Polish Word Sketches
21 Pages
English
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Polish Word Sketches

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Learn all about the services we offer
21 Pages
English

Description

  • fiche de synthèse - matière potentielle : a word
  • leçon - matière potentielle : verb
Polish Word Sketches Adam Radziszewski Adam Kilgarriff Robert Lew Wrocław University of Technology Lexical Computing Ltd, Adam Mickiewicz University University of Leeds Abstract Word sketches are one-page automatic, corpus-based summaries of a word's grammatical and collocational behaviour. They were first used in the production of the Macmillan English Dictionary (Rundell 2002).
  • tagger
  • polish tagger takipi
  • grammatical relation
  • sketch engine
  • psu mięso
  • metaphorical use
  • corpus
  • word
  • query-language
  • query language

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Language English

Exrait

Summary
Plan
Section 1. Algebraically closed elds, general notions
1. Notation.
2. A ne sets, ideals.
3. Zariski topology, irreducibility.
4. Coordinate ring, morphism.
5. Projective space and projective sets.
6. Rational maps of projective varieties.
7. functions.
8. Dimension.
9. Local properties.
10. Curves.
11. Hurwitz’s Formula.
1
All the notions mentioned in the rst section can be found in any
textbook in Algebraic Geometry: for example, in the books of Shafare-
vich ([Sha]), Hartshorne [Har], or in the survey of Danilov([Dan]).
The second section of the chapter contains more advanced topics. refs!
Sophisticated methods used in xx Chapter x are beyond the frame of
this introduction.
2. General background
2.1. Notation. Let k be an algebraically closed eld.
(1) k =knf0g:
N N(2) By A ( or, simply, A ) we denote the N-dimensionalx ;:::;x1 N
a ne space over k with coordinates x ;:::;x :1 N
(3) By k[x ;:::;x ] = k[x] we denote the ring of polynomials in1 N
N variables x ;:::;x with coe cients in k:1 N
(4) By k(x ;:::;x ) = k(x) we denote the eld of rational func-1 N
tions in N variables x ;:::;x with coe cients in k:1 N
(5) P stands for the restriction of a function P to a set X:
X
(6) I = hg ;:::;gi means that an ideal I is generated by its1 n
elements g ;:::;g :1 n
12
(7) I =hhg ;:::;gii is the minimal radical ideal containing I =1 n
hg ;:::;gi; hhIii is the ideal con I:1 n
(8) The degree of a polynomial P is denoted by degP .
(9) A B means \A is isomorphic toB" (whateverA andB are).=
2.2. A ne sets and ideals. Algebraic Geometry studies geom-
etry of sets de ned by polynomial equations. The building block of
Algebraic Geometry is the notion of a ne set .
NDefinition 2.1. An a ne subset X of A is the set of allx ;:::;x1 N
common zeros of a nite collection of polynomials, in other words, the
Nset of all points x2A such that
(2.1) P (x) = 0;:::;P (x) = 0;1 n
where P ;:::;P 2k[x]:1 n
The role of an a ne set in Algebraic Geometry is somewhat sim-
ilar to the role of a ball around a point in Analysis: it is a standard
neighbourhood endowed with a coordinate system. Every point inside
such a neighbourhood is de ned uniquely by the values of coordinates
x ;:::;x :1 N
Let I k[x] be a radical ideal. By Hilbert’s Basis Theorem (see,
e.g., [Lan, Chap. VI,x2]), there are elements g ;:::;g 2 k[x] such1 n
Nthat I =hg ;:::;gi: LetX (I) =fg = 0;:::;g = 0gA : Then,1 n 1 n
by de nition, it is an a ne set. Moreover, for any element g2I there
are h 2k[x] such that g =g h +:::g h ; hencei 1 1 n n
N N(2.2) X (I) =fx2A :g(x) = 08g2IgA
is the set of all common zeros of all elements of I: If I I ; then1 2
X (I )X (I ):1 2
NOn the other hand, to any set XA (not necessary a ne) one
can relate a radical ideal

I(X) =fp(x)2k[x] : p = 0gk[x]
X
of all polynomials vanishing at every point ofX: From the de nition it
follows that if YX; thenI(Y )I(X):
Definition 2.2. The setX =X (I(X)) is called the a ne closure
of X:
Lemma 2.3. The operations I!X (I) and X!I(X) enjoy the
following properties.
(1) X is an a ne subset.
(2) XX.
(3)I(X (I)) =hhIii:2. GENERAL BACKGROUND 3
(4) If X is a ne, then X =X (I(X)) =X:
(5) X is the smallest a ne subset containing X: This means that
if Y is an a ne subset such that YX, then YX:
(6) XY =)XY:
Proof. 1. Follows from the de nition of X (I):
2. X is the set of all common zeros of all elements ofI(X): On
the other hand, every element ofI(X) vanishes at any pointx2X by
de nition. Thus x2X:
3. Every element g2hhIii vanishes onX (I) and thus belongs to
I(X (I)): Let g2I(X (I)). Let I =hg ;:::;gi: Then g vanishes at1 n
all common zeros of g ;:::;g : According to Hilbert’s Nullstellensatz1 n
(see, e.g., [Lan, Chap. X,x2]), there exist a numberm and polynomials
h ;:::;h 2k[x] such that1 n
mg =g h +:::g h ;1 1 n n
mi.e., g 2I: Hence g2hhIii:
4. Assume that X is a ne and is de ned by equations (2.1). If
x2X; then all elements ofI(X) vanish at x: In particular, equations
(2.1) are ful lled. Thus x2X:
5. Assume that Y X and Y is a ne. Then I(Y )I(X) and
X (I(Y )) =Y by the previous item. ThusY =X (I(Y ))X (I(X)) =
X:
6. XY =)I(X)I(Y ) =)X (I(X))X (I(Y )):
Lemma 2.3 may be summarized as follows:
Although there is no one-to-one correspondence between ideals in
Nk[x] and subsets of A , the correspondence between radical ideals and
a ne subsets is one-to-one. Every a ne subset may be related in a
unique way to a radical ideal, and vice versa.
1Example 2.4. The a ne subsets of A are all nite subsets, ;; and
1A :
Example 2.5. (1)I(;) =k[x];X (k[x]) =; since; is the set
of common zeros of all polynomials.
N(2)I(A ) =f0g; X (f0g) = k[x] since the only polynomial van-
ishing everywhere is 0:
N(3) LetX =fag =f(a ;:::;a )gA be a point. ThenI(X) =1 N
NP
f (x a )hg is the maximal ideal of all the polynomialsi i i
1
vanishing at a:
1 1Example 2.6. If XA is an in nite set, then X =A .6
4
2 2 2Example 2.7. Let X =fx +y = 0g A . If char(k) = 2,
2 2thenI(X) =hx +yi: If char(k) = 2, thenI(X) =hx +yi: This
example shows that an a ne set is indeed a set of points: a ne sets
2X = fx = 0g and X = fx = 0g are the same. If we want to1 2
distinguish them, i.e., to count points with multiplicities, we have to
consider ideals (even non radical) as the main tool. This way leads to
the notion of scheme (see Section ??).
2.3. Zariski topology, irreducibility. In the case k = C the
Na ne space C is endowed with the topology induced by the Euclidean
topology of Complex Plane. In this topology the property to be con-
tinuous is a local one: a function may be continuous at one point and
discontinuous at another. In this subsection we will describe another
topology, re ecting the algebraic nature of our functions. It will be well
de ned for any eld, and only algebraic functions will be continuous
relative to this topology. Since a polynomial is a global object (a -
nite number of parameters de ne a polynomial everywhere), continuity
becomes a global property.
NLemma 2.8. (1) If X ;X A ; thenI(X [X ) =I(X )\1 2 1 2 1
I(X ):2
N(2) Assume thatfXg;2A; is a family of a ne subsets of A ;
and letI =I(X ) =hg ;:::;g i: LetJ =hhg ; 1in ; 2Aii. 1; n ; i;
By Hilbert’s Basis Theorem (see, e.g., [Lan, Ch. VI,x2]),J =
hh ;:::;hi for some h 2 k[x]: Let Y =fh = 0;:::;h =1 n j 1 nT
N0gA : Then Y = X :
2A
Corollary 2.9. The union of two a ne sets is an a ne set. The
intersection of any family of a ne sets is an a ne set.
In light of Corollary 2.9, one can de ne a new topology on an a ne
space.
NDefinition 2.10. The Zariski topology on A is de ned by the
Nfollowing rule: a subset XA is closed if and only if it is a ne.
Further on, unless explicitly stated otherwise, we will use the Zariski
Ntopology. As in any topology, every subset ofA is endowed with the
Zariski topology induced from the ambient space.
A Zariski analogue of a connected closed subset is an irreducible
a ne set.
NDefinition 2.11. An a ne set XA is irreducible if for any
two closed subsets Y ;Y X the equality Y [Y = X implies that1 2 1 2
either X =Y or X =Y :1 26
2. GENERAL BACKGROUND 5
2Example 2.12. The setfxy = 0gA is reducible, one can take
Y = fx = 0g and Y = fy = 0g: On the other hand, Y and Y1 2 1 2
are irreducible: the line cannot be a union of two nite subsets (see
Example 2.4).
NTheorem 2.13. Any a ne set X A can be represented as a
nite union X =[Y of its irreducible subsets. Up to the order of thei
sets Y , this representation is unique, provided that Y 6 Y for i =j:i i j
Theorem 2.14. The closure of a non-empty open subset of an
Na ne irreducible set XA coincides with X: Every two non-empty
Nopen subsets of an a ne irreducible set X A have a
intersection.
Proof. IfU is open,Z =XnU is closed andX =U[Z: Therefore,
X =X =U[Z: Since both U and Z are closed and X is irreducible,
eitherZ =X (hence,U =;), orU =X: Any setV such thatU\V =;
is contained in Z: Thus we have V = Z: Hence V cannot be an open
subset of X; as already shown.
Note that the second statement of Theorem 2.14 shows that the
Zariski topology is not a Hausdor one.
NCertainly the property of a closed set XA to be irreducible is
re ected in the properties of the corresponding ideal: I(X) is prime.
Every polynomial p(x) is continuous in the Zariski topology since
1 1p (a) =fx :p(x) =ag is an a ne set for any a2A k: Letk =C;=
NXC , and letX denote the closure ofX in the Euclidean topology.c
Every polynomial p2I(X) is continuous in the topology
and hence vanishes at every point x2X : Thus x2X; i.e., X X:c c
2.4. Coordinate ring, morphism. Another algebraic object re-
lated to an a ne set X is the coordinate ring k[X] = k[x]=I(X):
Let : k[x]! k[X] be the natural projection. If P ;P 2 k[x], then1 2
(P ) =(P ) if and only ifP P 2I(X); i.e., the restrictions ofP1 2 1 2 1
andP toX coincide. Roughly speaking,k[X] is the ring of restrictions2
to X of all the polynomials de ned on the ambient space.
2 3 2Example 2.15. LetX =fx y = 0gA andk =C: We wantx;y
to computeC[X]: The easiest way to do it is to nd one standard repre-
sentative in every equivalence class and de ne ring operations between
them.
2 3 2 31. Finding a representative. SinceI(X) =hx yi; x = y onP
i j 2 3X: Let p(x;y) = a xy 2C[x;y]: We may change x to y in everyij6
summand. Thus the polynomialp de nes the same element of C[X] as
X X XX
j+3m j+3mq(x;y) = a xy + a y =xq (y) +q (y):ij ij 1 2
i=2m+1 j i=2m j
(Here the rst (second) sum contains all monomials of p in which x
enters in odd (even) power, respectively.)
2. Showing that each class contains exactly one representative of