Singularities and dimension of varieties
In all the examples below, the variety is irreducible and the given polynomials
generate its ideal. V denotes the variety in each example.
Example 1. y 2 − x3 = 0 in A2 (C).
J = (−3x2
2y)
The points where rankJP = 0 are the solutions of y 2 − x3 =
−3x2 = 2y = 0, so x = y = 0. Therefore rankJP = 0 and
dim TP V = 2 at P = (0, 0), and rankJP = 1 and dim TP V = 1
at all other points of V , since rankJP = 1 can only be 0 or 1,
because JP is a 1 × 2 matrix.
Hence dim V = 1, Sing V = {(0, 0)}.
Example 2. y 2 − x3 − x2 = 0 in A2 (C).
J = (−3x2 − 2x 2y)
The points where rankJP = 0 are the solutions y 2 − x3 − x2 =
−3x2 −2x = 2y = 0, so necessarily y = 0. −3x2 −2x = 0 gives
x = 0 or x = −2/3, but (x, y) = (−2/3, 0) does not satisfy
y 2 − x3 − x2 = 0. Therefore rankJP = 0 and dim TP V = 2
at P = (0, 0), and rankJP = 1 and dim TP V = 1 at all other
points of V . Hence dim V = 1 and Sing V = {(0, 0)}.
Example 3. −x4 + 2x2 y 2 + y 4 − 2x2 + 2y 2 − 1 = 0 in A2 (C).
J = (−4x3 −4x+4xy 2
4x2 y+4y 3 +4y) = (−4x(1+x2 −y 2 ) 4y(1+x2 +y 2 ))
Let f = −x4 + 2x2 y 2 + y 4 − 2x2 + 2y 2 − 1. The points where rankJP = 0
are the points where f = fx = fy = 0. By the factorisation of fx and fy , we
need to consider four cases.
1. x = y = 0, f (0, 0) = −1, so this is not a solution.
2. x = 1 + x2 + y 2 = 0, so y = ±i, but f (0, ±i) = −2,
so these are not solutions either.
3. 1 + x2 − y 2 = y = 0, so x = ±i. f (±i, 0) = 0, so
(x, y) = (i, 0) and (x, y) = (−i, 0) are solutions.
4. 1+x2 −y 2 = 1+x2 +y 2 = 0, then from the difference
of the two equations we get y 2 = 0, so y = 0 and we
get the same solutions as in the previous case.
Therefore rankJP = 0 and dim TP V = 2 at P = (±i, 0), and rankJP = 1
and dim TP V = 1 at all other points of V . Hence dim V = 1 and Sing V =
{(i, 0), (−i, 0)}.
Example 4. x2 + y 2 − z 2 = 0 in A3 (C).
J = (2x 2y − 2z)
The points where rankJP = 0 are the solutions of
x2 + y 2 − z 2 = 2x = 2y = −2z = 0, so x = y =
z = 0. Therefore rankJP = 0 and dim TP V = 3 at
P = (0, 0, 0), and rankJP = 1 and dim TP V = 2 at
all other points of V .
Hence dim V = 2 and Sing V = {(0, 0, 0)}.
Example 5. x2 + y 2 + z 2 − 1 = 0 in A3 (C).
J = (2x 2y 2z)
The points where rankJP = 0 are the solutions of
x2 + y 2 + z 2 − 1 = 2x = 2y = 2z = 0, but x = y =
z = 0 do not satisfy x2 + y 2 + z 2 − 1 = 0, so there
is no solution. Therefore there is no P ∈ V where
rankJP = 0, so rankJP = 1 and dim TP V = 2 at
every P ∈ V .
Hence dim V = 2 and Sing V = ∅.
Example 6. x2 + y 2 + z 2 − 1 = x2 + y 2 + y = 0 in A3 (C).
2x
2y
2z
J=
.
2x 2y + 1 0
Let f = x2 + y 2 + z 2 − 1 and g = x2 + y 2 + y. The points where rankJP = 0
are the points where all the entries of J vanish and f = g = 0. As 2y and
2y + 1 cannot both be 0, there are no such points.
Next we consider the points where rankJP ≤ 1,
these are the points of V where the 2 × 2 minors
of J vanish, i. e., 2x = −4xz = −2(2y + 1)z =
f = g = 0. Substituting x = 0 into g gives y 2 +
y = 0, so y = 0 or y = −1. Hence 2y + 1 6= 0
and −2(2y + 1)z = 0 implies z = 0. There are
two possibilities, (x, y, z) = (0, 0, 0) and (x, y, z) =
(0, −1, 0). f (0, 0, 0) = −1, so (0, 0, 0) ∈
/ V , but
(0, −1, 0) ∈ V and all the 2 × 2 minors are 0 there.
Therefore rankJP = 1 and dim TP V = 2 at P = (0, −1, 0), and rankJP = 2
and dim TP V = 1 at all other points of V . Hence dim V = 1 and Sing V =
{(0, −1, 0)}.