inflection points and canonical forms of non-singular cubic curves

inflection points and canonical forms of non-singular cubic curves

In this entry, we shall investigate the

inflection points

of

non-singular

complex

cubic curves

and show that the
equations of such curves can always be put in various

canonical forms

. The reason for lumping these two subjects
in a single entry is that, being related, it is more
efficient to develop them both at the same time than it would
be to treat either one in isolation — on the one hand,
putting the equation of the curve in a canonical form
simplifies calculations relating to inflection points whilst,
on the other hand, information about inflection points allows
one to construct canonical forms.

We shall consider our curves as

projective curves

and describe
them with homogenous equations; i.e. a curve is a locus in

ℂ

⁢

ℙ

2

where

F

⁢

(

x

,

y

,

z

)

=

0

for some third-
order homogenous

polynomial

F

. For the purpose at hand, the
term “inflection point” may be taken to mean a point on the
curve where the

tangent

intersects

the curve with

multiplicity

3 — a point on the curve will have this property if and only
if it is a zero of the

Hessian

.

We begin by presenting a crude canonical form. With further

refinement

, this will become the Weierstrass canonical form.

Theorem 1

.

Suppose

C

is a non-singular complex cubic curve and

P

a point
on that curve. Then we may choose homogenous coordiantes in which
the coordiantes of

P

are

(

0

,

0

,

1

)

and equation of the curve
looks as follows:

y

⁢

z

2

+

a

⁢

x

3

+

b

⁢

x

2

⁢

y

+

c

⁢

x

⁢

y

2

+

d

⁢

y

3

+

e

⁢

x

2

⁢

z

=

0

Proof.

The most general equation of a cubic is as follows:

A

⁢

X

3

+

B

⁢

X

2

⁢

Y

+

C

⁢

X

2

⁢

Z

+

D

⁢

X

⁢

Y

2

+

E

⁢

X

⁢

Y

⁢

Z

+

F

⁢

X

⁢

Z

2

+

G

⁢

Y

3

+

H

⁢

Y

2

⁢

Z

+

K

⁢

Y

⁢

Z

2

+

L

⁢

Z

3

=

0

Requiring that the curve

pass through

(

0

,

0

,

1

)

means that

L

=

0

. Since our curve is not

singular

, it has a tangent
at all points, in particular, a tangent through

P

. We
may ask that the equation of the tangent through

P

be

Y

=

0

, which requires that

F

=

0

. Since our curve is
not singular at

(

0

,

0

,

1

)

, both

F

and

K

cannot be zero
so, without loss of generality, we may take

K

=

1

.

We may then make the following linear transform:

X

=

x

Y

=

y

Z

=

z

-

E

⁢

x

/

2

-

H

⁢

y

/

2

Upon doing so, our equation assumes the form

y

⁢

z

2

+

a

⁢

x

3

+

b

⁢

x

2

⁢

y

+

c

⁢

x

⁢

y

2

+

d

⁢

y

3

+

e

⁢

x

2

⁢

z

=

0

where the coefficients

a

,

b

,

c

,

d

,

e

are defined as follows:

a

=

A

-

C

⁢

E

/

2

b

=

B

-

C

⁢

H

/

2

-

E

2

/

4

c

=

D

-

E

⁢

H

/

2

d

=

G

-

H

2

/

4

e

=

C

∎

Next, we turn our attention to the Hessian to begin our
investigation of inflection points.

Theorem 2

.

If

C

is a non-singular complex cubic curve with homogenous
equation

F

⁢

(

x

,

y

,

z

)

=

0

, then the Hessian of

F

does not equal
a

multiple

of

F

.

Proof.

By the foregoing

theorem

, we know that there exists
a system of

coordinates

in which

F

⁢

(

x

,

y

,

z

)

=

y

⁢

z

2

+

a

⁢

x

3

+

b

⁢

x

2

⁢

y

+

c

⁢

x

⁢

y

2

+

d

⁢

y

3

+

e

⁢

x

2

⁢

z

.

Let us examine the Hessian at the point

(

0

,

0

,

1

)

:

H

⁢

(

0

,

0

,

1

)

=

|

2

⁢

e

0

0

0

0

2

0

2

0

|

=

8

⁢

e

If

e

≠

0

, then the Hessian cannot me a multiple of

F

because

F

⁢

(

0

,

0

,

1

)

=

0

. Assume then that

e

=

0

. In that
case, we must have

a

≠

0

because, if

a

=

0

, then

F

would factor as a

product

of

y

and a quadratic polynomial,
which is impossible because the curve

C

is assumed not
to be singular.

1

1

Were

F

⁢

(

x

,

y

,

z

)

=

y

⁢

Q

⁢

(

x

,

y

,

z

)

for
some quadratic polynomial

Q

, then the

gradient

of

F

would vanish at the

intersection

of the line

y

=

0

and
the conic

Q

⁢

(

x

,

y

,

z

)

=

0

, so the

variety

described by

F

⁢

(

x

,

y

,

z

)

=

0

would be singular at the point(s) of
intersection.

Evaluating the Hessian on the line

y

=

0

,

H

⁢

(

x

,

0

,

z

)

=

|

6

⁢

a

⁢

x

2

⁢

b

⁢

x

0

2

⁢

b

⁢

x

2

⁢

c

⁢

x

2

⁢

z

0

2

⁢

z

0

|

=

24

⁢

a

⁢

x

⁢

z

2

Since

a

cannnot be zero, this does not vanish identically
and is clearly not a non-zero multiple of

F

⁢

(

x

,

0

,

z

)

.
∎

This theorem tells us that

H

⁢

(

x

,

y

,

z

)

=

0

is the equation of
a cubic curve distinct from

C

. Hence, we know that there
must be at least one and at most nine point of intersection
of these two curves, i.e. at least one and at most nine
inflection points. Knowing that inflection points exist, we
can improve our canonical form by choosing

P

as an
inflection point. With a little tidying up, this will give us
Weierstrass’ canonical form.

Theorem 3

.

Suppose

C

is a non-singular complex cubic curve and

P

a point
of inflection of on that curve. Then we may choose homogenous
coordinates in which the coordiantes of

P

are

(

0

,

0

,

1

)

and
equation of the curve looks as follows:

y

⁢

z

2

+

4

⁢

x

3

+

g

2

⁢

x

⁢

y

2

+

g

3

⁢

y

3

=

0

Proof.

By our earlier result, we know that there exist
coordinates in which the equation of the curve is

Y

⁢

Z

2

+

a

⁢

X

3

+

b

⁢

X

2

⁢

Y

+

c

⁢

X

⁢

Y

2

+

d

⁢

Y

3

+

e

⁢

X

2

⁢

Z

=

0

and the coordinates of

P

are

(

0

,

0

,

1

)

. By the
calculation made in the proof of the previous theorem,
we know that

H

⁢

(

0

,

0

,

1

)

=

8

⁢

e

. Since

P

is a point
of inflection, this implies that

e

=

0

.

As noted in the previous proof, we cannot have

a

=

0

because that would make our curve singular. Hence,
we can divide by

a

1

/

3

to make the following coordinate

transformation

:

X

=

x

-

b

⁢

y

/

3

Y

=

a

⁢

y

/

4

Z

=

z

After canceling a factor of

a

, our equation becomes

y

⁢

z

2

+

4

⁢

x

3

+

g

2

⁢

x

⁢

y

2

+

g

3

⁢

y

3

=

0

,

where

g

2

=

1

4

⁢

a

⁢

c

-

7

12

⁢

b

2

g

3

=

11

432

⁢

b

3

+

1

48

⁢

a

⁢

b

⁢

c

+

1

16

⁢

a

2

⁢

d

∎

Using this canonical form, it is easy to show that we
have the

maximum

number of inflection points possible.

Theorem 4

.

A non-singular complex cubic curve has nine inflection points.

Proof.

Given an inflection point, the foregoing theorem tells us
that we can choose a

coordinate system

in which that point
has coordinates

(

0

,

0

,

1

)

and the equation of the curve
assumes the form

F

⁢

(

x

,

y

,

z

)

=

0

where

F

⁢

(

x

,

y

,

z

)

=

y

⁢

z

2

+

4

⁢

x

3

+

g

2

⁢

x

⁢

y

2

+

g

3

⁢

y

3

.

Computing the Hessian, we have

H

⁢

(

x

,

y

,

z

)

=

|

24

⁢

x

2

⁢

g

2

⁢

y

0

2

⁢

g

2

⁢

y

2

⁢

g

2

⁢

x

+

6

⁢

g

3

⁢

y

2

⁢

z

0

2

⁢

z

2

⁢

y

|

=

96

⁢

g

2

⁢

x

2

⁢

y

+

288

⁢

g

3

⁢

x

⁢

y

2

-

8

⁢

g

2

2

⁢

y

3

-

96

⁢

x

⁢

z

2

Computing its gradient at

(

0

,

0

,

1

)

, we find that

(

∂

⁡

H

∂

⁡

x

,

∂

⁡

H

∂

⁡

y

,

∂

⁡

H

∂

⁡

z

)

⁢

(

0

,

0

,

1

)

=

(

-

96

,

0

,

0

)

.

For comparison, we have

(

∂

⁡

F

∂

⁡

x

,

∂

⁡

F

∂

⁡

y

,

∂

⁡

F

∂

⁡

z

)

⁢

(

0

,

0

,

1

)

=

(

0

,

1

,

0

)

.

Thus, we see that the curves

H

⁢

(

x

,

y

,

z

)

=

0

and

F

⁢

(

x

,

y

,

z

)

=

0

are both smooth at

(

0

,

0

,

1

)

and
intersect transversely, hence they have intersection
multiplicity 1 there. Because we could choose our
coordinates to place any inflection point at

(

0

,

0

,

1

)

, this means that every intersection of
our curve with its Hessian has intersection
multiplicity 1. Since both

F

and

H

are of
third order, Bezout’s theorem implies that there
are nine distinct intersection points, i.e. nine
distinct points of inflection.
∎

Knowing that there exist multiple inflection points,
we will now cast the equation in a form which places
two inflection points at priveleged locations.

Theorem 5

.

Suppose

C

is a non-singular complex cubic curve and

P

and

Q

are points of inflection of on that curve. Then we
may choose

homogeneous coordinates

in which the coordinates
of

P

are

(

1

,

0

,

0

)

, the coordinates of

Q

are

(

0

,

1

,

0

)

and the equation of the curve looks as follows:

x

2

⁢

y

+

y

2

⁢

x

+

z

3

+

a

⁢

x

⁢

y

⁢

z

=

0

Proof.

Since

P

is a point of its inflection, the tangent to

C

at

P

intersects

C

with multiplicity 3 at

P

. Since

C

is a cubic, this means that this tangent cannot
intersect

C

at any other points; in particular, it
cannot pass through

Q

. Likewise, the tangent through

Q

cannot pass through

P

. Hence, the tangent at

P

and the tangent at

Q

must intersect at some third point

R

which is not

collinear

with

P

and

Q

.

We may choose our coordinates so as to place

P

at

(

1

,

0

,

0

)

,

Q

at

(

0

,

1

,

0

)

and

R

at

(

0

,

0

,

1

)

. The equation of our
curve is

F

⁢

(

x

,

y

,

z

)

=

0

, where

F

⁢

(

x

,

y

,

z

)

=

a

⁢

x

⁢

y

⁢

z

+

b

⁢

z

3

+

c

⁢

y

⁢

z

2

+

d

⁢

y

2

⁢

z

+

e

⁢

y

3

+

f

⁢

x

⁢

z

2

+

g

⁢

x

2

⁢

z

+

h

⁢

x

⁢

y

2

+

j

⁢

x

2

⁢

y

+

k

⁢

x

3

.

Since

F

⁢

(

1

,

0

,

0

)

=

k

and

F

⁢

(

0

,

1

,

0

)

=

e

, we must have

k

=

0

and

e

=

0

in order for

P

and

Q

to lie on

C

. The equation of the line

P

⁢

R

is

y

=

0

.
Restricting to this line, we have

F

⁢

(

x

,

0

,

z

)

=

b

⁢

z

3

+

f

⁢

z

2

+

g

⁢

x

2

⁢

z

.

Since

P

is an inflection point, the curve must have
third order contact with this

tangent line

, the

restriction

of

F

to this line should have a triple root at

P

,
i.e. should be a multiple of

z

3

. Hence,

f

=

0

and

g

=

0

.

Likewise, the equation of the line

Q

⁢

R

is

x

=

0

.
Restricting to this line, we have

F

⁢

(

0

,

y

,

z

)

=

b

⁢

z

3

+

c

⁢

y

⁢

z

2

+

d

⁢

y

2

⁢

z

.

Again, since

Q

is an inflection point, we must have
a triple root, so this quantity should be a multiple
of

z

3

, so we must have

c

=

0

and

d

=

0

.

Summarizing our progress so far, we have found that

F

⁢

(

x

,

y

,

z

)

=

a

⁢

x

⁢

y

⁢

z

+

b

⁢

z

3

+

h

⁢

x

⁢

y

2

+

j

⁢

x

2

⁢

y

.

Next, we note that

h

cannot be zero because that
would imply that

C

had a singularity at

Q

.
Likewise, we must not have

j

be zero because that
would mean a singularity at

P

. Also, were

b

=

0

,
we would have

F

⁢

(

x

,

y

,

z

)

=

x

⁢

y

⁢

(

z

+

h

⁢

y

+

j

⁢

x

)

which would
have a singularity at

R

. Thus, by rescaling

x

,

y

, and

z

if necessarry, we can take

h

=

j

=

b

=

1

, so our equation assumes the form

F

⁢

(

x

,

y

,

z

)

=

a

⁢

x

⁢

y

⁢

z

+

z

3

+

x

⁢

y

2

+

x

2

⁢

y

.

∎

Upon examining the form of the equation just derived
more closely, we learn an important geometric fact:

Theorem 6

.

Suppose that

C

is a non-singular complex cubic curve
and that

P

and

Q

are inflection points of

C

. Then
the line

P

⁢

Q

intersects

C

in a third point

R

, which
is also an inflection point.

Proof.

By the foregoing result, we know that we can write the
equation for

C

as

F

⁢

(

x

,

y

,

z

)

=

0

with

F

⁢

(

x

,

y

,

z

)

=

x

⁢

y

2

+

x

2

⁢

y

+

z

3

+

a

⁢

x

⁢

y

⁢

z

with

P

located at

(

1

,

0

,

0

)

and

Q

located at

(

0

,

1

,

0

)

.
The line

P

⁢

Q

then has equation

z

=

0

. Restricting

F

to this line,

F

⁢

(

x

,

y

,

0

)

=

x

⁢

y

2

+

x

2

⁢

y

=

x

⁢

(

x

+

y

)

⁢

y

, so

R

, the third point of intersection of

C

with

P

⁢

Q

, has
coordinates

(

1

,

-

1

,

0

)

.

Next, we compute the Hessian determinant at

R

:

H

⁢

(

1

,

-

1

,

0

)

=

|

-

1

0

-

a

0

1

a

-

a

a

0

|

=

0

Since it equals zero,

R

is an inflection point.
∎

Since our curve is of the third order, a line cannot
intersect more than three points. Hence, we see that
a line passing through two inflection points of a
non-singular complex cubic plane curve must pass through
exactly three inflection points. As it turns out, this
observation, together with the fact that there are exactly
nine inflection points suffices to determine the locations
of the inflection points up to

collineation

. However, rather
than pursuing this line of reasoning here, we will instead
cast the equation of the curve in a form which makes it easy
to locate all the inflection points and makes the

symmetry

of the curve apparent.

Theorem 7

.

Suppose

C

is a non-singular complex cubic curve. Then we
may choose homogeneous coordinates the equation of the curve
assumes the form

x

3

+

y

3

+

z

3

+

6

⁢

m

⁢

x

⁢

y

⁢

z

=

0

.

Proof.

We know that there exists a

straight line

L

which intersects

C

in exactly three points, all of which are inflection points
of

C

. Choose a homogenous coordinate system in which the
equation of

L

is

Z

=

0

. Since the three points of intersection
of

C

with

L

are distinct, we may place them at any three
locations; we shall choose

(

1

,

-

1

,

0

)

,

(

1

,

-

ρ

,

0

)

, and

(

1

,

-

ρ

2

,

0

)

, where

ρ

=

(

-

1

+

-

3

)

/

2

is a

primitive

cube root

of unity.

With this choice, the equation of

C

becomes

X

3

+

Y

3

+

D

⁢

X

2

⁢

Z

+

E

⁢

X

⁢

Y

⁢

Z

+

F

⁢

Y

2

⁢

Z

+

B

⁢

X

⁢

Z

2

+

C

⁢

Y

⁢

Z

2

+

A

⁢

Z

3

=

0

.

By making the change of variables

x

=

X

-

F

⁢

Z

/

3

y

′

=

Y

-

D

⁢

Z

/

3

we simplify the equation of the curve to

x

3

+

y

3

+

E

⁢

x

⁢

y

⁢

Z

+

b

⁢

x

⁢

Z

2

+

c

⁢

y

⁢

Z

2

+

a

⁢

Z

3

=

0

.

where

a

=

A

-

B

⁢

D

3

+

2

⁢

D

3

27

-

C

⁢

F

3

+

D

⁢

E

⁢

F

9

+

2

⁢

F

3

27

b

=

B

-

D

2

3

-

E

⁢

F

3

c

=

C

-

D

⁢

E

3

-

F

2

3

Note that this transformation leaves the coordinates of the three
points of intersection of

L

and

C

unchanged.

Next, we impose the requirement that the three points lying on

L

be
inflecton points of

C

. Setting

Z

to zero, the Hessian becomes

|

6

⁢

x

0

E

⁢

y

0

6

⁢

y

E

⁢

x

E

⁢

y

E

⁢

x

2

⁢

b

⁢

x

+

2

⁢

c

⁢

y

|

=

72

⁢

b

⁢

x

2

⁢

y

+

72

⁢

c

⁢

x

⁢

y

2

-

6

⁢

E

2

⁢

(

x

3

+

y

3

)

In order for the three points intersection of

L

and

C

to be
inflection points, we must have this be a multiple of

x

3

+

y

3

.
This requires that

b

and

c

both be zero.

Were

a

zero as well, our cubic would have a singularity at

(

0

,

0

,

1

)

.
Since

C

is assumed to not be singular, that means that

a

≠

0

, so
we can make the further transform

z

=

a

1

/

3

⁢

Z

m

=

a

-

1

/

3

⁢

E

/

6

to put the equation for

C

in the form

x

3

+

y

3

+

z

3

+

6

⁢

m

⁢

x

⁢

y

⁢

z

=

0

.
∎

While we have shown that we can transform the equation of any
non-singular curve into our canonical form, there remains the
possibility that a curve whose equation is expressible in
our canonical form may be singular or degenerate. We will
now examine this possibility.

Theorem 8

.

A equation

x

3

+

y

3

+

z

3

+

6

⁢

m

⁢

x

⁢

y

⁢

z

=

0

describes a singular
curve if and only if

8

⁢

m

3

+

1

=

0

.

Proof.

The curve described by the equation

f

⁢

(

x

,

y

,

z

)

=

0

is singular
if and only if there exists a point on the curve at which all
three

partial derivatives

of

f

go zero. Taking

derivatives

and doing a little bit of

algebraic

manipulation, this means
that, for the curve to be singular, there must be a non-trivial
solution to the system of equations

x

2

=

-

2

⁢

m

⁢

y

⁢

z

y

2

=

-

2

⁢

m

⁢

x

⁢

z

z

2

=

-

2

⁢

m

⁢

x

⁢

y

.

Multiplying the last three equations together, we obtain

x

2

⁢

y

2

⁢

z

2

=

-

8

⁢

m

3

⁢

x

2

⁢

y

2

⁢

z

2

=

0

, or

(

8

⁢

m

3

+

1

)

⁢

x

2

⁢

y

2

⁢

z

2

=

0

. If

8

⁢

m

3

+

1

≠

0

, the only
way for this to be satisfied is to have either

x

=

0

or

y

=

0

or

z

=

0

. However, suppose that

x

=

0

. Then, by the last two
equations, we would also have

y

=

0

and

z

=

0

, so the solution
would be

trivial

. Likewise, if

y

=

0

, then it follows that

x

=

0

and

z

=

0

; and, if

z

=

0

, it follows that

x

=

0

and

y

=

0

.
Hence, when

8

⁢

m

3

+

1

≠

0

, we only have the trivial solution,
so the curve is non-singular.

We will finish by explicitly showing that the curve degenerates
when

8

⁢

m

3

+

1

=

0

, i.e. when

m

=

-

1

/

2

or

m

=

-

ρ

/

2

or

m

=

-

ρ

2

/

2

. (As before,

ρ

is a primitive cube
root of unity.) In each of these cases, we can factor our
equation into a product of three linear terms:

x

3

+

y

3

+

z

3

-

3

⁢

x

⁢

y

⁢

z

=

(

x

+

y

+

z

)

⁢

(

x

+

ρ

⁢

y

+

ρ

2

⁢

z

)

⁢

(

x

+

ρ

2

⁢

y

+

ρ

⁢

z

)

x

3

+

y

3

+

z

3

-

3

⁢

ρ

⁢

x

⁢

y

⁢

z

=

ρ

2

⁢

(

x

+

y

+

ρ

⁢

z

)

⁢

(

x

+

ρ

⁢

y

+

z

)

⁢

(

ρ

⁢

x

+

y

+

z

)

x

3

+

y

3

+

z

3

-

3

⁢

ρ

2

⁢

x

⁢

y

⁢

z

=

ρ

⁢

(

x

+

y

+

ρ

2

⁢

z

)

⁢

(

x

+

ρ

2

⁢

y

+

z

)

⁢

(

ρ

2

⁢

x

+

y

+

z

)

Thus, when

8

⁢

m

3

+

1

=

0

, our curve degenerates into a

triangle

(whose vertices are the singularities).
∎

Having obtained this canonical form, it is quite easy to exhibit all
nine inflection points.

Theorem 9

.

Given a non-singular complex cubic plane curve, there
exists a coordinate system in which the inflection
points of that curve have the following coordinates:

(

1

,

-

ρ

,

0

)

(

1

,

-

1

,

0

)

(

1

,

-

ρ

2

,

0

)

(

-

ρ

,

0

,

1

)

(

-

1

,

0

,

1

)

(

-

ρ

2

,

0

,

1

)

(

0

,

1

,

-

ρ

)

(

0

,

1

,

-

1

)

(

0

,

1

,

-

ρ

2

)

.

As previously,

ρ

denotes a primitive third

root of unity

.

Proof.

For convenience, set

f

=

x

3

+

y

3

+

z

3

+

6

⁢

m

⁢

x

⁢

y

⁢

z

and
let

h

be

1

/

216

times the Hessian of

f

.
Computing, we have

h

=

1

216

⁢

|

6

⁢

x

6

⁢

m

⁢

z

6

⁢

m

⁢

y

6

⁢

m

⁢

z

6

⁢

y

6

⁢

m

⁢

x

6

⁢

m

⁢

y

6

⁢

m

⁢

x

6

⁢

z

|

=

(

2

⁢

m

3

+

1

)

⁢

x

⁢

y

⁢

z

-

m

2

⁢

(

x

3

+

y

3

+

z

3

)

Thus the Hessian is of the same form, but with

m

replaced by

-

(

2

⁢

m

3

+

1

)

/

6

⁢

m

2

. Next, we form two

combinations

of

f

ande

h

:

m

2

⁢

f

+

h

=

(

8

⁢

m

3

+

1

)

⁢

x

⁢

y

⁢

z

(

2

⁢

m

3

+

1

)

⁢

f

-

6

⁢

m

⁢

h

=

(

8

⁢

m

3

+

1

)

⁢

(

x

3

+

y

3

+

z

3

)

As we have just seen, the condition for the curve not to be
singular is exactly that

8

⁢

m

3

+

1

≠

0

, so we may cancel
to conclude that

x

3

+

y

3

+

z

3

=

0

and

x

⁢

y

⁢

z

=

0

.

The latter equation will be satisfied if either

x

=

0

or

y

=

0

or

z

=

0

. When

x

=

0

, the former equation reduces
to

y

3

+

z

3

=

0

, which gives us the solutions

(

0

,

1

,

-

ρ

)

,

(

0

,

1

,

-

1

)

,

(

0

,

1

,

-

ρ

2

)

Likewise, when

y

=

0

, it reduces to

x

3

+

z

3

=

0

with
the solutions

(

-

ρ

,

0

,

1

)

,

(

-

1

,

0

,

1

)

,

(

-

ρ

2

,

0

,

1

)

and, when

z

=

0

, it reduces to

x

3

+

y

3

=

0

, with
solutions

(

1

,

-

ρ

,

0

)

,

(

1

,

-

1

,

0

)

,

(

1

,

-

ρ

2

,

0

)

.

∎

Title

inflection points and canonical forms of non-singular cubic curves

Canonical name

InflectionPointsAndCanonicalFormsOfNonsingularCubicCurves

Date of creation

2014-01-16 16:32:44

Last modified on

2014-01-16 16:32:44

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

42

Author

rspuzio (6075)

Entry type

Feature

Classification

msc 14N10

Classification

msc 14H52

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