128
LECTURE 13.
TANGENT SPACE
Proof.
Obviously, it suffices to find an open subset
U
of
X
where
dim
K
T
(
X
)
x
=
dim
X
for all
x
∈
U
. Replacing
X
by an open affine set, we may assume that
X
is
isomorphic to an open subset of a hypersurface
V
(
F
)
⊂
A
n
(
K
)
for some irreducible
polynomial
F
(Theorem
4.10
of Lecture 4).
This shows that we may assume that
X
=
V
(
F
)
. For any
x
∈
X
, the tangent space
T
(
X
)
x
is given by one equation
∂F
∂Z
1
(
x
)
b
1
+
. . .
+
∂F
∂Z
n
(
x
)
b
n
= 0
.
Clearly, its dimension is equal to
n

1 = dim
X
unless all the coefficients are zeroes.
The set of common zeroes of the polynomials
∂F
∂Z
i
is a closed subset of
A
n
(
K
)
contained
in each hypersurface
V
(
∂F
∂Z
i
)
. Obviously,
∂F
∂Z
i
∈
(
F
)
unless it is equal to zero (compare
the degrees). Now the assertion follows from Lemma
13.13
.
Obviously, the assertion of the previous theorem is not true for a reducible set. To
see this it is sufficient to consider the union of two sets of different dimension. It is
easy to modify the statement to extend it to the case of reducible sets.
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 Spring '09
 wei
 Algebra, Geometry, Vector Space, Empty set, Manifold, Euclidean space

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