1. Introduction
To study smooth curves one intersects the curve with a general hyperplane and
studies the resulting finite set of points. The Harris’s lemma [2] about a set of points in
the uniform position has attracted much attention in algebraic geometry. This is a set of
points of a projective space such that any two subsets, each with the same cardinality, have
the same Hilbert function. We bring up the question of whether other uniformed position
properties remain unchanged when a curve is intersected by any hyperplane. To answer
this question, we use the notation ground form which was given by E. Noether [8], B.L.
van der Waerden [11] and specializations of modules and of graded modules which was
given by D. V. Nhi and N. V. Trung [6, 7]. In this article, we will prove the preservation
of some properties of the generic linear subspace sections of nondegenerate varieties by
specializations and ground forms of components of variety are conjugated.
Throughout this paper, Ω will be the universal field, which is algebraic and has an
infinite degree of transcendence over an infinite ground field K

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JOURNAL OF SCIENCE OF HNUE
Mathematical and Physical Sci., 2012, Vol. 57, No. 7, pp. 12-19
This paper is available online at
THE LINEAR SUBSPACE SECTION OF VARIETY BY SPECIALIZATIONS
Dam Van Nhi
School for Gifted Students, Hanoi National University of Education
Abstract. In this paper, we prove the preservation of some properties of the generic
linear subspace sections of nondegenerate varieties by specializations.
Keywords: Specialization, variety, absolutely irreducible.
1. Introduction
To study smooth curves one intersects the curve with a general hyperplane and
studies the resulting finite set of points. The Harris’s lemma [2] about a set of points in
the uniform position has attracted much attention in algebraic geometry. This is a set of
points of a projective space such that any two subsets, each with the same cardinality, have
the same Hilbert function. We bring up the question of whether other uniformed position
properties remain unchanged when a curve is intersected by any hyperplane. To answer
this question, we use the notation ground form which was given by E. Noether [8], B.L.
van der Waerden [11] and specializations of modules and of graded modules which was
given by D. V. Nhi and N. V. Trung [6, 7]. In this article, we will prove the preservation
of some properties of the generic linear subspace sections of nondegenerate varieties by
specializations and ground forms of components of variety are conjugated.
Throughout this paper, Ω will be the universal field, which is algebraic and has an
infinite degree of transcendence over an infinite ground field K.
2. Some results about specializations of modules
Let K be an infinite ground field and K be the algebraic closure of K. Denote
by u = (u1, u2, . . . , us), a system of s new indeterminates ui, which are algebraically
independent of K, K[u, x] and the polynomial ring K[u1, . . . , us, x1, . . . , xn]. We shall
Received September 28, 2012. Accepted October 5, 2012.
2000Mathematics Subject Classification: Primary 13A30, Secondary 13D45.
Contact Dam Van Nhi, e-mail address: damvannhi@yahoo.com
12
The linear subspace section of variety by specializations
regard the ui as parameters over K. Let f(u, x) =
m1∑
i1=0
. . .
mn∑
in=0
ai1...inu
d1
1 . . . u
ds
s x
i1
1 . . . x
in
n
be any polynomial inK[u, x]. Let α = (α1, . . . , αs) be any element ofΩ
s. The polynomial
f(α, x) =
m1∑
i1=0
. . .
mn∑
in=0
ai1...inα
d1
1 . . . α
ds
s x
i1
1 . . . x
in
n
is said to be a specialization of f with respect to the substitution u→ α.
We denote the polynomial rings in n variables x1, . . . , xn over K(u) and K(α) by
R = K(u)[x] and by Rα = K(α)[x], respectively. Each element a(u, x) of R can be
written in the form a(u, x) =
p(u, x)
q(u)
with p(u, x) ∈ K[u, x]andq(u) ∈ K[u] \ {0}. For
any α with q(α) 6= 0 we define
a(α, x) =
p(α, x)
q(α)
.
We shall say that a property holds for almost all α if it holds for all points of a
Zariski-open non-empty subset of Ωm. For convenience we will generally omit the phrase
"for almost all α" in the proofs of the results presented in this paper. Then we have the
following lemmas:
Lemma 2.1. [9, lemma 7] Let f(u, x) be a polynomial inK[u, x]. If f(u, x) is irreducible
in K(u)[x] then f(α, x) is also irreducible for almost all α.
Lemma 2.2. [9, lemma 8] If f(u, x) ∈ K(u)[x] is a power of an irreducible polynomial
in K(u)[x] then abvef(α, x) is the same power of an irreducible polynomial in K(α)[x]
for almost all α.
Lemma 2.3. [1, 2.Satz] If f(u, x) is an absolute irreducible polynomial in K[u, x] then
f(α, x) is also an absolute irreducible polynomial for almost all α.
Following [10] we define the specialization of an ideal I of a polynomial ring R =
K(u)[x] with respect to the substitution u −→ α as the ideal Iα of Rα = K(α)[x]
generated by elements of the set {f(α, x)| f(u, x) ∈ I ∩ k[u, x]}. This is obviously an
ideal of the polynomial ringRα. For almost all substitutions u −→ α, the ideal Iα inherits
most of the basic properties of I.
The specialization of ideals can be generalized to modules. Let F be a free
R-module of finite rank. The specialization Fα of F is a free Rα-module of the same
rank. Let φ : F −→ G be a homomorphism of free R-modules. We can represent φ by
a matrix A = (aij(u, x)) with respect to fixed bases of F and G. Let Aα = (aij(α, x)).
Then Aα is well-defined for almost all α. The specialization φα : Fα −→ Gα of φ is
given by the matrix Aα provided that Aα is well-defined. We note that the definition of
φα depends on the chosen bases of Fα and Gα.
13
Dam Van Nhi
Definition 2.1. [6] Let L be an R-module. Let F1
φ−→ F0 −→ L −→ 0 be a finite free
presentation of L. Let φα : (F1)α −→ (F0)α be a specialization of φ. We call Lα :=
Coker φα a specialization of L (with respect to φ).
If we choose a different finite free presentation F ′1 −→ F ′0 −→ L −→ 0 we may
get a different specialization L′α of L, but Lα and L
′
α are canonically isomorphic. Hence
Lα is uniquely determined up to isomorphisms and we have the result.
Lemma 2.4. [6, theorem 2.4] Let 0 −→ L −→ M −→ N −→ 0 be an exact sequence
of finitely generated R-modules. Then the sequence 0 −→ Lα −→ Mα −→ Nα −→ 0 is
exact for almost all α.
Let L be a graded R-module of dimension d. Let hL(t) denote the Hilbert
polynomial of L. Then we have:
Lemma 2.5. [7, Corollary 2.3] Let L be a graded R-module. Then, for almost all α, we
have hLα(t) = hL(t),
3. Preservation of some properties of hyperplane sections of
varieties by specializations
In this section, we are interested in the hyperplane sections of varieties.
Before reproving some results about hyperplane sections of varieties we recall the
notation of a ground form which is introduced in order to study the properties of points on
a variety. The concept of ground forms was formulated by E. Noether [8]. More accounts
will be found in W. Krull [4] or B. L. van der Waerden [11].
Denote by (u) = (uij) with i = 0, 1, . . . , n, and j = 0, 1, . . . , n, a system
of (n + 1)2 new indeterminates uij, are algebraically independent over K. We enlarge
K by adjoining (u). Consider the polynomial rings K(u)[x] = K(u)[x0, x1, . . . , xn]
and K(u)[y] = K(u)[y0, y1, . . . , yn]. The general linear transformation establishes an
isomorphism between K(u)[x] and K(u)[y] when in every polynomial of K(u)[y] the
substitution
yi =
n∑
j=0
uijxj , i = 0, 1, . . . , n,
is carried out and the inverse transformation
xi =
n∑
j=0
vijyj, i = 0, 1, . . . , n,
has its coefficients vij ∈ K(u). We get K(u)[x] = K(u)[y]. Every ideal I of K[x]
generates an ideal IK(u)[x], which is transformed by the above isomorphism into the
14
The linear subspace section of variety by specializations
ideal
I∗ = ({f(
n∑
j=0
v0jyj,
n∑
j=0
v1jyj, . . . ,
n∑
j=0
vnjyj)|f(x), x1, . . . , xn) ∈ I}).
We consider an unmixed d-dimenisional homogeneous ideal P ⊂ K[x]. Then, P is
transformed into the ideal P ∗ and the following ideal
P ∗ ∩K(u)[y0, y1, . . . , yd] = (F (u, y0, y1, . . . , yd))
is a principal ideal ofK(u)[y0, . . . , yd].We may suppose that F is normalized so as to be
a polynomial in uij and primitive in them. By a linear projective transformation, we can
choose F so that it is regular in y0. The form F (u, y0, . . . , yd) is called a ground form of P.
It was well known that the ground form F of a prime ideal P is an irreducible form, but P
is primary if and only if its ground form is a power of an irreducible form. We emphasize
that if P1 and P2 are distinct d-dimensional prime ideals, then the ground form of P1 is
not a constant multiple of the ground form of P2, and the ground form of a d-dimensional
ideal is a product of ground forms of d-dimensional primary componentes, see [4], Satz
3 and Satz 4.
Given any homogeneous ideal I of the standard grading polynomial ring K[x] =
K[x0, . . . , xn] with deg xi = 1. We now set R = K[x]/I =
⊕
t≥0Rt. The
Hilbert function of I, which is denoted by h(−; I), is defined as follows h(t; I) =
dimK Rt for all t ≥ 0. We make a number of simple observations which are needed
afterwards.
It is easy to check the following result:
Lemma 3.1. The Hilbert function is unchanged by projective inverse transformation. If
K∗ is an extension field ofK, then
h(t; I) = h(t; IK∗[x]) for all t ≥ 0.
Now we want to study the hyperplane section of nondegenerate varieties.
Definition 3.1. A variety V of Pn is nondegenerate if it does not lie in any hyperplane.
Definition 3.2. A variety V of Pn defined overK is said to be absolutely irreducible if it
is still irreducible when it is considered as a variety over any extension ofK.
Put a = I(V ) =
⊕
j≥1 aj. Note that V is nondegenerate if and only if a1 = 0 or
h(1; a) = n+1. Consider the intersectionWα = V ∩Hα of an irreducible nondegenerate
variety V with a hyperplane
Hα : ℓα = α0x0 + · · ·+ αnxn = 0, αi ∈ K∗.
15
Dam Van Nhi
Proposition 3.1. Let V be an irreducible nondegenerate variety of Pn with d = dimV >
1. Then Wα = V ∩ Hα is again an irreducible nondegenerate variety of Pn−1 with
dimWα = dim V − 1 for almost all α.
Proof. Denote Hv : ℓv = v0x0 + · · · + vnxn = 0 the general hyperplane. Since the
irreducibility of a variety is preserved by finite pure transcendental extension of ground
field, therefore V is still an irreducible variety of dimension d overK(v).We have I(V ∩
Hv) = (a, ℓv) and V ∩ Hv is an irreducible variety of dimension d − 1 by Bertini’s
theorem. Using a general transformation, the ground form of (a, ℓv) can be assumed as a
form F (u, v, y0, . . . , yd−1). By [9, theorem 6], F (u, α, y0, . . . , yd−1) is the ground form of
(a, ℓα) or of V ∩Hα. Since V ∩Hv is an irreducible variety, therefore F (u, v, x0, . . . , xd−1)
is a power of an irreducible form. By lemma 2.2, F (u, α, x0, . . . , xd−1) is also a power of
an irreducible form. Hence Wα is again an irreducible variety of dimension d − 1 for
almost all α.
Since V is nondegenerate, there is h(1; a) = n + 1. Because aK(v)[x] : ℓv = aK(v)[x],
therefore aK(α)[x] : ℓα = aK(α)[x] for almost all α. Hence
h(1; (aK(v)[x], ℓv)) = h(1; aK(v)[x])− h(0; aK(v)[x]) = n+ 1− 1 = n.
By lemma 2.5, h(1; (aK(α)[x], ℓα)) = h(1; aK(α)[x]) − h(0; aK(α)[x]) and we obtain
h(1; I(Wα)) = n.HenceWα = V ∩Hα is again an irreducible nondegenerate variety.
Assume that V ⊂ Pn is an irreducible variety of dimension d overK. Denote I(V )
in K[x] by p. Then pK(u)[x] ∩ K(u)[y] = (F (u; y0, . . . , yd)) is a principal ideal. Note
that if we consider V as an algebraic set over K then V may be a reducible set. Then we
have the following result.
Lemma 3.2. Let V be an irreducible variety of dimension d > 1 over K and Hα :
α0x0+α1x1+ · · ·+αnxn = 0, αi ∈ K. Suppose that V =
s⋃
i=1
Vi is a decomposition of V
into the irreducible varieties Vi over K. Then, every Vi ∩Hα is an absolutely irreducible
component of V ∩Hα over K for almost all α.
Proof. Denote Hv : ℓv = v0x0 + v1x1 + · · · + vnxn = 0 a general hyperplane, where
the vi are indeterminates. Assume that V =
s⋃
i=1
Vi is a decomposition of V into the
irreducible varieties Vi over K. Since Vi is an irreducible variety over K, therefore Vi
is an absolutely irreducible variety of dimension d > 1. Hence V ∩ Hv =
s⋃
i=1
Vi ∩ Hv,
where every Vi ∩ Hv is an absolutely irreducible variety by [3, X.13 theorem I]. Since
Vi ∩ Hv is an absolutely irreducible variety, therefore the ground form Fi(u, v, y) of
Pv = (I(Vi), ℓv)K(v)[x] is a power an absolutely irreducible form f(u, v, x), an example
16
The linear subspace section of variety by specializations
being Fi(u, v, y) = f(u, v, y)
s. Since the gound form of Pα is Fi(u, α, y) = f(u, α, y)
s
and f(u, α, y) is also an absolutely irreducible form by lemma 2.3, Vi ∩ Hα is an
irreducible variety over K. Hence, every Vi ∩ Hα is an absolutely irreducible variety
for almost all α.
Now we assume that V is a nondegenerate irreducible variety of dimension d over
K. Denote I(V ) by p. Suppose that the generic linear subspace Sun−d of dimension n− d
is given by the generic linear equations
Hiu : ℓiu = ui0x0 + ui1x1 + · · ·+ uinxn = 0, i = 1, 2, . . . , d.
Put ui for the set of all uij with j = 0, 1, . . . , n and i = 0, 1, . . . , d. It was
well known that a generic linear subspace Sun−d meets V in a finite set of points T,
and each of them is a generic point of V over K(u1, . . . , ud). The Harris’s lemma [2]
shows that T has the Uniform Position Property. That is, any two subsets of T with the
same cardinality have the same Hilbert function. The property about a set in the uniform
position has attracted much attention in algebraic geometry. Consider a linear subspace
Sαn−d of dimension n− d that is given by the linear equations
Hiα : ℓiα = αi0x0 + αi1x1 + · · ·+ αinxn = 0, ∀αij ∈ K, i = 1, 2, . . . , d.
The problem of concern in the following section is the existence of isomorphism
between round forms of components of the intersection V ∩ Sαn−d.
Assume that V ⊂ Pn is an irreducible variety of dimension d over K. Let (ξ) =
(ξ0, ξ1, . . . , ξn) be a generic point of V such thatK(ξ) is separable overK.Without loss of
generality, we may suppose that this is normalized so that ξ0 = 1. The uij are algebraically
independent over K(ξ1, . . . , ξn). We put λi = −
n∑
j=0
uijξj with i = 0, 1, . . . , d. Then
λ1, . . . , λd are algebraically independent over K(u), but λ0, λ1, . . . , λd are algebraically
dependent over K(u).We therefore have an equation F (u;λ0, . . . , λd) = 0.We note that
if (ξ′) is a conjugate of (ξ) then λ′0 = −
n∑
j=0
u0jξ
′
j is a conjugate of λ0 and (ξ) 6= (ξ′) if
and only if λ0 6= λ′0. Bertini’s theorem proves the following lemma:
Lemma 3.3. Let V ⊂ Pn be an irreducible variety of dimension d ≥ 1 and Sun−d a generic
linear subspace of dimension n − d. The intersection V ∩ Sun−d is an irreducible variety
of dimension 0 overK(u1, . . . , ud) or the ideal (pK(u1, . . . , ud)[x], ℓ1u, . . . , ℓdu) = Pu is
a 0-dimensional prime ideal ofK(u1, . . . , ud)[x].
The varieties of dimension 0 over a field consist of a finite number of points. We
shall denote by T = {qi = (λi0, λi1, . . . , λin)|i = 1, 2, . . . , s} the intersection V ∩H1u ∩
17
Dam Van Nhi
. . .∩Hdu.All qi are conjugate to each other overK(u) := K(u1, . . . , ud). OverK(u)[u0],
we define F (u0, u) to be the form
s∏
i=1
(u00λi0 + u01λi1 + · · ·+ u0nλin).
This form F (u0, u) is the ground form of V and is irreducible overK(u)[u0].
We now consider V as an algebraic set overK.We have a decomposition of V into
K-varieties Vi and V =
r⋃
i=1
Vi. Hence, over K(u) we obtain
V ∩H1u ∩ . . . ∩Hdu =
r⋃
i=1
[Vi ∩H1u ∩ . . . ∩Hdu].
Denote the ground form of Vi ∩H1u ∩ . . . ∩Hdu by Fi(u0, u) with i = 1, 2, . . . , r.
Then we have F (u0, u) =
r∏
i=1
Fi(u0, u) and each Fi(u0, u) is an irreducible form.
Lemma 3.4. The ground form Fi(u0, u) is absolutely irreducible and there are
isomorphisms φij(Fi(u0, u)) = Fj(u0, u) for all i, j = 1, . . . , r.
Proof. Denote Vi ∩ H1u ∩ . . . ∩ Hdu by V ∗i . By Bertini’s theorem, every variety V ∗i is
irreducible. Since K is algebraically closed and all uij ∈ Ω, therefore K(u) is also
algebraically closed in Ω by [9, lemma 4]. Hence V ∗i and it’s ground form Fi(u0, u)
is absolutely irreducible. Suppose that qi and qj are generic points of V
∗
i and V
∗
j ,
respectively. It is obvious that all of qi are generic points of the irreducible variety
V ∩ H1u ∩ . . . ∩ Hdu over K(u). By [5, I.5 Theorem 9], there is an isomorphism ϕij
such that ϕij(qi) = qj . Extend this isomorphism ϕij to an isomorphism δij of K(u)(qi)
ontoK(u)(qj), whose restriction toK(u) is an isomorphism, but not necessarily identity.
The isomorphism δij induce an isomorphism φij such that φij(Fi(u0, u)) = Fj(u0, u) for
all i, j = 1, . . . , r.
Consider the intersection V ∩Sαn−d. Then, we have V ∩H1α ∩ . . .∩Hdα =
r⋃
i=1
[Vi ∩
H1α∩ . . .∩Hdα] and the ground form of V ∩H1α∩ . . .∩Hdα is F (u0, α), and the ground
form of Vi ∩H1α ∩ . . . ∩Hdα is Fi(u0, α), see [4].
Corollary 3.1. Every variety Vi ∩ H1α ∩ . . . ∩ Hdα and it’s ground form Fi(u0, α) are
absolutely irreducible for almost all α.
Proof. By lemma 3.2, the variety Vi∩H1α∩ . . .∩Hdα is absolutely irreducible for almost
all α. Because the ground form Fi(u0, u) is absolutely irreducible by lemma 3.4, therefore
Fi(u0, α) is absolutely irreducible for almost all α by lemma 2.3.
18
The linear subspace section of variety by specializations
The following result follows from lemma 2.4, lemma 3.4 and corollary 2.2:
Theorem 3.1. For almost all α, there are isomorphisms Φij(Fi(u0, α)) =
Fj(u0, α), ∀i, j = 1, . . . , r.
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