# Algebraic Geometry

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孙晟昊 shsun@mail.tsinghua.edu.cn

- 答疑：Fri 9-10 am, online
- HW (40%) + Final (60%)

## References

- Harris, GTM133
- Shafarevich, Basic AG, Vol I
- Mumford, AG I: Complex Proj. Var.
- Hartshorne, GTM52, Chap I.

## Preliminaries

### Constructions from rings

Finitely generated: f.g.
$\newcommand{\aqty}[1]{\left\langle#1\right\rangle}$

- f.g. $A$-module: $A\aqty{\alpha_1,\cdots,\alpha_n}$, linear combinations with coefficients in $A$.

- f.g. $A$-algebra: $A[\alpha_1,\cdots,\alpha_n]$, polynomials over $A$.

- f.g. field extension $L/K$: $L = K(\alpha_1,\cdots,\alpha_n)$: fraction of poly's $P(\cdots)/Q(\cdots)$.

### Integral element

$A\subset B$ rings, $b \in B$ **integral over $A$**:

$$
  b^n + a_1 b^{n-1} + \cdots + a_n = 0
  \quad\text{in $B$,}\quad
  a_i \in A
  \label{eq:integralElement}
$$

**i.e.** $b$ is a root of some _monic_ ($a_0 = 1$) poly. over $A$.

**e.g.** $\sqrt{2}\in\mathbb{R}$ integral over $\mathbb{Z}$: $(\sqrt{2})^2 - 2 = 0$.

**Idea:** We are extending $A$ with roots of its poly's. [Wiki:](https://en.wikipedia.org/wiki/Integral_element) If $A, B$ are fields, then the notions of "integral over" and of an "integral extension" are precisely "algebraic over" and "**algebraic extensions**" in field theory (since the root of any polynomial is the root of a monic polynomial).

**Lem.** $A\subset B$ integral domain (整环), $b \in B$ integral over $A$ <br/>
$\Longleftrightarrow$ $b\cdot W \subset W$, for some fintely generated (f.g.) non-zero sub-$A$-module $W\in B$.

**Pf.**

- $\Longrightarrow$: Take $W = A[b] = A\aqty{1,b,b^2,\cdots,b^{n-1}}$, $b^n$ can be reduced using \eqref{eq:integralElement}.

- $\Longleftarrow$: $b\cdot W \subset W$ implies that:

  $$
    \sum_j \pqty{
      b\,\delta_{ij} - a_{ij}
    }\,b_j
    = 0,
  \quad
    \det \pqty{
      b\,\delta_{ij} - a_{ij}
    } = 0
  $$

  The determinant is a poly. in the form of \eqref{eq:integralElement}. Here $b_j$'s are the generators: $W = A\aqty{b_1, b_2, \cdots, b_n} \subset B$.

**Prop.** $A\subset B$ int. dom., $b,c\in B$ int. elements, then so is $(b+c)$, or $(b\cdot c)$.

**Pf.** Use **Lem.**; construct $W = A[b,c]$.

**Comment:** This is non-trivial! Consider finding the minimal poly. of $\sqrt{2} + \sqrt[3]{5} + i$ over $\mathbb{Q}$. **Prop.** in fact gives an upper limit of the poly. deg.

**Thm. (Hilbert's Nullstellensatz --- zero-locus-theorem, algebraic version)**

If:
- $L/k$ field ext., also!
- $L$ is a f.g. $k$-alg. as well, $L = k[\alpha_1,\cdots,\alpha_n]$.

**i.e.** $
  L = k[\alpha_1,\cdots,\alpha_n]
  = k(\alpha_1,\cdots,\alpha_n)
$.

Then $L/k$ is an _algebraic & finite ext._

**Concepts:** field ext.
- algebraic: all elements are roots, e.g. $\mathbb{C}/\mathbb{R}$.
- transcendental: non-roots, e.g. $\mathbb{R}/\mathbb{Q}$.

All transcendental exts are of $\infty$ deg. Hence all finite exts are algebraic. But

**Pf.**

- It suffices to show that each $\alpha_i$ is alg. over $k$.

- Induction on $n$ (the number of generators).

- By contradiction: for $\alpha = T$ transcendental, consider $A = k[T, \tfrac{1}{f(T)}] \subset k(T)$ --- this is called _localization_: by manually adding inverses, the domain is restricted / localized around "non-zero" regions.

  - We can check that $A$ is a field;

  - However, $A$ is not a field, since there are $\infty$-ly many irreducible poly. in $k[T]$, which can be shown using Euclid's arguments, similar to the proof that there are $\infty$-ly many primes.

## Intro

Geometry: $X$, $\mathbb{R}$-algebra:

$$
  \mathbb{R} \longrightarrow
  C(X) = \Bqty{\,
    f\colon X\to\mathbb{R}
  \,}
$$

- $X$: Top., $C^0(X)$
- $X$: Diff., $C^\infty(X)$
- $X$: complex, $\mathcal{H}(X)$
- $X$: alg. variety, $\mathcal{O}(X)$: polynomials on $X$.

Consider a _surjective_ $\mathbb{R}$-_algebra hom._

$$
  \mop{ev}_x\colon
  \ C(X) \longto \mathbb{R},\quad
  f\longmapsto f(x)
$$

$m_x\subset C(X)$ a _maximal ideal:_ $
  m_x = \{f \,|\, f(x) = 0\}
$. We have thus represented a pt. with an algebraic object.

> Note: the kernel of ring hom. is an ideal, just like the kernel of group hom. is a normal subgroup. If we quotient out the ideal (or normal subgroup), then the isom. descends to an isom. to the image. These are the so-called [**isomorphism theorems**](https://en.wikipedia.org/wiki/Isomorphism_theorems).
>
> Furthermore, for _commutative_ rings, if the image is a field, then the kernel (ideal) is maximal.

If $X$ compact Hausdorff, we have a **_bijection_**:

$$
  \mop{ev}\colon
  \ X \longto
  \Hom_{\mathbb{R}-alg.} \pqty{
    C(X),\mathbb{R}
  }, \quad
  x\longmapsto \mop{ev}_x
$$

**Moral:** Recover geometry $X$ from algebra hom's.

$$
  \mop{Specm}\big(C(X)\big)
  = \Big\{
      \text{all max ideals in $C(X)$}
    \Big\}
$$

To study geometry, we can consider various structs:

- $S^7$, $\mathbb{R}^4$ diffeo structs
- Torus complex structs

For $(x,y) \in \mathbb{C}$, consider:

$$
  y^2 = x^3 + 1
  \quad\text{vs}\quad
  y^2 = x^3 - x
$$

diffeo, but not biholomorphic (neither isom. as $\mathbb{C}$-alg. varieties).

### Affine Space

Field $k = \bar{k}$ (assume algebraic closed for simplicity); affine space: $\mathbb{A}_k^n$.
As a set, $
  \mathbb{A}_k^n
  = k^n
$; however we'd like to allow all poly. functions:

$$
  f\colon\ k^n \to k,\quad
  \forall\ f(T) \in k\bqty{
    T_1,\cdots,T_n
  } =\colon A
$$

**Idea.** Define geometry (alg. var.) from the allowed functions; recover $\mathbb{A}_k^n$ from poly. ring $A$. Similarly, we use $\mop{ev}_a$;

$$
  m_a = \ker (\mop{ev}_a)
  = \Bqty{
    f\in A
    \,|\,
    f(a) = 0
  }
$$

In fact, _any_ max ideal $m\subset A$ takes the form:

$$
  m = \pqty{
    T_1 - a_1,\cdots,T_n-a_n
  }
  = \ker(\mop{ev}_a) = m_a
$$

Why? Think of Taylor expansion around the root $a$. The proof of this follows from _Nullstellensatz_: $A/m$ is a field, and a f.g. $k$-alg., therefore it is a finite extension of $k$; but $k = \bar{k}$ so $A/m = k$.

**Affine (sub)varieties:** solution set of a system of poly. equations.

However, this is a redundant description since many systems can lead to the same solution set. How to remove such redundancies? Use the _**ideal**_ $I\subset A$ generated by the system of eqns. This gives the _common zero locus_:

$$
  X = Z(I)\subset \mathbb{A}^n,\quad
  Z(I) = \Bqty{
    a\in\mathbb{A}^n
    \,|\,
    f(a) = 0,\ \forall\ f\in I
  }
$$

Similarly, we have $1:1$ identifications:
- pts. in $Z(I)$
- $k$-alg. hom. $A/I \to k$, i.e. $\Hom_{k-\mathrm{alg.}}(A/I,k)$.
- max ideal $m\subset A/I$, or equivalently $I\subset m \subset A$. This one relies on $k = \bar{k}$ and the Nullstellensatz.

Conversely, given some arbitrary $X\subset \mathbb{A}^n$, def. its ideal:

$$
  I(X) = \Bqty{
    f\in A
    \,\Big|\,
    f|_X \equiv 0
  }
$$

We can then define the _affine coord. ring_ with $f \equiv g\ \mop{mod}\, I(X)$:

$$
  k[X] \mathbin{:=} A/I(X)
$$


It is illuminating to go back and forth: $I \to X = Z(I) \to I(X)$, and see if $I \overset{?}{=} I(X)$. In fact, we have:
- $I(X) \supset I$; e.g. for $I = (T^2) \subset k[T]$, we have $X = Z(I) = \{a = 0\}$, $I(X) = (T) \supsetneq I$. However, in this case we do have this funny result:

  $$
    \sqrt{I} = \sqrt{(T^2)} = (T) = I(X)
  $$

  We are taking the "square root" of $I$; this can be formalized as the [**radical**](https://en.wikipedia.org/wiki/Radical_of_an_ideal) of $I$: $\mop{rad} I \equiv \sqrt{I}$.
- $I(X) = \sqrt{I(X)}$: always radical --- $f^N(a) = 0\ \To\ f(a) = 0$, a field does not contain nilpotent element.
- In fact, for $k = \bar{k}$, we always have $I(X) = \sqrt{I}$. This is the **Geometric Nullstellensatz**. This leads to the fact that:

  $$
    \Hom_{k-\mathrm{alg.}}(A/I,k)
    \cong
    \Hom_{k-\mathrm{alg.}}(k[X],k),\quad
    k[X] = A/I(X)
  $$

  i.e. $A/I\to k$ always descends through $A/I(X) = k[X]$:

  $$
    A/I
    \longto A/I(X)
    \longto k
  $$

### Localization

**Idea.** By adding allowed functions, we shrink the space (removing pts on which the functions cannot be defined).

- $\mathbb{A}^1_k \leadsto k[T]$
- $\mathbb{A}^1_k - \{0\} \leadsto k[T, \tfrac{1}{T}]$
- $\mathbb{A}^1_k - \{0,1\} \leadsto k[T, \tfrac{1}{T}, \tfrac{1}{T-1}] = k[T, \frac{1}{T(T-1)}]$

Notation: $S = (f) \subset A$ some subset closed under multiplication, we can define $S^{-1} A = A[\frac{1}{f}]$. Furthermore, if $S = A - p$, where $p$: some prime ideal, then we denote:

$$
  A_p \mathbin{:=} S^{-1} A
$$

With this notation, we can write down the algebraic description of "a small neighborhood around $0\in \mathbb{A}^1_k$" --- in this case we want to remove (almost) every point except the origin, i.e.

$$
  \Bqty{
    \frac{f(T)}{g(T)} \,\bigg|\,
    g(0) \ne 0,\ \text{or}\ g(T) \notin (T)
  }
  = S^{-1} k[T]
  = k[T]_{(T)}, \\
  p = (T),\quad
  S = k[T] - p
$$

### Tangent Space

**e.g.** Plane curve $C = Z(f) \subset \mathbb{A}^2$, tangent line $T_x(C)$? For intrinsic def, use _**derivation**_ which satisfies the Leibnitz's rule:

$$
  \pd\colon\ C^\infty(X)
  \to k,\quad
$$

However, $C^\infty$ might be too much: e.g. $\mathbb{C}$ mfd is very _rigid_; for compact $\mathbb{C}$ mfd, global holomorphic functions are always constant (by [maximum modulus principle](https://en.wikipedia.org/wiki/Maximum_modulus_principle)); i.e. _topology_ of the mfd is a very strong constraint (too strong than what we'd like). Therefore, we shall consider "locally holomorphic" functions.

For plane curves defined by $f = 0$, locally defined polynomials can be neatly written down with the help of localization:

$$
  \Big( A/(f) \Big)_{m_p},\quad
  m_p = (x_1 - p_1, \cdots, x_n - p_n)
$$

We shall then consider the following derivations which spans $\mathrm{T}_p C$:

$$
  \pd\colon
  \Big( A/(f) \Big)_{m_p}
  \longto k,
$$

Before continuing, let us first consider $A_{m_p}$ without restricting to the curve, or equivalently, taking $f \mathbin{:=} 0$. e.g.
$
  T_{p=0}(\mathbb{R}^n) = \mathbb{R}\aqty{
    \pdv{}{T_1},\cdots,\pdv{}{T_n}
  }
  = \Bqty{
    v^i \pdv{}{T_i}
  } \cong \mbb{R}^n
$.
$\pd$-action on $A_{m_0}$ is induced from its action on $A$, and is completely fixed by the Leibnitz's rule. e.g. $\pd\,(1) = \pd\,(1\cdot 1) = 2\,\pd\,(1) = 0$, etc. We have:

$$
  T_0 \mathbb{A}^n
  = \Bqty{ \pd\colon A_{m_0} \to k }
  \cong \Bqty{ \pd\colon A \to k }
  \cong k^n
  \label{eq:tangent_derivation_vector}
$$

**Rmk.** We see that there is no need for localization (to a point $p$) to define the tangent space on a $\mathbb{R}$-mfd; why? We have partition of 1 (unity) for a $\mathbb{R}$-mfd, so that local functions can always be stitched together to make a global function. This is not true for $\mathbb{C}$-mfds.

Back to the curve $C\colon f = 0$, we have:

$$
  A \longto A_{m_p}
  \longto A_{m_p} / (f)
  \cong \Big( A/(f) \Big)_{m_p}
$$

Note that localization commutes with the quotient. $\pd$ on $\Big( A/(f) \Big)_{m_p}$ is contrained by the quotient: $\pd f = 0$. By \eqref{eq:tangent_derivation_vector}, we have:

$$
  T_p C
  \cong \Bqty{
    v\in k^2
    \,\bigg|\,
    \pd_v f = v^i \pdv{f}{x^i} = 0
  }
$$

Note that:

- smooth pt $\leadsto$ $T_p C$: a tangent line (tangent);
- singular pt $\leadsto$ $T_p C \cong k^2$: the whole plane, $\pdv{f}{x} = \pdv{f}{y} = 0$.

e.g.

![](img/punctured_line.png)

![](img/singular_curve.png)

e.g (affine) twisted cubic curve: $(t,t^2,t^3)$, or

$$
  I = \{f \,\big|\, f(t,t^2,t^3) = 0\}
  = (y - x^2, z - x^3)
  = (y - x^2, z - xy)
$$

### Zariski Top.

**Def.** closed sets are $Z(I),\ I \subset A$ ideals; $Z(0) = \mathbb{A}^n,\ Z(1) = \varnothing$.

- $Z(I_1) \cup Z(I_2) = Z(I_1 I_2)$, by the property of _prime_ ideals (if $f_1f_2 \in I$, then $f_1\in I$ or $f_2\in I$; max ideals are always prime for comm. rings); or think of products of equations;

- $\bigcap_i Z(I_i) = Z\pqty{\sum_i I_i}$, think of a system of equations;

**e.g.** closed subsets in $\mathbb{A}^1$ are finite pts and $\mathbb{A}^1$ itself (余有限拓扑). Note that open sets are very large in such topology, hence this is a very _coarse_ topology.

There is a natural top. basis: $U_f = \mathbb{A}^n - Z(f),\ \forall\ f\in A$. By definition, any open set can be written as:

$$
  \mathbb{A}^n - Z(I)
  = \mathbb{A}^n
    - \bigcap_{f\in I} Z(f)
  = \bigcup_{f\in I} \big(
      \mathbb{A}^n
      - Z(f)
    \big)
  = \bigcup_{f\in I} U_f
$$

Note that $f\in A$ irreducible poly. $\To$ $(f) \subset A$ prime ideal; this motivates us to define that:

**Def. irreducible set (algebraic):** $X$ irreducible if $I(X) \subset A$ prime ideal. This is related to the geometric (topological) notion of irreducibility:

**Prop. irreducible set (geometric):** $X$ irreducible _iff_ $X \ne Z(I_1) \cup Z(I_2)$, i.e. $X$ can not be decomposed into two proper closed sets.

In the spirit of localization, we can def. (local) regular functions:

$$
  \mcal{O}(X) = \Bqty{
    f \,\bigg|\,
    \forall\ p\in X,
    \ \exists\ U_p \subset X,
    \ \text{s.t.}
    \ f\big|_{U_p} = \frac{g}{h},
    \ h\big|_{U_p} \ne 0
  }
$$

e.g.

- $\frac{1}{T}$ regular on $\mathbb{A}^1 - \{0\}$
- $\frac{1}{T^2 + 1}$ regular on $\mathbb{A}^1_{\mathbb{R}}$

**Rmk.** a different perspective: a resolution without localization is to start with _function fields_ on $X$, e.g. meromorphic functions on a $\mathbb{C}$-mfd.

Consider:

$$
  k[X] \mathbin{:=} A/I(X)
  \hookrightarrow \mathcal{O}(X)
$$

**Q.** Surjection?

LHS: some poly.'s, nonzero on $X$; RHS: locally-defined fractions. This is clearly not true, for $X$: arbitrary set, e.g. $X = \mathbb{A}^1 - \{0\}$. However, what if:

- $X = Z(I)$: a alg. var., or equivalently,
- $X$ closed in $\mathbb{A}_k$

This is still not true if $k \ne \bar{k}$, e.g. for $k = \mathbb{R}$ we have $\frac{1}{T^2 + 1} \in \mathcal{O}(x)$.

Surprisingly, if $k = \bar{k}$, we do have $A/I(X) \cong \mathcal{O}(X)$, i.e. local fractions are all restrictions of global poly's. In fact, we have:

- pt-wise: $\mathbb{A} \cong \mop{Specm} A$, or

  $$
    \Big\{
      a\in \mathbb{A}
    \Big\}
    \longleftrightarrow
    \Big\{
      \text{max. ideals}
      \ m_a \subset A
    \Big\}
  $$


- Affine varieties:

  $$
    \Big\{
      X = Z(I) \subset \mathbb{A}^n_k
    \Big\}
    \longleftrightarrow
    \Big\{
      \text{rad. ideals}
      \ I \subset A
    \Big\}
  $$

It is easy to check that:

$$
  X = Z(I)\mapsto I(X)\mapsto Z(I(X))
$$

is an identity map, since $I(X) \supset I$, $X\subset Z(I(X)) \subset Z(I) = X$. What about the other way around? i.e.

$$
  I = \sqrt{I}
  \mapsto Z(I) =\colon X
  \mapsto I(X) = I(Z(I))
$$

This is non-trivial, and due to the Nullstellensatz.

### Geometric Nullstellensatz

$$
  I\pqty{Z(I)} = \sqrt{I}
$$

Obvious: $I\pqty{Z(I)} \supset \sqrt{I}$.

**Special case:** $Z(I) = \varnothing\ \To\ I = A$. Otherwise, by alg. Nullstellensatz there is some max ideal $m$ s.t. $I \subset m \subset A$, and $Z(I) \supset Z(m) = \{a\} \ne \varnothing$.

**General case:** reduced to special case! For $f\in I(Z(I))$, consider "embedding" into higher dimensional space:
$A \hookrightarrow A' = k[T_1,\cdots,T_n,T_{n+1}]$, then we can construct:

$$
  J = I\cdot A' + \Big(
      T_{n+1} f(T_1,\cdots,T_n) - 1
    \Big)
$$

Clearly $Z(J) = \varnothing$, thus by the special case $J = A'$.

This is _the trick of Rabinowitsch_. Intuitively, we embed $Z(I)\hookrightarrow Z(I)\times \mathbb{A}^1$, but then "removes" $Z(I)$ (and possibly more) by imposing a "parabolic" constraint: $T_{n+1} f(T_1,\cdots,T_n) = 1$. The resulting $Z(J)$ is clearly empty. This is localization in disguise!

$$
  A'/(T_{n+1}f - 1) = A[\tfrac{1}{f}]
$$

Now we have $A'/J = 0 = A[\tfrac{1}{f}] / (I\cdot A[\tfrac{1}{f}])$, i.e. $1 = \sum a_i b_i \in A[\tfrac{1}{f}]$. Equivalently,

$$
  f^N = \sum a_i b'_i \in I \subset A
$$

Thus $f\in \sqrt{I}$. $\ \blacksquare$