _{n,n+1}: \mathbb P^n \to \mathbb P^{n+1}$
be the map defined by
$\varphi _{n,n+1}(a_0, \ldots , a_n)=(a_0, \ldots , a_n, 0)$
. Then,
$$ \begin{align*}r(K[n]/I_X) = r(K[n+1]/I_{\varphi_{n,n+1}(X)}).\end{align*} $$
Proof. We prove the proposition by induction on s. If
$s=1$
, then
$X=\{(P_1, m_1)\}$
,
$I_X=\wp _1^{m_1}\subset K[n]$
and
$\varphi _{n,n+1}(X)=\{(\varphi _{n,n+1}(P_1), m_1)\}$
,
$I_{\varphi _{n,n+1}(X)}=\mathfrak q_1^{m_1}\subset K[n+1]$
. By [Reference Davis, Geramita and Geramita3, Corollary 2.3],
$r(K[n]/\wp _1^{m_1})=m_1-1$
and
$r(K[n+1]/\mathfrak q_1^{m_1})=m_1-1$
. Hence, the proposition is true for
$s=1$
.
With
$s\ge 2$
, we suppose that the proposition is true for
$s-1$
. Put
$$ \begin{align*}Y=\{(P_1, m_1), \ldots, (P_{s-1}, m_{s-1})\}.\end{align*} $$
Then,
$I_Y=\bigcap _{i=1}^{s-1}\wp _i^{m_i}$
,
$I_X=I_Y \cap \wp _{s}^{m_s}$
,
$I_{\varphi _{n,n+1}(Y)}=\bigcap _{i=1}^{s-1} \mathfrak q_i^{m_i}$
and
$I_{\varphi _{n,n+1}(X)}=I_{\varphi _{n,n+1}(Y)}\cap \mathfrak q_{s}^{m_s}$
. By [Reference Catalisano, Trung and Valla2, Lemma 1],
$$ \begin{align*}r(K[n]/I_X) = \max\{m_s-1, r(K[n]/I_Y), r(K[n]/(I_Y+\wp_s^{m_s})) \} \end{align*} $$
and
$$ \begin{align*}r(K[n+1]/I_{\varphi_{n,n+1}(X)}) = \max\{m_s-1, r(K[n+1]/I_{\varphi_{n,n+1}(Y)}), r(K[n+1]/(I_{\varphi_{n,n+1}(Y)}+\mathfrak q_{s}^{m_s})) \}.\end{align*} $$
By [Reference Catalisano, Trung and Valla2, Corollary 2],
$r(K[n]/I_X) \ge m_1+m_2-1$
and
$r(K[n+1]/I_{\varphi _{n,n+1}(X)}) \ge m_1+m_2-1$
. Thus,
$$ \begin{align*}r(K[n]/I_X) = \max\{r(K[n]/I_Y), r(K[n]/(I_Y+\wp_s^{m_s})) \}\end{align*} $$
and
$$ \begin{align*}r(K[n+1]/I_{\varphi_{n,n+1}(X)}) = \max\{r(K[n+1]/I_{\varphi_{n,n+1}(Y)}), r(K[n+1]/(I_{\varphi_{n,n+1}(Y)}+\mathfrak q_{s}^{m_s})) \}.\end{align*} $$
From the inductive assumption,
$$ \begin{align*}r(K[n]/I_Y) = r(K[n+1]/I_{\varphi_{n,n+1}(Y)}),\end{align*} $$
and from Proposition 3.1,
$$ \begin{align*}r(K[n]/(I_Y+\wp_s^{m_s}))= r(K[n+1]/(I_{\varphi_{n,n+1}(Y)}+\mathfrak q_{s}^{m_s})).\end{align*} $$
Therefore,
$r(K[n]/I_X) = r(K[n+1]/I_{\varphi _{n,n+1}(X)})$
as desired.
Theorem 3.3. Let
$X=\{(P_1, m_1), \ldots , (P_s, m_s)\}$
be a set of fat points in
$\mathbb P^n$
, where
$s\ge 2$
and
$m_1 \ge m_2\ge \cdots \ge m_s$
. Let
$\varphi _{n,m}: \mathbb P^n \to \mathbb P^{m}$
be the map defined by
$\varphi _{n,m}(a_0, \ldots , a_n)=(a_0, \ldots , a_n, 0, \ldots , 0)$
. Then,
$$ \begin{align*}r(K[n]/I_X) = r(K[m]/I_{\varphi_{n,m}(X)}).\end{align*} $$
Proof. We prove the theorem by induction on
$m-n$
. Consider
$\varphi _{n,n+1}: \mathbb P^n \to \mathbb P^{n+1}$
and
$X\subset \mathbb P^n$
. From Proposition 3.2,
$$ \begin{align*}r(K[n]/I_X) = r(K[n+1]/I_{\varphi_{n,n+1}(X)}).\end{align*} $$
Consider
$\varphi _{n+1,m}: \mathbb P^{n+1}\to \mathbb P^{m}$
and
$\varphi _{n,n+1}(X)\subset \mathbb P^{n+1}$
. By the inductive assumption,
$$ \begin{align*}r(K[n+1]/I_{\varphi_{n,n+1}(X)}) = r(K[m]/I_{\varphi_{n+1,m}(\varphi_{n,n+1}(X))}).\end{align*} $$
Since
$\varphi _{n,m}=\varphi _{n+1,m} \circ \varphi _{n,n+1}$
,
$$ \begin{align*}\kern-12pt r(K[n]/I_X)=r(K[n+1]/I_{\varphi_{n,n+1}(X)}) =r(K[m]/I_{\varphi_{n+1,m}(\varphi_{n,n+1}(X))}) =r(K[m]/I_{\varphi_{n,m}(X)}).\\[-33pt] \end{align*} $$
The following result is the conjecture of Franceschini and Lorenzini (see [Reference Franceschini and Lorenzini6, page 180, line 1]); it is an extension of [Reference Franceschini and Lorenzini6, Theorem 4.1].
Theorem 3.4. Let
$X=\{(P_1, m_1), \ldots , (P_s, m_s) \} \subset \mathbb P^n$
be a set of fat points whose support is contained in a linear subspace of dimension r, where
$s\ge 2$
and
$m_1 \ge m_2\ge \cdots \ge m_s$
. Assume that
$P_i=(a_{i0}, \ldots , a_{ir}, 0, \ldots , 0)$
,
$i=1, \ldots , s$
. Put
$P_{i}^{\mathbb P^r}=(a_{i0}, \ldots , a_{ir})$
,
$i=1, \ldots , s$
,
$X^{\mathbb P^r}=\{(P_1^{\mathbb P^r}, m_1), \ldots , (P_s^{\mathbb P^r}, m_s)\}$
,
$I=\bigcap _{i=1}^s \wp _i^{m_i}, \ I^{0, \mathbb P^r}= I \cap K[r]$
and
$\sigma (X)=r(X)+1.$
Also denote
$\sigma (X)$
by
$\sigma (I)$
. Then,
$$ \begin{align*}\sigma(I^{0, \mathbb P^r})=\sigma(I).\end{align*} $$
Proof. Consider the map
$\varphi _{r,n}: \mathbb P^r \to \mathbb P^{n}$
defined by
$(a_0, \ldots , a_r)\mapsto (a_0, \ldots , a_r, 0, \ldots , 0)$
. From Theorem 3.3,
$$ \begin{align*}r(K[r]/I_{X^{\mathbb P^r}})=r(K[n]/I_{\varphi_{r,n}(X^{\mathbb P^r})}).\end{align*} $$
However,
$\varphi _{r,n}(X^{\mathbb P^r})=X$
and
$I_{\varphi _{r,n}(X^{\mathbb P^r})}=I_X$
, so
$$ \begin{align*}r(K[r]/I_{X^{\mathbb P^r}})=r(K[n]/I_X),\end{align*} $$
that is,
$r(X^{\mathbb P^r})=r(X).$
This implies
$$ \begin{align*}\sigma(X^{\mathbb P^r})=\sigma(X).\end{align*} $$
We see that
$I=\bigcap _{i=1}^s\wp _i^{m_i}$
is the defining ideal
$I_X$
of
$X \subset \mathbb P^n$
and
$I^{0, \mathbb P^r}=\bigcap _{i=1}^s \wp _i^{m_i} \cap K[r]$
is the defining ideal
$I_{X^{\mathbb P^r}}$
of
$X^{\mathbb P^r} \subset \mathbb P^r$
(see Corollary 2.4). Therefore,
$$ \begin{align*} \sigma(I^{0, \mathbb P^r})=\sigma(I).\\[-33pt] \end{align*} $$
Acknowledgement
We thank the referee for helpful suggestions and a careful reading of the paper.
