proof of the weak Nullstellensatz

proof of the weak Nullstellensatz

Let

K

be an algebraically closed field, let

n

≥

0

, and let

I

be
an ideal in the

polynomial ring

K

⁢

[

x

1

,

…

,

x

n

]

. Suppose

I

is
strictly smaller than

K

⁢

[

x

1

,

…

,

x

n

]

. Then

I

is contained in a

maximal ideal

M

of

K

⁢

[

x

1

,

…

,

x

n

]

(note that we don’t have to
accept Zorn’s lemma to find such an

M

, since

K

⁢

[

x

1

,

…

,

x

n

]

is

Noetherian

by Hilbert’s basis theorem), and the

quotient ring

L

=

K

⁢

[

x

1

,

…

,

x

n

]

/

M

is a field. We view

K

as a

subfield

of

L

via the

natural
homomorphism

K

↪

L

, and we denote the images of

x

1

,

…

,

x

n

in

L

by

x

¯

1

,

…

,

x

¯

n

. Let

{

t

1

,

…

,

t

m

}

be a transcendence basis of

L

over

K

; it is
finite since

L

is

finitely generated

as a

K

-

algebra

. Now

L

is
an

algebraic extension

of

K

⁢

(

t

1

,

…

,

t

m

)

. By multiplying the

minimal polynomial

of

x

¯

i

over

K

⁢

(

t

1

,

…

,

t

m

)

by a
suitable element of

K

⁢

[

t

1

,

…

,

t

m

]

for each

i

, we obtain
non-zero

polynomials

f

i

∈

K

⁢

[

t

1

,

…

,

t

m

]

⁢

[

X

]

with the
property that

f

i

⁢

(

x

¯

i

)

=

0

in

L

:

f

i

=

c

i

,

0

+

c

i

,

1

X

+

⋯

+

c

i

,

d

i

X

d

i

(

1

≤

i

≤

n

)

for certain

integers

d

i

>

0

and polynomials

c

i

,

j

∈

K

⁢

[

t

1

,

…

,

t

m

]

with

c

i

,

d

i

≠

0

. Since

K

is

algebraically
closed

(hence

infinite

), we can choose

u

1

,

…

,

u

n

∈

K

such that

c

i

,

d

i

⁢

(

u

1

,

…

,

u

m

)

≠

0

for all

i

. We define a

homomorphism

ϕ

:

K

⁢

[

t

1

,

…

,

t

m

]

⟶

K

by taking

ϕ

to be the

identity

on

K

and sending

t

j

to

u

j

.
Let

N

be the kernel of this homomorphism. Then

ϕ

can be
extended to the

localization

K

⁢

[

t

1

,

…

,

t

m

]

N

of

K

⁢

[

t

1

,

…

,

t

m

]

. Since

c

i

,

d

i

∉

N

for all

i

, the

x

¯

i

are integral over this ring. Since

K

is algebraically
closed, the

extension theorem

for

ring homomorphisms

implies that

ϕ

can be extended to a homomorphism

ϕ

:

(

K

⁢

[

t

1

,

…

,

t

m

]

N

)

⁢

[

x

¯

1

,

…

,

x

¯

n

]

=

L

⟶

K

.

Because

L

is an

extension field

of

K

and

ϕ

is the identity on

K

, we see that

ϕ

is actually an

isomorphism

, that

m

=

0

, and
that

N

is the

zero ideal

of

K

. Now let

a

1

=

ϕ

⁢

(

x

¯

1

)

,

…

,

a

n

=

ϕ

⁢

(

x

¯

n

)

. Then for all polynomials

f

in the
ideal

I

we started with, the fact that

f

∈

M

implies

f

⁢

(

a

1

,

…

,

a

n

)

=

ϕ

⁢

(

f

⁢

(

x

1

,

…

,

x

n

)

+

M

)

=

0

.

We conclude that the

zero set

V

⁢

(

I

)

of

I

is not empty.

Title

proof of the weak Nullstellensatz

Canonical name

ProofOfTheWeakNullstellensatz

Date of creation

2013-03-22 15:27:43

Last modified on

2013-03-22 15:27:43

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

4

Author

pbruin (1001)

Entry type

Proof

Classification

msc 13A10

Classification

msc 13A15

Generated on Fri Feb 9 17:20:40 2018 by

LaTeXML