Advanced Algebra II Apr. 27, 2007

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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.

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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.

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For an affine variety XinAⁿ, 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.

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For an affine variety XinAⁿ, 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 ).

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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.

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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.

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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 dimAⁿ= n.

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

However, it’s not trivial to prove that dimAⁿ= n.

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

However, it’s not trivial to prove that dimAⁿ= n.

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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)}.

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Definition

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

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Definition

We define

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

ht(I ) := inf{ht(p)|I ⊂ p}.

We define

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Definition

We define

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

ht(I ) := inf{ht(p)|I ⊂ p}.

We define

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Example

Let R := k[x₁, ..., 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 Aⁿ_k.

Then ht(p) is nothing but the codimension of V(p) in Aⁿ_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}.

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Example

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Example

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Example

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Example

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Example

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}.

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Example

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}.

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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.

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Theorem

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

(where tr.d._kR := tr.d._kF , where F is the quotient field of R.)

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We first claim that dimR ≤ tr.d.kR.

To see this, it suffices to show that if p ( q ∈ Spec(R), then tr.d._kR/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 {α₁, ..., α_r} then it gives an relation among {β1, ..., βr} via the homomorphism ϕ, which is absurd.

Hence we have shown that tr.d._kR/p ≥ tr.d._kR/q.

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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 {y₁, ..., yr} ⊂ R.

Let S = k[y1, ..., yr] − {0}. We are going to localize w.r.t S . Write R/p as k[α₁, ..., α_r, α_{r +1}, ..., α_n],

then

S⁻¹R/S⁻¹p= ¯S⁻¹(R/p) = k(α₁, ..., α_r)[α_{r +1}, ..., α_n]. We claim that S⁻¹R/S⁻¹pis a field, hence S⁻¹p is a maximal ideal.

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

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

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then

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

Write R/p as k[α₁, ..., α_r, α_{r +1}, ..., α_n], then

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then

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then

S⁻¹R/S⁻¹p= ¯S⁻¹(R/p) = k(α₁, ..., α_r)[α_{r +1}, ..., α_n].

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

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then

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then

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then

S⁻¹p ( S⁻¹q ( S⁻¹R.

This is the required contradiction.

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To see the claim, let F = k(α₁, ..., α_n) be the quotient field of R/p.

Since {α₁, ..., 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.

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The next step is to show that dimR ≥ tr.d._kR.

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

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

0 ⊂ (y₁) ⊂ (y₁, y₂)... ⊂ (y₁, ..., y_r). This completes the proof.

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

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More precisely, let r := tr.d._kR, then there exist algebraic independent {y₁, ..., y_r} ⊂ R and R is integral over k[y₁, ..., y_r].

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

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

And also, dimk[y₁, ..., y_r] ≥ r because clearly we have a chain of prime ideals of length n,

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

And also, dimk[y₁, ..., y_r] ≥ r because clearly we have a chain of prime ideals of length n,

0 ⊂ (y₁) ⊂ (y₁, y₂)... ⊂ (y₁, ..., y_r).

This completes the proof.

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Theorem

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

htp + dimR/p = dimR.

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We first reduce R to polynomial rings.

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

dim R/p = dim k[x₁, ..., x_r]/q.

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

q∩ k[y₁, ..., y_r] = (y₁, ..., y_t) where t = htq. We prove this when ht = 1.

Pick any y1 6= 0 ∈ q. If y₁ = x₁^e+ g (x1, ..., xn) with degx1(g ) < e. Then x₁ is integral over R⁰:= k[y₁, x₂, ..., x_n]. Since (y₁) ⊂ q ∩ R⁰ and ht(y1) = ht(q ∩ R⁰) = htq = 1. So we have (y1) = q ∩ R⁰. The remaining part is similar to the proof of Noether’s

normalization theorem.

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q∩ k[y₁, ..., y_r] = (y₁, ..., y_t) where t = htq.

We prove this when ht = 1.

Pick any y1 6= 0 ∈ q. If y₁ = x₁^e+ g (x1, ..., xn) with degx1(g ) < e. Then x₁ is integral over R⁰:= k[y₁, x₂, ..., x_n]. Since (y₁) ⊂ q ∩ R⁰ and ht(y1) = ht(q ∩ R⁰) = htq = 1. So we have (y1) = q ∩ R⁰. The remaining part is similar to the proof of Noether’s

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

Then x₁ is integral over R⁰:= k[y₁, x₂, ..., x_n]. Since (y₁) ⊂ q ∩ R⁰ and ht(y1) = ht(q ∩ R⁰) = htq = 1. So we have (y1) = q ∩ R⁰.

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

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

Then x₁ is integral over R⁰:= k[y₁, x₂, ..., x_n]. Since (y₁) ⊂ q ∩ R⁰ and ht(y1) = ht(q ∩ R⁰) = htq = 1. So we have (y1) = q ∩ R⁰. The remaining part is similar to the proof of Noether’s

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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.

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A projective n-space, denoted Pⁿ is defined to be the set of equivalence classes of (n + 1) tuples (a₀, ..., 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 (a₀: ... : a_n).

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

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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 Pⁿ an open covering of n + 1 copies of Aⁿ. It follows that every projective variety can be covered by affine varieties.

To this end, we can simply consider

ıj : Aⁿ→ Pⁿ by ıj(a1, ..., an) = [a1, ..., aj −1, 1, aj, .., an].

On the other hand, let Hj be the hyperplane xj = 0 in Pⁿ. Then we have

pj : Pⁿ−H_j → Aⁿ by pj[a0, ..., an] = (a0/aj, ..., aj −1/aj, aj +1/aj, ..., an/aj).

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Example

We have seen that an elliptic curve E can be maps to C² by the Weierstrass functions. Compose with ı2, we have a map

ϕ : E → P². The defining equation in C² is y² = 4x³− g₂x − g₃. While the defining equation in P² is the homogenized equation y²z = 4x³− g₂xz²− g₃z³

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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).

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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.

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Definition

Let Y be an affine variety in Aⁿ. 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[x₁, ..., x_n].

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

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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?

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Proposition

If we identified k with A¹, 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.

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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.

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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 ).

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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 ).