4. RATIONAL MAPS
9
(Week 3: 2 classes)
Now we will show an important fact, which guarantees that we do not lose any information to
consider the coordinate rings instead of affine varieties.
Proposition 3.6. The contravariant functor
α : (affine varieties)→ (f.g. integral domains over k)
by X 7→ A(X) is an equivalence of categories.
Proof. Given f ∈ Hom(X, Y ), it induces f ∗ ∈ Hom(A(Y ), A(X)) by pullback g ∈ A(Y ) to
g ◦ f ∈ A(X).
Conversely, given h : A(Y ) → A(X), how can we find f such that f ∗ = h? Suppose A(Y ) =
k[y1 , . . . , yn ]/IY and A(X) = k[x1 , . . . , xm ]/IX . Then let f¯i = h(ȳi ), and take any preimage fi of
f¯i . We have a morphism
f˜ = (f1 , . . . , fn ) : Am → An
Now we show f˜ maps X into Y , that is to show, ∀g ∈ IY , ∀P ∈ X, we have g(f˜(P )) = 0.
g(f˜(P )) = g(f1 (P ), . . . , fn (P )) = g(h(ȳ1 )(P ), . . . , h(ȳn )(P )) = h(g(ȳ1 , . . . , ȳn ))(P ) = 0
because h maps IY to IX . The third equality we use the fact that h is a homomorphism of algebras.
Then α is a fully faithful functor, to show it gives an equivalence of categories, we need to
show any f.g. integral domains over k actually occurs as the coordinate ring of an affine variety,
i.e. k[x1 , . . . , xm ]/I for a prime ideal I, but this is obvious.
4. Rational maps
One main difference between algebraic geomtry and differential geometry(or topology) is that
it is more rigid:
Lemma 4.1. Let X and Y be two varieties, suppose two morphisms ϕ, ψ : X → Y coincide on
a open subset U ⊂ X, then ϕ = ψ.
Proof. W.l.o.g, we can assume Y = Pn . Consider the map
ϕ × ψ : X → Pn × Pn
When restrict to U , the image is in the diagonal ∆ ⊂ Pn × Pn . Since ∆ is a closed subset (defined
by {xi yj − xj yi |0 ≤ i < j ≤ n}) and (ϕ × ψ) is continuous, (ϕ × ψ)−1 (∆) is a closed subset of X
containing U , hence is X. This implies ϕ = ψ. (note in the proof, “any nonempty open subset is
dense” is not true in differential geometry.)
Definition 4.2. Let X, Y be varieties. A rational map ϕ : X 99K Y is an equivalence of
pairs hU, ϕU i where U is a nonempty open subset of X, ϕU is a morphism from U to Y , and where
hU, ϕU i and hV, ϕV i are equivalent if ϕU = ϕV on U ∩ V .
The rational map ϕ : X 99K Y is dominant if for some pair hU, V i the image is dense in Y .
The lemma implies that the relation is indeed an equivalence relation: if ϕU = ϕV on U ∩ V
and ϕV = ϕW on V ∩ W , then ϕU = ϕW on U ∩ V ∩ W , therefore by the lemma the equality still
holds on U ∩ W .
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1. VARIETIES
Definition 4.3. A birational map ϕ : X 99K Y between varieties is a rational map which
admits an inverse rational map: ψ : Y 99K X such that ψ ◦ ϕ = idX and ϕ ◦ ψ = idY as rational
maps. In this case, we say X and Y are birational.
Now observe that a rational dominant map ϕ : X 99K Y induces a field morphism ϕ∗ : K(Y ) →
K(X). Indeed, an element g in K(Y ) is a regular function on an open subset V of Y . Suppose ϕ|U
is a morphism, then substituting U by U ∩ ϕ−1 V , we can assume ϕ : U → V is a morphism. Then
ϕ∗ (g) is a regular function on U , hence is a rational function on X.
The main theorem of this section:
Theorem 4.4. The contravariant functor X 7→ K(X), ϕ 7→ ϕ∗ gives an equivalence between
following two categories:
A. Category of varieties and dominant rational maps.
B. Category of finitely generated field extensions of the base field k.
Proof. We construct an inverse functor from B to A. For each f.g. field extension K =
k(y1 , . . . , yn ) where y1 , . . . , yn are not necessarily algebraically independent, consider k[y1 , . . . , yn ].
This is a quotient ring of the polynomial ring k[z1 , . . . , zn ], hence is the coordinate ring of some
vareity Y ⊆ An .
For a field morphism θ : K(Y ) → K(X), we need to define a dominant rational map ϕ : X 99K
Y . we may assume Y is affine in An , whose coordinates are y1 , . . . , yn . If fi = θ(yi ) are regular
on X for all 1 ≤ i ≤ n, then ϕ = (f1 , . . . , fn ) defines the desired map. If fi are not all regular,
replacing X by a nonempty open affine subset U on which all fi are regular. Then we have a ring
homomorphism A(Y ) → A(U ), by Proposition 3.6 it induces a morphism U → Y , which extends
to a rational map X 99K Y .
In the above proof we need a very useful fact to reduce a problem on variety to a problem on
affine variety:
Proposition 4.5. On any variety Y , there is a base for the topology consisting of affine open
subsets, i.e., ∀P ∈ Y and a open subset U 3 P , there is an affine open subset V s.t. P ∈ V ⊆ U .
Proof. Reduce to the case Y = U ⊆ An . If Y is affine there is nothing to prove. Otherwise
define Z = Ȳ − Y which is a closed subset of An , hence is defined by an ideal I. There is f ∈ I s.t.
f (P ) 6= 0. Then we use the following fact:
Fact: affine variety−Z(f ) =affine variety.
Actually, suppose X ⊆ An is defined by the ideal I, then X − Z(f ) can be embedded into
by
1
)
(x1 , . . . , xn ) 7→ (x1 , . . . , xn ,
f (x1 , . . . , xn )
An+1
which is defined by the ideal I + (1 − f xn+1 ).
Remark : We are slow in this week since the presentation on homeworks took longer than I
expect. For next week, only two problems instead of three will be presented.
4. RATIONAL MAPS
11
Extra material. I want to digress a little bit from the main theme and talk about Gröbner bases.
The reason is because some students are concerning about how to compute a set of generators for
the kernel of a ring homomorphism.
First we briefly review Gröbner basis. I highly suggest reading Hal Schenck’s book “Computational algebraic geometry”, and also learn to use Macaulay 2.
1. Fix a polynomial ring k[x1 , . . . , xn ], and fix an order on the monomials. Given a polynomial
f , in(f ) is the lagest monomial appears in f , called the initial monomial of f .
P
P
ai >
bi , or if
Using graded reverse lexicographic: xa11 . . . xann > xb11 . . . xbnn if
P Example:
P
ai =
bi and the right most nonzero entry of (a1 − b1 , . . . , an − bn ) is negative.
x31 x32 > x1 x42 , x31 x52 x23 > x41 x2 x53 . Then in(x31 x52 x23 − x41 x2 x53 ) = x31 x52 x23 .
2. Definition of Gröbner basis.
Definition 4.6. A subset {f1 , . . . , fn } of an ideal I called a Gröbner basis for I if the ideal
< in(f )|f ∈ I >=< in(f1 ), . . . , in(fn ) >.
3. S-pairs. For f, g monic polynomials, define
S(f, g) =
LCM (in(f ), in(g))
LCM (in(f ), in(g))
f−
g
in(f )
in(g)
4. The algorithm to find a Gröbner bases is not hard to describe–Buchberger algorithm.
Start from G = {f1 , . . . , fk }. Repeat adding into G the remainder S of S(f, g) modulo
G for each pair of elements in G, until no more nonzero element can be added.
5. To find a set of generators for the kernel of
(f1 , . . . , fn ) : k[x1 , . . . , xm ] → k[y1 , . . . , yn ],
consider the ideal (y1 − f1 , . . . , yn − fn ) in the larger ring k[x1 , . . . , xm , y1 , . . . , yn ]. Fix any order
that xi < yj , ∀i, j. Find its Gröbner basis. The elements in Gröbner basis which contains only x
variables form a Gröbner basis for the kernel.
Example: Find a set of generators for kerk[x, y, z] → k[t] given by (x, y, z) 7→ (t3 , t4 , t5 ).
Fix Lex order t > z > y > x. Start from G = {t5 − z, t4 − y, t3 − x}.
S(t5 − z, t4 − y) = ty − z, and G = {t5 − z, t4 − y, t3 − x, ty − z};
S(t5 − z, t3 − x) = t2 x − z, and G = {t5 − z, t4 − y, t3 − x, t2 x − z, ty − z};
S(t5 − z, t2 x − z) = t3 z − zx = z(t3 − x) ≡ 0 mod(G), G unchanged;
S(t5 − z, ty − z) = t4 z − zy = z(t4 − y) ≡ 0 mod(G), G unchanged;
S(t4 − y, t3 − x) = tx − y, G = {t5 − z, t4 − y, t3 − x, t2 x − z, ty − z, tx − y};
S(t5 − z, tx − y) = t4 y − zx = (ty − z)t3 + z(t3 − x) ≡ 0, G unchanged;
S(t4 − y, t2 x − z) = t2 z − yx, G = {t5 − z, t4 − y, t3 − x, t2 z − yx, t2 x − z, ty − z, tx − y};
S(t5 − z, t2 z − yx) = t3 yx − z 2 = z 2 − yx2 , G = {t5 − z, t4 − y, t3 − x, t2 z − yx, t2 x − z, ty −
z, tx − y, z 2 − yx2 };
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1. VARIETIES
S(t5 − z, z 2 − yx2 ) = t5 yx2 − z 3 ≡ 0;
S(t4 − y, t2 z − yx) = t2 yx − zy = y(t2 x − z) ≡ 0;
S(t4 − y, ty − z) = t3 z − y 2 ≡ zx − y 2 , G = {t5 − z, t4 − y, t3 − x, t2 z − yx, t2 x − z, ty − z, tx −
y, z 2 − yx2 , zx − y 2 };
S(t5 − z, zx − y 2 ) = t5 y 2 − z 2 x ≡ 0;
S(t4 − y, tx − y) = t3 y − yx ≡ 0;
S(t4 − y, z 2 − yx2 ) = t4 yx2 − z 2 y ≡ 0;
S(t4 − y, zx − y 2 ) = t4 y 2 − zyx ≡ 0;
S(t3 − x, t2 z − yx) = tyx − zx ≡ 0;
S(t3 − x, ty − z) = t2 z − yx ≡ 0;
S(t3 − x, tx − y) = t2 y − x2 ≡ tz − x2 , G = {t5 − z, t4 − y, t3 − x, t2 z − yx, t2 x − z, tz − x2 , ty −
z, tx − y, z 2 − yx2 , zx − y 2 };
S(t3 − x, t2 x − z) = tz − x2 ≡ 0;
S(t3 − x, z 2 − yx2 ) = t3 yx2 − z 2 x ≡ 0;
S(t3 − x, zx − y 2 ) = t3 y 2 − zx2 ≡ 0;
S(t2 x − z, ty − z) = tx − y ≡ 0;
...
Finially we get G = {t5 − z, t4 − y, t3 − x, t2 z − yx, t2 x − z, tz − x2 , ty − z, tx − y, z 2 − yx2 , zy −
x3 , zx − y 2 }. So the last three form Gröbner basis for the kernel.
As you noticed, this computation is not suitable for human. (In this example we can use much
simpler discussion by hand.) That is why I encourage you to run the following code in Macaulay 2.
S=QQ[t]
R=QQ[x,y,z]
F=map(S,R,{t^3,t^4,t^5})
ker(F)
--the following is the output-2
2
2
3
o48 = ideal (y - x*z, x y - z , x - y*z)