Advanced Algebra II Apr. 27, 2007

Definition

For a Noetherian topological space X , the dimension of X , denoted dimX ,

is defined to be the supremum (=maximum) of the length of chain of closed subvarieties.

Definition

For a Noetherian topological space X , the dimension of X ,

denoted dimX , is defined to be the supremum (=maximum) of the length of chain of closed subvarieties.

For an affine variety XinA^{n}, the polynomial functions A restrict to
X is nothing but the homomorphism π : A → A/I (X ).

The ring A/I (X ) is called the coordinate ring of X , denoted A(X ). One can recover the geometry of X from A(X ) by considering Spec(A(X )), which consist of prime ideals in A(X ).

One can give the Zariski topology on Spec(A(X )) which is closely related to the Zariski topology on X .

This is actually the construction of affine scheme. And affine variety can be viewed as a nice affine scheme.

For an affine variety XinA^{n}, the polynomial functions A restrict to
X is nothing but the homomorphism π : A → A/I (X ). The ring
A/I (X ) is called the coordinate ring of X , denoted A(X ).

One can recover the geometry of X from A(X ) by considering Spec(A(X )), which consist of prime ideals in A(X ).

One can give the Zariski topology on Spec(A(X )) which is closely related to the Zariski topology on X .

This is actually the construction of affine scheme. And affine variety can be viewed as a nice affine scheme.

For an affine variety XinA^{n}, the polynomial functions A restrict to
X is nothing but the homomorphism π : A → A/I (X ). The ring
A/I (X ) is called the coordinate ring of X , denoted A(X ).

One can recover the geometry of X from A(X ) by considering Spec(A(X )), which consist of prime ideals in A(X ).

One can give the Zariski topology on Spec(A(X )) which is closely related to the Zariski topology on X .

This is actually the construction of affine scheme. And affine variety can be viewed as a nice affine scheme.

For an affine variety XinA^{n}, the polynomial functions A restrict to
X is nothing but the homomorphism π : A → A/I (X ). The ring
A/I (X ) is called the coordinate ring of X , denoted A(X ).

One can recover the geometry of X from A(X ) by considering Spec(A(X )), which consist of prime ideals in A(X ).

^{n}, the polynomial functions A restrict to
X is nothing but the homomorphism π : A → A/I (X ). The ring
A/I (X ) is called the coordinate ring of X , denoted A(X ).

Exercise

The coordinate ring of an affine variety is a domain and a finitely generated k-algebra.

Conversely, a domain which is a finitely generated k-algebra is a coordinate ring of an affine variety.

Exercise

The coordinate ring of an affine variety is a domain and a finitely generated k-algebra.

Conversely, a domain which is a finitely generated k-algebra is a coordinate ring of an affine variety.

One can also similarly define the Krull dimension or simply

dimension to be the supremum of length of chain of prime ideals of a ring.

It’s easy to see that for an algebraic set X , then
dimX = dimA(X ).
However, it’s not trivial to prove that dimA^{n}= n.

One can also similarly define the Krull dimension or simply

dimension to be the supremum of length of chain of prime ideals of a ring.

It’s easy to see that for an algebraic set X , then dimX = dimA(X ).

However, it’s not trivial to prove that dimA^{n}= n.

One can also similarly define the Krull dimension or simply

dimension to be the supremum of length of chain of prime ideals of a ring.

It’s easy to see that for an algebraic set X , then dimX = dimA(X ).

However, it’s not trivial to prove that dimA^{n}= n.

In this section, we are going to explore dimension theory a little bit.

Let first recall the definition.

Definition

Let R be a ring and p ∈ Spec(R).

We define

ht(p) := sup{n|p_{n}( ... ( p0 = p}.
And for an ideal I C R, we define

ht(I ) := inf{ht(p)|I ⊂ p}. We define

dimR := sup{ht(p)|p ∈ Spec(R)}.

In this section, we are going to explore dimension theory a little bit.

Let first recall the definition.

Definition

Let R be a ring and p ∈ Spec(R).

We define

ht(p) := sup{n|p_{n}( ... ( p0 = p}.

And for an ideal I C R, we define

ht(I ) := inf{ht(p)|I ⊂ p}. We define

dimR := sup{ht(p)|p ∈ Spec(R)}.

In this section, we are going to explore dimension theory a little bit.

Let first recall the definition.

Definition

Let R be a ring and p ∈ Spec(R).

We define

ht(p) := sup{n|p_{n}( ... ( p0 = p}.

And for an ideal I C R, we define

ht(I ) := inf{ht(p)|I ⊂ p}.

We define

dimR := sup{ht(p)|p ∈ Spec(R)}.

In this section, we are going to explore dimension theory a little bit.

Let first recall the definition.

Definition

Let R be a ring and p ∈ Spec(R).

We define

ht(p) := sup{n|p_{n}( ... ( p0 = p}.

And for an ideal I C R, we define

ht(I ) := inf{ht(p)|I ⊂ p}.

We define

dimR := sup{ht(p)|p ∈ Spec(R)}.

Example

Let R := k[x_{1}, ..., x_{n}] with k algebraically closed. We will see later
that dimR is n.

Let p be a prime ideal, then V(p) defines a variety in A^{n}_{k}.

Then ht(p) is nothing but the codimension of V(p) in A^{n}_{k} because
there is a one-to-one correspondence between prime ideals and
subvarieties.

Let I C k[x1, ..., x_{n}] be an ideal, then V(I ) is not necessarily
irreducible.

We usually define the dimension of V(I ) to be the dimension of irreducible component of maximal dimension.

That is, we are looking for

dimV(I ) := max{dimY | Y is an irreducible component in V(I )}. Hence the codimension of V(I ) corresponds to inf{ht(p)|I ⊂ p}.

Example

Let R := k[x_{1}, ..., x_{n}] with k algebraically closed. We will see later
that dimR is n.

Let p be a prime ideal, then V(p) defines a variety in A^{n}_{k}.

Then ht(p) is nothing but the codimension of V(p) in A^{n}_{k} because
there is a one-to-one correspondence between prime ideals and
subvarieties.

Let I C k[x1, ..., x_{n}] be an ideal, then V(I ) is not necessarily
irreducible.

We usually define the dimension of V(I ) to be the dimension of irreducible component of maximal dimension.

That is, we are looking for

dimV(I ) := max{dimY | Y is an irreducible component in V(I )}. Hence the codimension of V(I ) corresponds to inf{ht(p)|I ⊂ p}.

Example

Let R := k[x_{1}, ..., x_{n}] with k algebraically closed. We will see later
that dimR is n.

Let p be a prime ideal, then V(p) defines a variety in A^{n}_{k}.

Then ht(p) is nothing but the codimension of V(p) in A^{n}_{k} because
there is a one-to-one correspondence between prime ideals and
subvarieties.

Let I C k[x1, ..., x_{n}] be an ideal, then V(I ) is not necessarily
irreducible.

We usually define the dimension of V(I ) to be the dimension of irreducible component of maximal dimension.

That is, we are looking for

dimV(I ) := max{dimY | Y is an irreducible component in V(I )}. Hence the codimension of V(I ) corresponds to inf{ht(p)|I ⊂ p}.

Example

Let R := k[x_{1}, ..., x_{n}] with k algebraically closed. We will see later
that dimR is n.

Let p be a prime ideal, then V(p) defines a variety in A^{n}_{k}.

^{n}_{k} because
there is a one-to-one correspondence between prime ideals and
subvarieties.

Let I C k[x1, ..., x_{n}] be an ideal, then V(I ) is not necessarily
irreducible.

That is, we are looking for

Example

Let R := k[x_{1}, ..., x_{n}] with k algebraically closed. We will see later
that dimR is n.

Let p be a prime ideal, then V(p) defines a variety in A^{n}_{k}.

^{n}_{k} because
there is a one-to-one correspondence between prime ideals and
subvarieties.

Let I C k[x1, ..., x_{n}] be an ideal, then V(I ) is not necessarily
irreducible.

That is, we are looking for

Example

Let R := k[x_{1}, ..., x_{n}] with k algebraically closed. We will see later
that dimR is n.

Let p be a prime ideal, then V(p) defines a variety in A^{n}_{k}.

^{n}_{k} because
there is a one-to-one correspondence between prime ideals and
subvarieties.

Let I C k[x1, ..., x_{n}] be an ideal, then V(I ) is not necessarily
irreducible.

That is, we are looking for

dimV(I ) := max{dimY | Y is an irreducible component in V(I )}.

Hence the codimension of V(I ) corresponds to inf{ht(p)|I ⊂ p}.

Example

Let R := k[x_{1}, ..., x_{n}] with k algebraically closed. We will see later
that dimR is n.

Let p be a prime ideal, then V(p) defines a variety in A^{n}_{k}.

^{n}_{k} because
there is a one-to-one correspondence between prime ideals and
subvarieties.

Let I C k[x1, ..., x_{n}] be an ideal, then V(I ) is not necessarily
irreducible.

That is, we are looking for

dimV(I ) := max{dimY | Y is an irreducible component in V(I )}.

Hence the codimension of V(I ) corresponds to inf{ht(p)|I ⊂ p}.

One has the following property immediately by the correspondence we’ve been built up.

Proposition
1. htp = dimR_{p}.

2. dimR/I + ht(I ) ≤ dimR.

Theorem

Let R be a finitely generated domain over a field k. Then
tr.d._{k}R = dimR.

(where tr.d._{k}R := tr.d._{k}F , where F is the quotient field of R.)

We first claim that dimR ≤ tr.d.kR.

To see this, it suffices to show that if p ( q ∈ Spec(R), then
tr.d._{k}R/p
tr.d.kR/q.

Let {β1, ..., βr} be a transcendental basis of R/q.

Then it lifts to {α1, ..., αr} which is algebraically independent in R/p.

This is because there is a surjective homomorphism ϕ : R/p → R/q.

If there is an algebraic relation among {α_{1}, ..., α_{r}} then it gives an
relation among {β1, ..., βr} via the homomorphism ϕ, which is
absurd.

Hence we have shown that tr.d._{k}R/p ≥ tr.d._{k}R/q.

We first claim that dimR ≤ tr.d.kR.

To see this, it suffices to show that if p ( q ∈ Spec(R), then
tr.d._{k}R/p
tr.d.kR/q.

Let {β1, ..., βr} be a transcendental basis of R/q.

Then it lifts to {α1, ..., αr} which is algebraically independent in R/p.

This is because there is a surjective homomorphism ϕ : R/p → R/q.

If there is an algebraic relation among {α_{1}, ..., α_{r}} then it gives an
relation among {β1, ..., βr} via the homomorphism ϕ, which is
absurd.

Hence we have shown that tr.d._{k}R/p ≥ tr.d._{k}R/q.

We first claim that dimR ≤ tr.d.kR.

To see this, it suffices to show that if p ( q ∈ Spec(R), then
tr.d._{k}R/p
tr.d.kR/q.

Let {β1, ..., βr} be a transcendental basis of R/q.

Then it lifts to {α1, ..., αr} which is algebraically independent in R/p.

This is because there is a surjective homomorphism ϕ : R/p → R/q.

If there is an algebraic relation among {α_{1}, ..., α_{r}} then it gives an
relation among {β1, ..., βr} via the homomorphism ϕ, which is
absurd.

Hence we have shown that tr.d._{k}R/p ≥ tr.d._{k}R/q.

We first claim that dimR ≤ tr.d.kR.

To see this, it suffices to show that if p ( q ∈ Spec(R), then
tr.d._{k}R/p
tr.d.kR/q.

Let {β1, ..., βr} be a transcendental basis of R/q.

Then it lifts to {α1, ..., αr} which is algebraically independent in R/p.

This is because there is a surjective homomorphism ϕ : R/p → R/q.

_{1}, ..., α_{r}} then it gives an
relation among {β1, ..., βr} via the homomorphism ϕ, which is
absurd.

Hence we have shown that tr.d._{k}R/p ≥ tr.d._{k}R/q.

We first claim that dimR ≤ tr.d.kR.

To see this, it suffices to show that if p ( q ∈ Spec(R), then
tr.d._{k}R/p
tr.d.kR/q.

Let {β1, ..., βr} be a transcendental basis of R/q.

Then it lifts to {α1, ..., αr} which is algebraically independent in R/p.

This is because there is a surjective homomorphism ϕ : R/p → R/q.

_{1}, ..., α_{r}} then it gives an
relation among {β1, ..., βr} via the homomorphism ϕ, which is
absurd.

Hence we have shown that tr.d._{k}R/p ≥ tr.d._{k}R/q.

We first claim that dimR ≤ tr.d.kR.

To see this, it suffices to show that if p ( q ∈ Spec(R), then
tr.d._{k}R/p
tr.d.kR/q.

Let {β1, ..., βr} be a transcendental basis of R/q.

Then it lifts to {α1, ..., αr} which is algebraically independent in R/p.

This is because there is a surjective homomorphism ϕ : R/p → R/q.

_{1}, ..., α_{r}} then it gives an
relation among {β1, ..., βr} via the homomorphism ϕ, which is
absurd.

Hence we have shown that tr.d._{k}R/p ≥ tr.d._{k}R/q.

We first claim that dimR ≤ tr.d.kR.

To see this, it suffices to show that if p ( q ∈ Spec(R), then
tr.d._{k}R/p
tr.d.kR/q.

Let {β1, ..., βr} be a transcendental basis of R/q.

Then it lifts to {α1, ..., αr} which is algebraically independent in R/p.

This is because there is a surjective homomorphism ϕ : R/p → R/q.

_{1}, ..., α_{r}} then it gives an
relation among {β1, ..., βr} via the homomorphism ϕ, which is
absurd.

Hence we have shown that tr.d._{k}R/p ≥ tr.d._{k}R/q.

Assume now that tr.d.kR/p = tr.d.kR/q. Then {α1, ..., αr} is an basis.

Lift to R, we have an algebraically independent set
{y_{1}, ..., yr} ⊂ R.

Let S = k[y1, ..., yr] − {0}. We are going to localize w.r.t S .
Write R/p as k[α_{1}, ..., α_{r}, α_{r +1}, ..., α_{n}],

then

S^{−1}R/S^{−1}p= ¯S^{−1}(R/p) = k(α_{1}, ..., α_{r})[α_{r +1}, ..., α_{n}].
We claim that S^{−1}R/S^{−1}pis a field, hence S^{−1}p is a maximal
ideal.

However, note that S ∩ p = S ∩ q = ∅. It follows that S^{−1}p and
S^{−1}q are prime ideals of S^{−1}R.

S^{−1}p ( S^{−1}q ( S^{−1}R.
This is the required contradiction.

Assume now that tr.d.kR/p = tr.d.kR/q. Then {α1, ..., αr} is an basis.

Lift to R, we have an algebraically independent set
{y_{1}, ..., yr} ⊂ R.

Let S = k[y1, ..., yr] − {0}. We are going to localize w.r.t S .
Write R/p as k[α_{1}, ..., α_{r}, α_{r +1}, ..., α_{n}],

then

S^{−1}R/S^{−1}p= ¯S^{−1}(R/p) = k(α_{1}, ..., α_{r})[α_{r +1}, ..., α_{n}].
We claim that S^{−1}R/S^{−1}pis a field, hence S^{−1}p is a maximal
ideal.

However, note that S ∩ p = S ∩ q = ∅. It follows that S^{−1}p and
S^{−1}q are prime ideals of S^{−1}R.

S^{−1}p ( S^{−1}q ( S^{−1}R.
This is the required contradiction.

Assume now that tr.d.kR/p = tr.d.kR/q. Then {α1, ..., αr} is an basis.

Lift to R, we have an algebraically independent set
{y_{1}, ..., yr} ⊂ R.

Let S = k[y1, ..., yr] − {0}. We are going to localize w.r.t S .

Write R/p as k[α_{1}, ..., α_{r}, α_{r +1}, ..., α_{n}],
then

S^{−1}R/S^{−1}p= ¯S^{−1}(R/p) = k(α_{1}, ..., α_{r})[α_{r +1}, ..., α_{n}].
We claim that S^{−1}R/S^{−1}pis a field, hence S^{−1}p is a maximal
ideal.

However, note that S ∩ p = S ∩ q = ∅. It follows that S^{−1}p and
S^{−1}q are prime ideals of S^{−1}R.

S^{−1}p ( S^{−1}q ( S^{−1}R.
This is the required contradiction.

Assume now that tr.d.kR/p = tr.d.kR/q. Then {α1, ..., αr} is an basis.

Lift to R, we have an algebraically independent set
{y_{1}, ..., yr} ⊂ R.

Let S = k[y1, ..., yr] − {0}. We are going to localize w.r.t S .
Write R/p as k[α_{1}, ..., α_{r}, α_{r +1}, ..., α_{n}],

then

^{−1}R/S^{−1}p= ¯S^{−1}(R/p) = k(α_{1}, ..., α_{r})[α_{r +1}, ..., α_{n}].
We claim that S^{−1}R/S^{−1}pis a field, hence S^{−1}p is a maximal
ideal.

^{−1}p and
S^{−1}q are prime ideals of S^{−1}R.

S^{−1}p ( S^{−1}q ( S^{−1}R.
This is the required contradiction.

Assume now that tr.d.kR/p = tr.d.kR/q. Then {α1, ..., αr} is an basis.

Lift to R, we have an algebraically independent set
{y_{1}, ..., yr} ⊂ R.

_{1}, ..., α_{r}, α_{r +1}, ..., α_{n}],

then

S^{−1}R/S^{−1}p= ¯S^{−1}(R/p) = k(α_{1}, ..., α_{r})[α_{r +1}, ..., α_{n}].

We claim that S^{−1}R/S^{−1}pis a field, hence S^{−1}p is a maximal
ideal.

^{−1}p and
S^{−1}q are prime ideals of S^{−1}R.

S^{−1}p ( S^{−1}q ( S^{−1}R.
This is the required contradiction.

Assume now that tr.d.kR/p = tr.d.kR/q. Then {α1, ..., αr} is an basis.

Lift to R, we have an algebraically independent set
{y_{1}, ..., yr} ⊂ R.

_{1}, ..., α_{r}, α_{r +1}, ..., α_{n}],

then

S^{−1}R/S^{−1}p= ¯S^{−1}(R/p) = k(α_{1}, ..., α_{r})[α_{r +1}, ..., α_{n}].

We claim that S^{−1}R/S^{−1}pis a field, hence S^{−1}p is a maximal
ideal.

^{−1}p and
S^{−1}q are prime ideals of S^{−1}R.

S^{−1}p ( S^{−1}q ( S^{−1}R.
This is the required contradiction.

Assume now that tr.d.kR/p = tr.d.kR/q. Then {α1, ..., αr} is an basis.

Lift to R, we have an algebraically independent set
{y_{1}, ..., yr} ⊂ R.

_{1}, ..., α_{r}, α_{r +1}, ..., α_{n}],

then

S^{−1}R/S^{−1}p= ¯S^{−1}(R/p) = k(α_{1}, ..., α_{r})[α_{r +1}, ..., α_{n}].

We claim that S^{−1}R/S^{−1}pis a field, hence S^{−1}p is a maximal
ideal.

^{−1}p and
S^{−1}q are prime ideals of S^{−1}R.

S^{−1}p ( S^{−1}q ( S^{−1}R.
This is the required contradiction.

Assume now that tr.d.kR/p = tr.d.kR/q. Then {α1, ..., αr} is an basis.

Lift to R, we have an algebraically independent set
{y_{1}, ..., yr} ⊂ R.

_{1}, ..., α_{r}, α_{r +1}, ..., α_{n}],

then

S^{−1}R/S^{−1}p= ¯S^{−1}(R/p) = k(α_{1}, ..., α_{r})[α_{r +1}, ..., α_{n}].

We claim that S^{−1}R/S^{−1}pis a field, hence S^{−1}p is a maximal
ideal.

^{−1}p and
S^{−1}q are prime ideals of S^{−1}R.

S^{−1}p ( S^{−1}q ( S^{−1}R.

This is the required contradiction.

To see the claim, let F = k(α_{1}, ..., α_{n}) be the quotient field of R/p.

Since {α_{1}, ..., α_{r}} is a transcendental basis, we have that α_{j} is
algebraic over K := k(α1, ..., αr) for all r < j ≤ n.

For algebraic element, it’s clear that ring extension is a field
extension. Thus F = K (α_{r +1}, ..., α_{n}) = K [α_{r +1}, ..., α_{n}]. The claim
follows.

To see the claim, let F = k(α_{1}, ..., α_{n}) be the quotient field of R/p.

Since {α_{1}, ..., α_{r}} is a transcendental basis, we have that α_{j} is
algebraic over K := k(α1, ..., αr) for all r < j ≤ n.

For algebraic element, it’s clear that ring extension is a field
extension. Thus F = K (α_{r +1}, ..., α_{n}) = K [α_{r +1}, ..., α_{n}]. The claim
follows.

To see the claim, let F = k(α_{1}, ..., α_{n}) be the quotient field of R/p.

Since {α_{1}, ..., α_{r}} is a transcendental basis, we have that α_{j} is
algebraic over K := k(α1, ..., αr) for all r < j ≤ n.

For algebraic element, it’s clear that ring extension is a field
extension. Thus F = K (α_{r +1}, ..., α_{n}) = K [α_{r +1}, ..., α_{n}]. The claim
follows.

The next step is to show that dimR ≥ tr.d._{k}R.

This follows from Noether normalization lemma, that is, reduce to polynomial ring.

More precisely, let r := tr.d._{k}R, then there exist algebraic
independent {y_{1}, ..., y_{r}} ⊂ R and R is integral over k[y_{1}, ..., y_{r}].
We have dimR = dimk[y1, ..., yr] since it’s an integral extension.
And also, dimk[y_{1}, ..., y_{r}] ≥ r because clearly we have a chain of
prime ideals of length n,

0 ⊂ (y_{1}) ⊂ (y_{1}, y_{2})... ⊂ (y_{1}, ..., y_{r}).
This completes the proof.

The next step is to show that dimR ≥ tr.d._{k}R.

This follows from Noether normalization lemma, that is, reduce to polynomial ring.

More precisely, let r := tr.d._{k}R, then there exist algebraic
independent {y_{1}, ..., y_{r}} ⊂ R and R is integral over k[y_{1}, ..., y_{r}].
We have dimR = dimk[y1, ..., yr] since it’s an integral extension.
And also, dimk[y_{1}, ..., y_{r}] ≥ r because clearly we have a chain of
prime ideals of length n,

0 ⊂ (y_{1}) ⊂ (y_{1}, y_{2})... ⊂ (y_{1}, ..., y_{r}).
This completes the proof.

The next step is to show that dimR ≥ tr.d._{k}R.

This follows from Noether normalization lemma, that is, reduce to polynomial ring.

More precisely, let r := tr.d._{k}R, then there exist algebraic
independent {y_{1}, ..., y_{r}} ⊂ R and R is integral over k[y_{1}, ..., y_{r}].

We have dimR = dimk[y1, ..., yr] since it’s an integral extension.
And also, dimk[y_{1}, ..., y_{r}] ≥ r because clearly we have a chain of
prime ideals of length n,

0 ⊂ (y_{1}) ⊂ (y_{1}, y_{2})... ⊂ (y_{1}, ..., y_{r}).
This completes the proof.

The next step is to show that dimR ≥ tr.d._{k}R.

This follows from Noether normalization lemma, that is, reduce to polynomial ring.

More precisely, let r := tr.d._{k}R, then there exist algebraic
independent {y_{1}, ..., y_{r}} ⊂ R and R is integral over k[y_{1}, ..., y_{r}].

We have dimR = dimk[y1, ..., yr] since it’s an integral extension.

And also, dimk[y_{1}, ..., y_{r}] ≥ r because clearly we have a chain of
prime ideals of length n,

0 ⊂ (y_{1}) ⊂ (y_{1}, y_{2})... ⊂ (y_{1}, ..., y_{r}).
This completes the proof.

The next step is to show that dimR ≥ tr.d._{k}R.

This follows from Noether normalization lemma, that is, reduce to polynomial ring.

More precisely, let r := tr.d._{k}R, then there exist algebraic
independent {y_{1}, ..., y_{r}} ⊂ R and R is integral over k[y_{1}, ..., y_{r}].

We have dimR = dimk[y1, ..., yr] since it’s an integral extension.

And also, dimk[y_{1}, ..., y_{r}] ≥ r because clearly we have a chain of
prime ideals of length n,

0 ⊂ (y_{1}) ⊂ (y_{1}, y_{2})... ⊂ (y_{1}, ..., y_{r}).

This completes the proof.

Theorem

Let R be a domain which is finitely generated over k. Let p be a prime ideal, then

htp + dimR/p = dimR.

We first reduce R to polynomial rings.

By Noether normalization theorem, there exists {x_{1}, ..., x_{r}} with
r = tr.d.kR = dim R such that R is integral over k[x1, ..., xr]. Let
q:= p ∩ k[x_{1}, ..., x_{r}], then htq = htp and

dim R/p = dim k[x_{1}, ..., x_{r}]/q.

Next we claim that there exists {y1, ..., yr} ⊂ k[x_{1}, ..., xr] so that
k[x1, ..., xr] is integral over k[y1, ..., yr] and

q∩ k[y_{1}, ..., y_{r}] = (y_{1}, ..., y_{t}) where t = htq.
We prove this when ht = 1.

Pick any y1 6= 0 ∈ q. If y_{1} = x_{1}^{e}+ g (x1, ..., xn) with degx1(g ) < e.
Then x_{1} is integral over R^{0}:= k[y_{1}, x_{2}, ..., x_{n}]. Since (y_{1}) ⊂ q ∩ R^{0}
and ht(y1) = ht(q ∩ R^{0}) = htq = 1. So we have (y1) = q ∩ R^{0}.
The remaining part is similar to the proof of Noether’s

normalization theorem.

We first reduce R to polynomial rings.

By Noether normalization theorem, there exists {x_{1}, ..., x_{r}} with
r = tr.d.kR = dim R such that R is integral over k[x1, ..., xr]. Let
q:= p ∩ k[x_{1}, ..., x_{r}], then htq = htp and

dim R/p = dim k[x_{1}, ..., x_{r}]/q.

Next we claim that there exists {y1, ..., yr} ⊂ k[x_{1}, ..., xr] so that
k[x1, ..., xr] is integral over k[y1, ..., yr] and

q∩ k[y_{1}, ..., y_{r}] = (y_{1}, ..., y_{t}) where t = htq.

We prove this when ht = 1.

Pick any y1 6= 0 ∈ q. If y_{1} = x_{1}^{e}+ g (x1, ..., xn) with degx1(g ) < e.
Then x_{1} is integral over R^{0}:= k[y_{1}, x_{2}, ..., x_{n}]. Since (y_{1}) ⊂ q ∩ R^{0}
and ht(y1) = ht(q ∩ R^{0}) = htq = 1. So we have (y1) = q ∩ R^{0}.
The remaining part is similar to the proof of Noether’s

normalization theorem.

We first reduce R to polynomial rings.

By Noether normalization theorem, there exists {x_{1}, ..., x_{r}} with
r = tr.d.kR = dim R such that R is integral over k[x1, ..., xr]. Let
q:= p ∩ k[x_{1}, ..., x_{r}], then htq = htp and

dim R/p = dim k[x_{1}, ..., x_{r}]/q.

Next we claim that there exists {y1, ..., yr} ⊂ k[x_{1}, ..., xr] so that
k[x1, ..., xr] is integral over k[y1, ..., yr] and

q∩ k[y_{1}, ..., y_{r}] = (y_{1}, ..., y_{t}) where t = htq.

We prove this when ht = 1.

Pick any y1 6= 0 ∈ q. If y_{1} = x_{1}^{e}+ g (x1, ..., xn) with degx1(g ) < e.

Then x_{1} is integral over R^{0}:= k[y_{1}, x_{2}, ..., x_{n}]. Since (y_{1}) ⊂ q ∩ R^{0}
and ht(y1) = ht(q ∩ R^{0}) = htq = 1. So we have (y1) = q ∩ R^{0}.

The remaining part is similar to the proof of Noether’s normalization theorem.

We first reduce R to polynomial rings.

_{1}, ..., x_{r}} with
r = tr.d.kR = dim R such that R is integral over k[x1, ..., xr]. Let
q:= p ∩ k[x_{1}, ..., x_{r}], then htq = htp and

dim R/p = dim k[x_{1}, ..., x_{r}]/q.

_{1}, ..., xr] so that
k[x1, ..., xr] is integral over k[y1, ..., yr] and

q∩ k[y_{1}, ..., y_{r}] = (y_{1}, ..., y_{t}) where t = htq.

We prove this when ht = 1.

Pick any y1 6= 0 ∈ q. If y_{1} = x_{1}^{e}+ g (x1, ..., xn) with degx1(g ) < e.

Then x_{1} is integral over R^{0}:= k[y_{1}, x_{2}, ..., x_{n}]. Since (y_{1}) ⊂ q ∩ R^{0}
and ht(y1) = ht(q ∩ R^{0}) = htq = 1. So we have (y1) = q ∩ R^{0}.
The remaining part is similar to the proof of Noether’s

normalization theorem.

Theorem ( Krull’s Hauptidealsatz)

Let R be a finitely generated k-algebra. Let f ∈ R be a non-zero nor a unit. Then every minimal prime containing f has height 1.

A projective n-space, denoted P^{n} is defined to be the set of
equivalence classes of (n + 1) tuples (a_{0}, ..., a_{n}), with not all zero.

Where the equivalent relation is (a0, ..., an) ∼ (λa0, ..., λan) for all
λ 6= 0. We usually write the equivalence class as [a0, ..., an] or
(a_{0}: ... : a_{n}).

One can first consider P^{n} as a quotient of A^{n+1}− {(0, ..., 0)}. Let
π : A^{n+1}− {(0, ..., 0)} → P^{n} be the quotient map. And we can
topologize P^{n} by the quotient topology of Zariski topology. Then
one sees that for a closed set Y ⊂ P^{n}, π^{−1}(Y ) corresponds to a
homogeneous ideal I C k[x0, ..., xn].

We have the similar correspondence between projective algebraic sets and homogeneous radical ideals. There is an ieal need to be excluded, the irrelevant maximal ideal, (x0, .., xn).

Another important description is to give P^{n} an open covering of
n + 1 copies of A^{n}. It follows that every projective variety can be
covered by affine varieties.

To this end, we can simply consider

ıj : A^{n}→ P^{n} by ıj(a1, ..., an) = [a1, ..., aj −1, 1, aj, .., an].

On the other hand, let Hj be the hyperplane xj = 0 in P^{n}. Then
we have

pj : P^{n}−H_{j} → A^{n} by pj[a0, ..., an] = (a0/aj, ..., aj −1/aj, aj +1/aj, ..., an/aj).

Example

We have seen that an elliptic curve E can be maps to C^{2} by the
Weierstrass functions. Compose with ı2, we have a map

ϕ : E → P^{2}. The defining equation in C^{2} is y^{2} = 4x^{3}− g_{2}x − g_{3}.
While the defining equation in P^{2} is the homogenized equation
y^{2}z = 4x^{3}− g_{2}xz^{2}− g_{3}z^{3}

In general, the equations in affine spaces and projective spaces are corresponding by homogenization and dehomoenization.

By a variety, we mean affine, quasi-affine, projective or

quasi-projective variety. (More generally, an abstract variety can be defined as an integral separated scheme of finite type over an algebraically closed field k).

So far, we have defined affine varieties and projective varieties. In order to study them more carefully, we need to know the morphism between them. In the category of varieties, we consider ”algebraic functions”, that is, function expressible as polynomial or quotient of polynomial, at least locally.

Definition

Let Y be an affine variety in A^{n}. A function f : Y → k is regular
at a point p if there is an open set U ⊂ Y such that f = ^{g}_{h} for
some g , h ∈ k[x_{1}, ..., x_{n}].

We say that f is regular on Y if f is regular at every point of Y .

The definition also works if Y is quasi-affine.

Exercise

How to define regular function on projective varieties so that it’s compatible with affine covering?

Proposition

If we identified k with A^{1}, then a regular function f : Y → k is
continuous.

In fact, Zariski topology on Y can be viewed as the coarsest topolopy which makes regular function continuous.

Proof.

It suffices to show that f^{−1}(a) is closed for all a ∈ k. Moreover,
f^{−1}(a) is closed if f^{−1}(a) ∩ U_{i} is closed in U_{i} for an open covering
{U_{i}}.

By definition, there is an open covering {Ui} such that on each U_{i},
f = ^{g}_{h} for some polynomial g , h. Thus on U_{i}, f^{−1}(a) = V(g − ah),
which is closed. So we are done.

An important consequence is Proposition

Let f , g be regular function on a variety X . If f = g on a non-empty open set U, then f = g .

Proof.

V(f − g ) is a closed set containing U. So X = V(f − g ) ∪ U^{c}. But
X is irreducible and U^{c} 6= X . Thus X = V(f − g ).

Now introduce some more notion on functions.

Definition

Let Y be a variety. We denote by O(Y ) the ring of regular functions.

If p ∈ Y is a point, we define the local ring of p on Y , denoted by
O_{p,Y}, to be the germs of regular functions near p. More precisely,
we consider (U, f ) with U 3 p an open set containing p and f a
regular function on U. (U, f ) and (V , g ) are said to be equivalent
of there is an open set W ⊂ U ∩ V containing p such that
f |_{W} = g |_{W}. A germ of regular function near p is an equivalent
class of (U, f ).