Introduction to Linear Algebraic Groups by itlpw9937

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									                Introduction to Linear Algebraic Groups


                                          Stephen Donkin

                                              University of York


                                       November 19, 2009




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   1 / 21
Lecture 6




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   2 / 21
Lecture 6

Hilbert’s Nullstellensatz (i) Let A be a finitely generated k -algebra.
Then every maximal ideal of A has codimension 1.
(ii) Let (V , A) be an affine variety. Let I be an ideal of A with I = A.
Then V(I) = ∅ (i.e. the ideal I has a “zero point").




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   2 / 21
Lecture 6

Hilbert’s Nullstellensatz (i) Let A be a finitely generated k -algebra.
Then every maximal ideal of A has codimension 1.
(ii) Let (V , A) be an affine variety. Let I be an ideal of A with I = A.
Then V(I) = ∅ (i.e. the ideal I has a “zero point").

By a maximal ideal of A I mean an ideal I such that I = A and such that the
only ideal of A properly containing I is A itself.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   2 / 21
Lecture 6

Hilbert’s Nullstellensatz (i) Let A be a finitely generated k -algebra.
Then every maximal ideal of A has codimension 1.
(ii) Let (V , A) be an affine variety. Let I be an ideal of A with I = A.
Then V(I) = ∅ (i.e. the ideal I has a “zero point").

By a maximal ideal of A I mean an ideal I such that I = A and such that the
only ideal of A properly containing I is A itself.

If I say that f is a regular function on an affine variety V I just mean that f
belongs to the coordinate algebra k [V ].




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   2 / 21
Lecture 6

Hilbert’s Nullstellensatz (i) Let A be a finitely generated k -algebra.
Then every maximal ideal of A has codimension 1.
(ii) Let (V , A) be an affine variety. Let I be an ideal of A with I = A.
Then V(I) = ∅ (i.e. the ideal I has a “zero point").

By a maximal ideal of A I mean an ideal I such that I = A and such that the
only ideal of A properly containing I is A itself.

If I say that f is a regular function on an affine variety V I just mean that f
belongs to the coordinate algebra k [V ].

Application Suppose that V is an affine variety and f is a regular function
on V which is never zero. Then the function 1/f is also regular. To see this
take I = k [V ] · f . Then V(I) = ∅ so that I = k [V ], i.e. f is invertible.


Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   2 / 21
From the Nullstellensatz one may deduce the strong Nullstellensatz.
First some additional terminology. Let A be a commutative ring. An
element f ∈ A is nilpotent if f n = 0 for some positive integer n. It is an
exercise to show that the set of all nilpotent elements forms an ideal.
This is called the nilradical of A and we denote it N(A).




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   3 / 21
From the Nullstellensatz one may deduce the strong Nullstellensatz.
First some additional terminology. Let A be a commutative ring. An
element f ∈ A is nilpotent if f n = 0 for some positive integer n. It is an
exercise to show that the set of all nilpotent elements forms an ideal.
This is called the nilradical of A and we denote it N(A).

The ring A is called reduced if N(A) = 0. For example there are no
non-zero nilpotent elements in Z so N(Z) = {0}, i.e. Z is reduced.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   3 / 21
From the Nullstellensatz one may deduce the strong Nullstellensatz.
First some additional terminology. Let A be a commutative ring. An
element f ∈ A is nilpotent if f n = 0 for some positive integer n. It is an
exercise to show that the set of all nilpotent elements forms an ideal.
This is called the nilradical of A and we denote it N(A).

The ring A is called reduced if N(A) = 0. For example there are no
non-zero nilpotent elements in Z so N(Z) = {0}, i.e. Z is reduced.
                                               √
More generally, if I is an ideal of A we write I for the set of all f ∈ A
such that f n √ I for some positive integer √ From the definitions it
              ∈                             n.
follows that I is the ideal of A such that I/I = N(A/I).




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   3 / 21
Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   4 / 21
Now, given a closed set Z in an affine variety V we write N (Z ) for the
ideal of all functions f ∈ k [V ] that vanish on Z .




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   4 / 21
Now, given a closed set Z in an affine variety V we write N (Z ) for the
ideal of all functions f ∈ k [V ] that vanish on Z .

What happens if we start with an ideal I of k [V ] then we take V(I), the
closed set consisting of the points at which all elements of I vanish,
and then we take N (V(I))? Clearly this ideal contains I but how much
bigger can it be?




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   4 / 21
Now, given a closed set Z in an affine variety V we write N (Z ) for the
ideal of all functions f ∈ k [V ] that vanish on Z .

What happens if we start with an ideal I of k [V ] then we take V(I), the
closed set consisting of the points at which all elements of I vanish,
and then we take N (V(I))? Clearly this ideal contains I but how much
bigger can it be?

The Strong Nullstellensatz (i) Let A be a finitely generated
k -algebra. Then the intersection of the maximal ideals of A is the
nilradical of A.
(ii) Let (V , A) be an affine variety and let I be an ideal of A. Then
                         √
N (V(I)) is the radical I = {f ∈ A | f r ∈ I for some r ≥ 0}.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   4 / 21
This effectively produces a complete dictionary between finitely
generated reduced algebras over k and affine algebraic varieties.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   5 / 21
This effectively produces a complete dictionary between finitely
generated reduced algebras over k and affine algebraic varieties.

On the one hand, if A = k [V ], for some variety V then A is reduced (if
f n = 0 then for x ∈ V we have f n (x) = f (x)n = 0, so f (x) = 0. Hence
f = 0.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   5 / 21
This effectively produces a complete dictionary between finitely
generated reduced algebras over k and affine algebraic varieties.

On the one hand, if A = k [V ], for some variety V then A is reduced (if
f n = 0 then for x ∈ V we have f n (x) = f (x)n = 0, so f (x) = 0. Hence
f = 0.

One the other hand suppose A is just some finitely generated reduced
commutative k -algebra. Suppose A has generators a1 , a2 , . . . , an .
Consider V = An . Then k [V ] = k [X1 , . . . , Xn ], the free polynomial
algebra in the coordinate functions X1 , . . . , Xn .




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   5 / 21
This effectively produces a complete dictionary between finitely
generated reduced algebras over k and affine algebraic varieties.

On the one hand, if A = k [V ], for some variety V then A is reduced (if
f n = 0 then for x ∈ V we have f n (x) = f (x)n = 0, so f (x) = 0. Hence
f = 0.

One the other hand suppose A is just some finitely generated reduced
commutative k -algebra. Suppose A has generators a1 , a2 , . . . , an .
Consider V = An . Then k [V ] = k [X1 , . . . , Xn ], the free polynomial
algebra in the coordinate functions X1 , . . . , Xn .




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   5 / 21
We define θ : k [X1 , . . . , Xn ] → A to be the k -algebra homomorphism
taking Xi to ai . Then θ is surjective with kernel I, say. Thus θ induces
an isomorphism from k [V ]/I to A. Let Z = V(I), the closed set in V
defined by I. Now Z is an affine variety with coordinate algebra k [V ]/J,
where J is the set of f ∈ √ ] such that f |Z = 0. This means that
                  √           k [V
J = N (V(I)) = I. Now I/I = N(k [V ]/I) = 0 (because k [V ]/I is
isomorphic to A hence reduced). So J = I. Hence k [Z ] is isomorphic
to k [V ]/I and hence to A. So every commutative finitely generated
reduced k -algebra is (isomorphic to) the coordinate algebra of some
affine variety.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   6 / 21
One may make this correspondence a bit more canonical and precise
as follows.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   7 / 21
One may make this correspondence a bit more canonical and precise
as follows.

Remark Let A be a finitely generated k -algebra which is reduced (i.e. 0 is
the only nilpotent element). We write Var(A) for the set Homk −alg (A, k ). We
have a natural map θ : A → Map(Var(A), k ), given by θ(a)(α) = α(a), for
α ∈ Var(A). By the Nullstellensatz, this map is injective. Identifying A with
an algebra of functions on Var(A) via θ we thus obtain an affine algebraic
variety (Var(A), A). It follows now from Exercise 6 of Chapter 1 that the
category of affine varieties is anti-equivalent to the category of finitely
generated reduced k -algebras.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   7 / 21
Dominant Morphisms




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   8 / 21
Dominant Morphisms

A continuous map φ : X → Y is called dominant if Im(f ) is dense in Y .




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   8 / 21
Dominant Morphisms

A continuous map φ : X → Y is called dominant if Im(f ) is dense in Y .

Exercise Show that a morphism of affine varieties φ : X → Y is dominant
if and only if the comorphism φ : k [Y ] → k [X ] is injective.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   8 / 21
Dominant Morphisms

A continuous map φ : X → Y is called dominant if Im(f ) is dense in Y .

Exercise Show that a morphism of affine varieties φ : X → Y is dominant
if and only if the comorphism φ : k [Y ] → k [X ] is injective.

We will call a dominant morphism of varieties φ : X → Y finite if k [X ] is an
integral extension of Im(φ ).




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   8 / 21
Dominant Morphisms

A continuous map φ : X → Y is called dominant if Im(f ) is dense in Y .

Exercise Show that a morphism of affine varieties φ : X → Y is dominant
if and only if the comorphism φ : k [Y ] → k [X ] is injective.

We will call a dominant morphism of varieties φ : X → Y finite if k [X ] is an
integral extension of Im(φ ).

In the prove of the following Lemma we shall use the result (proved in the
appendix) that if A is a subring of B with B integral over A then for each
prime ideal P of A there exists a prime ideal Q of B “over P", meaning that
A Q = P.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   8 / 21
We will also use the following observation. Let V be a variety and
x ∈ V . Then N = {f ∈ k [V ] | f (x) = 0} is an ideal of codimension 1
(i.e. dim k [V ]/N = 1), in fact N = Ker( x ) and x : k [V ] → k induces an
isomorphism k [V ]/N → k . Conversely, given an ideal M of k [V ] of
codimension 1, we have k [V ] = M ⊕ k and we may define a
homomorphism α : k [V ] → k , by α(m + λ) = λ. Since V is a variety,
we have α = x , for some x ∈ V and it follows that
M = Ker( x ) = {f ∈ k [V ] | f (x) = 0}.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   9 / 21
We will also use the following observation. Let V be a variety and
x ∈ V . Then N = {f ∈ k [V ] | f (x) = 0} is an ideal of codimension 1
(i.e. dim k [V ]/N = 1), in fact N = Ker( x ) and x : k [V ] → k induces an
isomorphism k [V ]/N → k . Conversely, given an ideal M of k [V ] of
codimension 1, we have k [V ] = M ⊕ k and we may define a
homomorphism α : k [V ] → k , by α(m + λ) = λ. Since V is a variety,
we have α = x , for some x ∈ V and it follows that
M = Ker( x ) = {f ∈ k [V ] | f (x) = 0}.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   9 / 21
Lemma Let φ : V → W be a dominant finite morphism of affine
varieties. Then φ is surjective and takes closed sets to closed sets.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   10 / 21
Lemma Let φ : V → W be a dominant finite morphism of affine
varieties. Then φ is surjective and takes closed sets to closed sets.

Proof First we will check that φ(V ) = W . Let A = k [V ] and let
B = Im(φ ). Let y ∈ W and let N = {f ∈ k [W ] | f (y ) = 0}. Note that
V(N) = {y }. Then φ (N) is an ideal of codimension 1 in B, in particular
φ (N) is a prime ideal of B. Hence there exists a prime ideal P of A such that
P   B = φ (N).




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   10 / 21
Lemma Let φ : V → W be a dominant finite morphism of affine
varieties. Then φ is surjective and takes closed sets to closed sets.

Proof First we will check that φ(V ) = W . Let A = k [V ] and let
B = Im(φ ). Let y ∈ W and let N = {f ∈ k [W ] | f (y ) = 0}. Note that
V(N) = {y }. Then φ (N) is an ideal of codimension 1 in B, in particular
φ (N) is a prime ideal of B. Hence there exists a prime ideal P of A such that
P   B = φ (N).

Let M be a maximal ideal of A containing P. Then M has codimension 1 in A
(by the Nullstellensatz) and φ (N) ≤ M.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   10 / 21
Let x ∈ V be the element such that Ker( x ) = M. Then we have
n(φ(x)) = φ (n)(x) = x (φ (n)) = 0, for n ∈ N. Hence
φ(x) ∈ V(N) = {y }. Hence φ(x) = y , and φ is onto.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   11 / 21
Let x ∈ V be the element such that Ker( x ) = M. Then we have
n(φ(x)) = φ (n)(x) = x (φ (n)) = 0, for n ∈ N. Hence
φ(x) ∈ V(N) = {y }. Hence φ(x) = y , and φ is onto.

Now if Z is a closed set in X and let T be the closure of φ(Z ) in Y .
Then one checks that the restriction φ0 : Z → T is dominant and finite.
So by the above φ0 maps Z onto T , i.e., φ(Z ) is closed. So φ takes
closed sets to closed sets.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   11 / 21
The next result shows that a dominant map is “approximately a finite
map".




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   12 / 21
The next result shows that a dominant map is “approximately a finite
map".

Proposition Let φ : V → W be a dominant morphism of irreducible
affine varieties. Then there exists 0 = g ∈ k [W ] such that the restriction
Vφ (g) → Wg factorizes as Vφ (g) → Wg × An → Wg , for some n ≥ 0,
where the first map is a finite dominant morphism and the second map is
projection.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   12 / 21
The next result shows that a dominant map is “approximately a finite
map".

Proposition Let φ : V → W be a dominant morphism of irreducible
affine varieties. Then there exists 0 = g ∈ k [W ] such that the restriction
Vφ (g) → Wg factorizes as Vφ (g) → Wg × An → Wg , for some n ≥ 0,
where the first map is a finite dominant morphism and the second map is
projection.

To prove this one puts B = φ (k [W ]). Then make the ring of fractions of A
with denominators non-zero elements of B. This produces a finitely generated
algebra over F , the field of fractions of B. Then one uses Noether
Normalization.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   12 / 21
It has an important corollary.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   13 / 21
It has an important corollary.

Corollary (Chevalley’s Theorem) Let φ : V → W be a dominant
morphism of irreducible varieties. Then Im(φ) contains a non-empty
open set of W .




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   13 / 21
It has an important corollary.

Corollary (Chevalley’s Theorem) Let φ : V → W be a dominant
morphism of irreducible varieties. Then Im(φ) contains a non-empty
open set of W .

Proof We factorize φ on a principal open set Vf → Wg × An → Wg , as
above. The first map is finite so onto by the Lemma above. The second map
(projection) is clearly onto. Hence Wg = φ(Vf ) ⊆ Im(φ), so Im(φ) contains
the open set Wg of W .




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   13 / 21
We have an application to algebraic groups.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   14 / 21
We have an application to algebraic groups.

Proposition 2 Let φ : G → H be a morphism of algebraic groups.
Then the image of φ is closed.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   14 / 21
We have an application to algebraic groups.

Proposition 2 Let φ : G → H be a morphism of algebraic groups.
Then the image of φ is closed.
                                         n                                      n
Proof We can write G =                       xi G0 . So φ(G) =                      φ(xi )φ(G0 ). So its
                                       i=1                                 i=1
enough to prove that φ(G0 ) is closed, i.e., we may assume that G is
connected. Replacing H by the closure of φ(G) we can assume that φ is
dominant. Now φ(G) contains an open set V , say.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups             November 19, 2009   14 / 21
But then φ(G) =                       φ(g)V is open, i.e., φ(G) is open. Hence
                             g∈G
              hφ(G) is open. So the complement of φ(G) is open and φ(G)
h∈H\φ(G)
is closed.




Stephen Donkin (University of York)      Introduction to Linear Algebraic Groups   November 19, 2009   15 / 21
But then φ(G) =                       φ(g)V is open, i.e., φ(G) is open. Hence
                             g∈G
              hφ(G) is open. So the complement of φ(G) is open and φ(G)
h∈H\φ(G)
is closed.




Stephen Donkin (University of York)      Introduction to Linear Algebraic Groups   November 19, 2009   15 / 21
But then φ(G) =                       φ(g)V is open, i.e., φ(G) is open. Hence
                             g∈G
              hφ(G) is open. So the complement of φ(G) is open and φ(G)
h∈H\φ(G)
is closed.

Dimension and Tangent Spaces




Stephen Donkin (University of York)      Introduction to Linear Algebraic Groups   November 19, 2009   15 / 21
But then φ(G) =                       φ(g)V is open, i.e., φ(G) is open. Hence
                             g∈G
              hφ(G) is open. So the complement of φ(G) is open and φ(G)
h∈H\φ(G)
is closed.

Dimension and Tangent Spaces

  A topological space X has dimension n (see A5) if there is a strictly
ascending chain of closed irreducible sets X0 ⊂ X1 ⊂ · · · ⊂ Xn , of length n,
and none longer.




Stephen Donkin (University of York)      Introduction to Linear Algebraic Groups   November 19, 2009   15 / 21
If X is an affine variety then a strictly ascending chain of closed
irreducible subsets X0 ⊂ X1 · · · ⊂ · · · ⊂ Xr gives rise to a strictly
ascending chain of prime ideals N (Xr ) ⊂ · · · ⊂ N (X1 ) ⊂ N (X0 ) of k [X ]
and, conversely, a strictly chain of prime ideals P0 ⊂ P1 ⊂ · · · ⊂ Pr
gives rise to a strictly ascending chain of closed irreducible subsets
V(Pr ) ⊂ · · · ⊂ V(P1 ) ⊂ V(P0 ).




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   16 / 21
The dimension of X is therefore the dimension dim k [X ] of the ring
k [X ], as discussed in A5. In particular the dimension is finite and, if X
is irreducible, is equal to the transcendence degree of the field of
fractions of k [X ] over k , see A5 Proposition 1. Of course, from the
definition, if X is irreducible and Y is a proper closed subset of X then
we have dim Y < dim X . This property will be frequently used in what
follows.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   17 / 21
  There is also a local notion of dimension, that of the dimension of the
tangent space Tx (V ), of a point x in an affine variety V , as a k vector
space. In this section we see how these different notions are related.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   18 / 21
  There is also a local notion of dimension, that of the dimension of the
tangent space Tx (V ), of a point x in an affine variety V , as a k vector
space. In this section we see how these different notions are related.

    Let (V , A) be an affine variety. For x ∈ V we define Tx (V ) to be the
space of all linear maps α : A → k such that
α(fg) = f (x)α(g) + α(f )g(x), for all f , g ∈ A. We get
α(1) = α(12 ) = α(1) + α(1) so that α(1) = 0 and, by linearity,
α(c.1) = 0 for all c ∈ k . Let g1 , . . . , gr be k -algebra generators of A
and let Z be the k -span of g1 , . . . , gr . Now if α ∈ Tx (V ) vanishes on
f , g ∈ A then clearly α vanishes on fg.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   18 / 21
Thus we get that α vanishes on the algebra generated by g1 , . . . , gr ,
which is A, and hence α = 0. Thus restriction Tx (V ) → Z ∗ is injective,
where Z ∗ = Homk (Z , k ) is the dual of Z . Thus we get that dim Tx (X ) is
finite dimensional in fact we have:




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   19 / 21
Thus we get that α vanishes on the algebra generated by g1 , . . . , gr ,
which is A, and hence α = 0. Thus restriction Tx (V ) → Z ∗ is injective,
where Z ∗ = Homk (Z , k ) is the dual of Z . Thus we get that dim Tx (X ) is
finite dimensional in fact we have:

(1) If k [V ] may be generated by r elements then we have
dim Tx (V ) ≤ r for all x ∈ V .




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   19 / 21
Example We take V = An so that k [V ] = k [X1 , . . . , Xn ]. Let
                                                                      ∂f
a = (a1 , . . . , an ) ∈ V . We define αi : k [V ] → k by αi (f ) =        |a , the
                                                                      ∂Xi
formal partial derivative of f = f (X1 , . . . , Xn ) evaluated at the point a ∈ An ,
for 1 ≤ i ≤ n. It is easy to see that the functions α1 , . . . , αn are linearly
independent elements of Ta (An ) and hence form a basis.




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   20 / 21
   Suppose that φ : V → W is a morphism of varieties and that x ∈ V .
For each α ∈ Tx (V ) the composite α ◦ φ is an element of Tφ(x) (V ). In
this way we get a linear map dφx : Tx (V ) → Tφ(x) (W ), called the
differential of φ, defined by dφx (α) = α ◦ φ .




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   21 / 21
   Suppose that φ : V → W is a morphism of varieties and that x ∈ V .
For each α ∈ Tx (V ) the composite α ◦ φ is an element of Tφ(x) (V ). In
this way we get a linear map dφx : Tx (V ) → Tφ(x) (W ), called the
differential of φ, defined by dφx (α) = α ◦ φ .

Exercise 4. Let V be an affine variety and let U be a principal open set.
Show that, for x ∈ U, the inclusion map U → V induces an isomorphism
Tx (U) → Tx (V ).




Stephen Donkin (University of York)   Introduction to Linear Algebraic Groups   November 19, 2009   21 / 21

								
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