Prologue

0.1 References

• Gathmann’s Algebraic Geometry notes[1].

0.2 Notation

• If a property $$P$$ is said to hold locally, this means that for every point $$p$$ there is a neighborhood $$U_p \ni p$$ such that $$P$$ holds on $$U_p$$.
Notation Definition
$$k[\mathbf{x}] = k[x_1, \cdots, x_n]$$ Polynomial ring in $$n$$ indeterminates
$$k(\mathbf{x}) = k(x_1, \cdots, x_n)$$ Rational function field in $$n$$ indeterminates
$$k(\mathbf{x}) = \left\{{f(\mathbf{x}) = p(\mathbf{x})/q(\mathbf{x}), {~\mathrel{\Big|}~}p,q, \in k[x_1, \cdots, x_{n}]}\right\}$$
$$\mathcal{U} \rightrightarrows X$$ An open cover
$$\mathcal{U} = \left\{{U_j {~\mathrel{\Big|}~}j\in J}\right\}, X = \bigcup_{j\in J}U_j$$
$$\Delta_X$$ The diagonal
$$\Delta_X \mathrel{\vcenter{:}}=\left\{{(x, x) {~\mathrel{\Big|}~}x\in X}\right\} \subseteq X\times X$$
$${\mathbb{A}}^n_{/k}$$ Affine $$n{\hbox{-}}$$space
$${\mathbb{A}}^n_{/k} \mathrel{\vcenter{:}}=\left\{{\mathbf{a} = {\left[ {a_1, \cdots, a_n} \right]} {~\mathrel{\Big|}~}a_j \in k}\right\}$$
$${\mathbb{P}}^n_{/k}$$ Projective $$n{\hbox{-}}$$space
$${\mathbb{P}}^n_{/k} \mathrel{\vcenter{:}}=\qty{k^n\setminus\left\{{0}\right\}}/x\sim \lambda x$$
$$V(J), V_a(J)$$ Variety associated to an ideal $$J {~\trianglelefteq~}k[x_1, \cdots, x_{n}]$$
$$V_a(J) \mathrel{\vcenter{:}}=\left\{{\mathbf{x}\in {\mathbb{A}}^n {~\mathrel{\Big|}~}f(\mathbf{x}) = 0,\, \forall f\in J}\right\}$$
$$I(S), I_a(S)$$ Ideal associated to a subset $$S \subseteq {\mathbb{A}}^n_{k}$$
$$I_a(S) \mathrel{\vcenter{:}}=\left\{{f\in k[x_1, \cdots, x_{n}] {~\mathrel{\Big|}~}f(\mathbf{x}) = 0\, \forall \mathbf{x}\in X}\right\}$$
$$A(X)$$ Coordinate ring of a variety
$$A(X) \mathrel{\vcenter{:}}= k[x_1, \cdots, x_{n}]/I(X)$$
$$V_p(J)$$ Projective variety of an ideal
$$V_p(J) \mathrel{\vcenter{:}}=\left\{{\mathbf{x} \in {\mathbb{P}}^n_{/k} {~\mathrel{\Big|}~}f(\mathbf{x}) = 0,\, \forall f\in J}\right\}$$
$$I_p(S)$$ Projective ideal (?)
$$I_p(S) \mathrel{\vcenter{:}}=\left\{{f\in k[x_1, \cdots, x_{n}] {~\mathrel{\Big|}~}f \text{ is homogeneous and } f(x) = 0\, \forall x\in S}\right\}$$
$$S(X)$$ Projective coordinate ring
$$S(X) \mathrel{\vcenter{:}}= k[x_1, \cdots, x_{n}]/ I_p(X)$$
$$f^h$$ Homogenization
$$f^h \mathrel{\vcenter{:}}= x_0^{\deg f} f\qty{{x_1 \over x_0}, \cdots, {x_n \over x_0}}$$
$$f^i$$ Dehomogenization
$$f^i \mathrel{\vcenter{:}}= f(1, x_1, \cdots, x_n)$$
$$J^h$$ Homogenization of an ideal
$$J^h \mathrel{\vcenter{:}}=\left\{{f^j {~\mathrel{\Big|}~}f\in J}\right\}$$
$$\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu$$ Projective closure of a subset
$$\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu \mathrel{\vcenter{:}}= V_p(J^h) \mathrel{\vcenter{:}}=\left\{{\mathbf{x} \in {\mathbb{P}}^n {~\mathrel{\Big|}~}f^h(\mathbf{x}) = 0\, \forall f\in X}\right\}$$
$$D(f)$$ Distinguished open set
$$D(f) \mathrel{\vcenter{:}}= V(f)^c = \left\{{x\in {\mathbb{A}}^n {~\mathrel{\Big|}~}f(x) \neq 0}\right\}$$
$$\mathcal{F}$$ Presheaf or a sheaf
$$f\in \mathcal{F}(U)$$ Section of a presheaf or sheaf
$$\underline{\mathbf{S}}$$ where $$S$$ is a set Locally constant functions valued in $$S$$

$$\mathcal{F}_p$$ Stalk of a sheaf
$$\mathcal{F}_p \mathrel{\vcenter{:}}=\left\{{(U, \phi) {~\mathrel{\Big|}~}p\in U \text{ open },\, \phi \in \mathcal{F}(U)}\right\}/\sim$$
where $$(U, \phi) \sim (U', \phi') \iff \exists p\in W \subset U\cap U' \text{ s.t. } { \left.{{\phi}} \right|_{{W}} } = { \left.{{\phi'}} \right|_{{W}} }$$
$$f\in \mathcal{F}_p$$ Germs at $$p$$
$${\mathcal{O}}_X$$ Structure sheaf
$${\mathcal{O}}_X \mathrel{\vcenter{:}}=\left\{{f:U\to k {~\mathrel{\Big|}~}U \subseteq X \text{ is open}, f \in k(\mathbf{x}) \text{ locally}}\right\}$$
$${\mathcal{O}}_X(U)$$ Regular functions on $$U$$
$${\mathcal{O}}_X(U) \mathrel{\vcenter{:}}=\left\{{f:U\to k {~\mathrel{\Big|}~}f \in k(\mathbf{x}) \text{ locally}}\right\}$$
$${\mathcal{O}}_{X, p}$$ Germs of Regular functions?

0.3 Summary of Important Concepts

• What is an affine variety?
• What is the coordinate ring of an affine variety?
• What are the constructions $$V({\,\cdot\,})$$ and $$I({\,\cdot\,})$$?
• What is the Nullstellensatz?
• What are the definitions and some examples of:
• The Zariski topology?
• Irreducibility?
• Connectedness?
• Dimension?
• What is the definition of a presheaf?
• What are some examples and counterexamples?
• What is the definition of sheaf?
• What are some examples?
• What are some presheaves that are not sheaves?
• What is the definition of $${\mathcal{O}}_X$$, the sheaf of regular functions?
• How does one compute $${\mathcal{O}}_X$$ for $$X = D(f)$$ a distinguished open?
• What is a morphism between two affine varieties?
• What is the definition of separatedness?
• What are some examples of spaces that are and are not separated?
• What is a projective space?
• What is a projective variety?
• What is the projective coordinate ring?
• How does one take the closure of an affine variety $$X$$ in projective space?
• What is completeness?
• What are some examples and counterexamples of complete spaces?

0.4 Useful Examples

0.4.1 Varieties

• $$V(x-p)$$ a point.
• $$V(x)$$ a coordinate axis
• $$V(xy) \subseteq {\mathbb{A}}^2$$ the coordinate axes
• $$V(xy-1) \subseteq {\mathbb{A}}^2$$ a hyperbola
• $$V(x_1^2 - x_2^2 - 1) \subseteq {\mathbb{A}}^2_{/{\mathbb{C}}}$$
• $${\mathbb{A}}^2\setminus\left\{{0}\right\}$$ is not an affine variety or a distinguished open

0.4.2 Presheaves / Sheaves

• $$C^\infty({\,\cdot\,}, {\mathbb{R}})$$, a sheaf of smooth functions
• $$C^0({\,\cdot\,}, {\mathbb{R}})$$, a sheaf of continuous functions
• $$\underline{{\mathbb{R}}}({\,\cdot\,})$$, the constant sheaf associated to $${\mathbb{R}}$$ (locally constant real-valued functions)
• $$\operatorname{Hol}({\,\cdot\,}, {\mathbb{C}})$$, a sheaf of holomorphic functions
• $$K_p$$ the skyscraper sheaf: \begin{align*} K_p(U) \mathrel{\vcenter{:}}= \begin{cases} k & p\in U \\ 0 & \text{else}. \end{cases} \end{align*}
• $${\mathcal{O}}_X({\,\cdot\,})$$, the sheaf of regular functions on $$X$$

0.5 The Algebra-Geometry Dictionary

Let $$k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$, we’re setting up correspondences:

Algebra Geometry
$$k[x_1, \cdots, x_{n}]$$ $${\mathbb{A}}^n_{/k}$$
Maximal ideals $$\mathfrak{m}={x_1 - a_1, \cdots, x_n - a_n}$$ Points $$\mathbf{a} \mathrel{\vcenter{:}}={\left[ {a_1, \cdots, a_n} \right]} \in {\mathbb{A}}^n$$
Radical ideals $$J = \sqrt{J} {~\trianglelefteq~}k[x_1, \cdots, x_{n}]$$ $$V(J)$$ the zero locus
Prime ideals $$\mathfrak{p}\in \operatorname{Spec}(k[x_1, \cdots, x_{n}])$$ Irreducible closed subsets
Minimal prime ideals of $$A(X)$$ Irreducible components of $$X$$
$$I(S)$$ the ideal of a set $$S \subseteq {\mathbb{A}}^n$$ a subset
$$I + J$$ $$V(I) \cap V(J)$$
$$\sqrt{I(V) + I(W)}$$ $$V\cap W$$
$$I \cap J, IJ$$ $$V(I) \cup V(J)$$
$$I(V) \cap I(W), \sqrt{I(V)I(W)}$$ $$V \cup W$$
$$I(V) : I(W)$$ $$\mkern 1.5mu\overline{\mkern-1.5muV\setminus W\mkern-1.5mu}\mkern 1.5mu$$
$$k[x_1, \cdots, x_{n}]/I(X)$$ $$A(X)$$ (Functions on $$X$$)
$$A(X)$$ a domain $$X$$ is irreducible
$$A(X)$$ indecomposable $$X$$ is connected
$$k{\hbox{-}}$$algebra morphism $$A(X)\to A(Y)$$ Morphisms of varieties $$X\to Y$$
Krull dimension $$n$$ (chaints of primes) Topological dimension $$n$$ (chains of irreducibles)
Integral domains $$S(X)$$ Irreducible projective varieties $$X$$

1 Intro and Motivation (Friday, August 21)

1.1 Coordinate Rings

General idea: functions in a coordinate ring $$R[x_1, \cdots, x_n]/I$$ will correspond to the geometry of the variety cut out by $$I$$.

• $$x^2 + y^2 - 1$$ defines a circle, say, over $${\mathbb{R}}$$

• $$y^2 = x^3-x$$ gives an elliptic curve:

• $$x^n+y^n-1$$: does it even contain a $${\mathbb{Q}}{\hbox{-}}$$point? (Fermat’s Last Theorem)

• $$x^2 + 1$$, which has no $${\mathbb{R}}{\hbox{-}}$$points.

• $$x^2 + y^2 + 1/{\mathbb{R}}$$ vanishes nowhere, so its ring of functions is not $${\mathbb{R}}[x, y] / \left\langle{x^2 + y^2 + 1}\right\rangle$$. The problem: $${\mathbb{R}}$$ is not algebraically closed.

• $$x^2 - y^2 = 0$$ over $${\mathbb{C}}$$ is not a manifold (no chart at the origin):

• $$x+y+1/{\mathbb{F}}_3$$, which has 3 points over $${\mathbb{F}}_3^2$$, but $$f(x, y) = (x^3 - x)(y^3-y)$$ vanishes at every point

• Not possible when algebraically closed. For example, is there a nonzero polynomial that vanishes on every point in $${\mathbb{C}}$$?

• $$V(f) = {\mathbb{F}}_3^2$$, so the coordinate ring is zero instead of $${\mathbb{F}}_3[x, y]/\left\langle{f}\right\rangle$$ This is addressed by scheme theory.

1.2 Harnack Curve Theorem

If $$f \in {\mathbb{R}}[x, y]$$ is of degree $$d$$, then1 \begin{align*} \pi_1 V(f) \subseteq {\mathbb{R}}^2 \leq 1 + {(d-1)(d-2) \over 2} \end{align*}

Take the curve \begin{align*} X = \left\{{(x, y, z) = (t^3, t^4, t^5) \in {\mathbb{C}}^3 {~\mathrel{\Big|}~}t\in {\mathbb{C}}}\right\} .\end{align*}

Then $$X$$ is cut out by three equations:

• $$y^2 = xz$$
• $$x^2 = yz$$
• $$z^2 = x^2 y$$

Show that the vanishing locus of the first two equations above is $$X\cup L$$ where $$L$$ is a line.

Compare to linear algebra: codimension $$d$$ iff cut out by exactly $$d$$ equations.

1.3 Connection to Riemann Surfaces

Given the Riemann surface \begin{align*} y^2 = (x-1)(x-2)\cdots(x-2n) ,\end{align*} how does one visualize its solution set?

On $${\mathbb{C}}$$ with some slits, you can consistently choose a square root of the RHS.

Away from $$x=1, \cdots, 2n$$, there are two solutions for $$y$$ given $$x$$.

After gluing along strips, obtain:

2 The Nullstellensatz (Tuesday, August 25)

2.1 Radicals, Degrees, and Affine Varieties

Given $$f\in k[x_1, \cdots, x_n]$$, we’ll denote by $$f(a)$$ the value of $$f$$ at the point $$(a_1, \cdots, a_n)$$. Let $$k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ and $$R$$ a ring containing ideals $$I, J$$. Recall the definition of the radical of an ideal:

The radical of an ideal $$I {~\trianglelefteq~}R$$ is defined as \begin{align*} \sqrt{I} = \left\{{r\in R {~\mathrel{\Big|}~}r^k\in I \text{ for some } k\in {\mathbb{N}}}\right\} .\end{align*}

Let \begin{align*} I &= (x_1, x_2^2) \subset {\mathbb{C}}[x_1, x_2] \\ &= \left\{{ f_1 x_1 + f_2 x_2 {~\mathrel{\Big|}~}f_1, f_2 \in {\mathbb{C}}[x_1, x_2]}\right\} \end{align*}

Then $$\sqrt{I} = (x_1, x_2)$$, since $$x_2^2 \in I \implies x_2 \in \sqrt{I}$$.

Define $$\deg(f)$$ as the largest value of $$i_1 + \cdots + i_n$$ such that the coefficient of $$\prod x_j ^{i_j}$$ is nonzero.

$$\deg(x_1 + x_2^2 + x_1 x_2^3) = 4$$

1. Affine $$n{\hbox{-}}$$space2 $${\mathbb{A}}^n = {\mathbb{A}}_k^n$$ is defined as \begin{align*} {\mathbb{A}}^n \mathrel{\vcenter{:}}=\left\{{(a_1, \cdots, a_n) {~\mathrel{\Big|}~}a_i \in k}\right\} \end{align*}

2. Let $$S\subset k[x_1, \cdots, x_n]$$ be a set of polynomials.3 Then define the affine variety of $$S$$ as \begin{align*} V(S) \mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{A}}^n {~\mathrel{\Big|}~}f(x) = 0}\right\} \subset {\mathbb{A}}^n \end{align*}

• Let $$f(x) = 0$$, then $${\mathbb{A}}^n = V\qty{\left\{{f}\right\}}$$ is an affine variety.
• Any point $$(a_1, \cdots, a_n)\in {\mathbb{A}}^n$$ is an affine variety, uniquely determined by \begin{align*} V(x_1 - a_1, \cdots, x_n - a_n) = \left\{{a_1, \cdots, a_n}\right\} \end{align*}
• For any finite set $$r_1, \cdots, r_k \in {\mathbb{A}}^1$$, there exists a polynomial $$f\in k[x_1]$$ whose roots are $$r_i$$.

We may as well assume $$S$$ is an ideal by taking the ideal it generates, \begin{align*} S\subseteq \left\langle{S}\right\rangle = \left\{{\sum g_i f_i {~\mathrel{\Big|}~}g_i \in k[x_1, \cdots, x_n],\, f_i\in S}\right\} .\end{align*}

\begin{align*} V(S) = V\qty{\left\langle{S}\right\rangle} .\end{align*}

It’s clear that $$V(\left\langle{S}\right\rangle) \subset V(S)$$. Conversely, if $$f_1, f_2$$ vanish at $$x\in {\mathbb{A}}^n$$, then $$f_1 + f_2$$ and $$gf_1$$ also vanish at $$x$$ for all $$g\in k[x_1, \cdots, x_n]$$. Thus $$V(S) \subset V(\left\langle{S}\right\rangle)$$.

2.2 Ideals and Properties of $$V({\,\cdot\,})$$

See for a review of properties of ideals.

\begin{align*} S_1 \subseteq S_2 &\implies V(S_1) \supseteq V(S_2) \tag{1}\\ V(S_1) \cup V(S_2) &= V(S_1 S_2) = V(S_1 \cap S_2) \tag{2} \\ \bigcap V(S_i) &= V\qty{\bigcup S_i} \tag{3} .\end{align*}

We thus have a map \begin{align*} V: \left\{{\text{Ideals in } k[x_1, \cdots, x_n]}\right\} \to \left\{{\text{Affine varieties in } {\mathbb{A}}^n_{/k}}\right\} .\end{align*}

Let $$X\subset {\mathbb{A}}^n$$ be any set, then the ideal of $$X$$ is defined as \begin{align*} I(X) \mathrel{\vcenter{:}}=\left\{{f\in k[x_1, \cdots, x_n] {~\mathrel{\Big|}~}f(x) = 0\, \forall x\in X}\right\} .\end{align*}

Let $$X$$ be the union of the $$x_1$$ and $$x_2$$ axes in $${\mathbb{A}}^2$$, then \begin{align*} I(X) = \left\langle{x_1 x_2}\right\rangle = \left\{{g x_1 x_2 {~\mathrel{\Big|}~}g\in k[x_1, x_2]}\right\} .\end{align*}

\begin{align*} X_1 \subset X_2 \implies I(X_1) \supset I(X_2) .\end{align*}

If $$f\in I(X_2)$$, then $$f(x) = 0$$ for all $$x\in X_2$$. Since $$X_1 \subset X_2$$, we have $$f(x) = 0$$ for all $$x\in X_1$$, so $$f\in I(X_2)$$.

\begin{align*} I(X) = \sqrt{I(X)} ,\end{align*} i.e. $$I(X)$$ is a radical ideal.

It’s clear that \begin{align*} I(X) \subset \sqrt{I(X)} \mathrel{\vcenter{:}}=\left\{{f\in k[x_1, \cdots, x_{n}] {~\mathrel{\Big|}~}f^k \in I(X)}\right\} \end{align*} since we can simply take $$k=1$$ in this definition.

For a fixed $$f\in k[x_1, \cdots, x_{n}]$$ and any $$k\in {\mathbb{N}}$$, \begin{align*} f(x)^k = 0\,\, \forall x\in X \implies f(x) = 0\,\, \forall x\in X .\end{align*}

Granting this claim, if $$f\in \sqrt{I(X)}$$ then $$f^k \in I(X)$$ and thus $$f\in I(X)$$, completing the proof.

Without loss of generality, we can take $$n = 1$$ and consider $$k[x_1, \cdots, x_{n}]$$ Toward a contradiction, fix a $$k$$ suppose $$f(x) \neq 0$$ but $$f(x)^k = 0$$. Then writing $$f(x) = \sum_{j=1}^d \alpha_j x^j$$ where $$d \mathrel{\vcenter{:}}=\deg(f)$$, we have $$\alpha_d \neq 0$$ and \begin{align*} f(x)^k = \alpha_d^k x^{dk} + \cdots .\end{align*} Equating coefficients, we have $$\alpha_d^k = 0$$ in the base field. But fields have no nonzero nilpotents, so we arrive at a contradiction.

These maps thus yield correspondences \begin{align*} \left\{{\substack{\text{Ideals in } k[x_1, \cdots, x_n]}}\right\} &\xrightarrow{V} \left\{{\substack{\text{Affine Varieties}}}\right\} \\ \left\{{\substack{\text{Radical Ideals}}}\right\} &\xleftarrow{I} \left\{{\substack{\text{Affine Varieties}}}\right\} .\end{align*}

We’ll find that if we restrict to radical ideals, this will yield a bijective correspondence.

2.3 The Nullstellensatz: Statement and Proof

1. For any affine variety $$X$$, \begin{align*} V(I(X)) = X .\end{align*}

2. For any ideal $$J \subset k[x_1, \cdots, x_n]$$, \begin{align*} I(V(J)) = \sqrt{J} .\end{align*}

Thus there is a bijection between radical ideals and affine varieties.

Recall the Hilbert Basis Theorem (): any ideal in a finitely generated polynomial ring over a field is again finitely generated. We need to show 4 inclusions, 3 of which are easy.

If $$x\in X$$ then $$f(x) = 0$$ for all $$f\in I(X)$$. So $$x\in V(I(X))$$, since every $$f\in I(X)$$ vanishes at $$x$$.

If $$f\in \sqrt{J}$$ then $$f^k \in J$$ for some $$k$$. Then $$f^k(x) = 0$$ for all $$x\in V(J)$$. So $$f(x) = 0$$ for all $$x\in V(J)$$. Thus $$f\in I(V(J))$$.

Need to now use that $$X$$ is an affine variety. Counterexample: $$X = {\mathbb{Z}}^2 \subset {\mathbb{C}}^2$$, then $$I(X) = 0$$. But $$V(I(X)) = {\mathbb{C}}^2 \not\subset {\mathbb{Z}}^2$$. By (b), $$I(V(J)) \supset \sqrt{J} \supset J$$. Since $$V({\,\cdot\,})$$ is order-reversing, taking $$V$$ of both sides reverses the containment. So $$V(I(V(J))) \subset V(J)$$, i.e. $$V(I(X)) \subset X$$.

Thus the hard direction that remains is (d),

\begin{align*} I(V(J)) \subset \sqrt{J} \end{align*}

We’ll need Noether Normalization (), which is perhaps more important than the Nullstellensatz!

Suppose $$k$$ is algebraically closed and uncountable4. Then the maximal ideals in $$k[x_1, \cdots, x_n]$$ are given by \begin{align*} \operatorname{mSpec}\qty{k[x_1, \cdots, x_{n}]} = \left\{{ \left\langle{x_1 - a_1, \cdots, x_n - a_n}\right\rangle {~\mathrel{\Big|}~}a_j\in k}\right\} .\end{align*}

Let $${\mathfrak{m}}$$ be a maximal ideal, then by the Hilbert Basis Theorem (), $${\mathfrak{m}}= \left\langle{f_1, \cdots, f_r}\right\rangle$$ is finitely generated. Let $$L = {\mathbb{Q}}[\left\{{c_i}\right\}]$$ where the $$c_i$$ are all of the coefficients of the $$f_i$$ if $$\operatorname{ch}(K) = 0$$, or $${\mathbb{F}}_p[\left\{{c_i}\right\}]$$ if $$\operatorname{ch}(k) = p$$. Then $$L\subset k$$. Define \begin{align*} {\mathfrak{m}}_0 = {\mathfrak{m}}\cap L[x_1, \cdots, x_n] \end{align*} Note that by construction, $$f_i \in {\mathfrak{m}}_0$$ for all $$i$$, and we can write $${\mathfrak{m}}= {\mathfrak{m}}_0 \cdot k[x_1, \cdots, x_n]$$.

$${\mathfrak{m}}_0$$ is a maximal ideal.

If it were the case that \begin{align*} {\mathfrak{m}}_0 \subsetneq {\mathfrak{m}}_0' \subsetneq L[x_1, \cdots, x_n] ,\end{align*} then \begin{align*} {\mathfrak{m}}_0\cdot k[x_1, \cdots, x_n] \subsetneq {\mathfrak{m}}_0'\cdot k[x_1, \cdots, x_n] \subsetneq k[x_1, \cdots, x_n] .\end{align*}

So far, we’ve constructed a smaller polynomial ring and a maximal ideal in it. Thus $$L[x_1, \cdots, x_n]/{\mathfrak{m}}_0$$ is a field that is finitely generated over either $${\mathbb{Q}}$$ or $${\mathbb{F}}_p$$. So $$L[x_1, \cdots, x_n]/{\mathfrak{m}}_0$$ is finite over some $${\mathbb{Q}}(t_1, \cdots, t_n)$$, and since $$k$$ is uncountable, there exists an embedding $${\mathbb{Q}}(t_1, \cdots, t_n) \hookrightarrow k$$.5

This extends to an embedding of $$\phi: L[x_1, \cdots, x_n]/{\mathfrak{m}}_0 \hookrightarrow k$$ since $$k$$ is algebraically closed. Letting $$a_i$$ be the image of $$x_i$$ under $$\phi$$, then $$f(a_1, \cdots, a_n) = 0$$ by construction, $$f_i \in (x_i - a_i)$$ implies that $${\mathfrak{m}}= (x_i - a_i)$$ by maximality.

3 More Nullstellensatz (Thursday, August 27)

3.1 Consequence of the Nullstellensatz

Recall Hilbert’s Nullstellensatz:

1. For any affine variety, $$V(I(X)) = X$$.

2. For any ideal $$J{~\trianglelefteq~}k[x_1, \cdots, x_n]$$, $$I(V(J)) = \sqrt{J}$$.

So there’s an order-reversing bijection \begin{align*} \left\{{\substack{\text{Radical ideals } k[x_1, \cdots, x_n]}}\right\} \mathrel{\operatorname*{\rightleftharpoons}_{V({\,\cdot\,})}^{I({\,\cdot\,})}} \left\{{\substack{\text{Affine varieties in } {\mathbb{A}}^n}}\right\} .\end{align*}

In proving $$I(V(J)) \subseteq \sqrt{J}$$, we needed Noether Normalization and an important theorem (): the maximal ideals of $$k[x_1, \cdots, x_n]$$ are of the form $$\left\langle{x-a_1, \cdots, x-a_n}\right\rangle$$.

If $$V(I)$$ is empty, then $$I = \left\langle{1}\right\rangle$$.

The only ideals that vanish nowhere are trivial.

This is because no common vanishing locus $$\implies$$ trivial ideal, so there’s a linear combination that equals 1. By contrapositive, suppose $$I\neq \left\langle{1}\right\rangle$$. By Zorn’s Lemma, these exists a maximal ideals $${\mathfrak{m}}$$ such that $$I \subset {\mathfrak{m}}$$. By the order-reversing property of $$V({\,\cdot\,})$$, $$V({\mathfrak{m}}) \subseteq V(I)$$. By the classification of maximal ideals, $${\mathfrak{m}}= \left\langle{x-a_1, \cdots, x-a_n}\right\rangle$$, so $$V({\mathfrak{m}}) = \left\{{a_1, \cdots, a_n}\right\}$$ is nonempty.

3.2 Proof of Remaining Part of Nullstellensatz

We now return to the remaining hard part of the proof of the Nullstellensatz:

\begin{align*} I(V(J)) \subseteq \sqrt{J} \end{align*}

Let $$f\in V(I(J))$$, we want to show $$f\in \sqrt{J}$$. Consider the ideal \begin{align*} \tilde J \mathrel{\vcenter{:}}= J + \left\langle{ft - 1}\right\rangle \subseteq k[x_1, \cdots, x_n, t] \end{align*}

$$f = 0$$ on all of $$V(J)$$ by the definition of $$I(V(J))$$.

However, if $$f=0$$, then $$ft-1 \neq 0$$, so \begin{align*} V(\tilde J) = V(G) \cap V(ft-1) = \emptyset \end{align*}

Applying the corollary $$\tilde J = (1)$$, so \begin{align*} 1 = \left\langle{ft-1}\right\rangle g_0(x_1, \cdots, x_n, t) + \sum f_i g_i(x_1, \cdots, x_n, t) \end{align*} with $$f_i \in J$$. Let $$t^N$$ be the largest power of $$t$$ in any $$g_i$$. Thus for some polynomials $$G_i$$, we have \begin{align*} f^N \mathrel{\vcenter{:}}=(ft-1) G_0(x_1, \cdots, x_n, ft) + \sum f_i G_i(x_1, \cdots, x_n, ft) \end{align*} noting that $$f$$ does not depend on $$t$$. Now take $$k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle$$, so $$ft=1$$ in this ring. This kills the first term above, yielding \begin{align*} f^N = \sum f_i G_i(x_1, \cdots, x_n, 1) \in k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle .\end{align*}

There is an inclusion \begin{align*} k[x_1, \cdots, x_n] \hookrightarrow k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle .\end{align*}

Since this is injective, this identity also holds in $$k[x_1, \cdots, x_n]$$. But $$f_i\in J$$, so $$f\in \sqrt{J}$$.

Why is the claim above true?

Consider $$k[x]$$. If $$J\subset k[x]$$ is an ideal, it is principal, so $$J = \left\langle{f}\right\rangle$$. We can factor $$f(x) = \prod_{i=1}^k (x-a_i)^{n_i}$$ and $$V(f) = \left\{{a_1, \cdots, a_k}\right\}$$. Then \begin{align*} I(V(f)) = \left\langle{(x-a_1)(x-a_2)\cdots(x-a_k)}\right\rangle = \sqrt{J} \subsetneq J ,\end{align*} so this loses information.

Let $$J = \left\langle{x-a_1, \cdots, x-a_n}\right\rangle$$, then $$I(V(J)) = \sqrt{J} = J$$ with $$J$$ maximal. Thus there is a correspondence \begin{align*} \left\{{\substack{\text{Points of } {\mathbb{A}}^n}}\right\} \iff \left\{{\substack{\text{Maximal ideals of }k[x_1, \cdots, x_n]}}\right\} .\end{align*}

\begin{align*} I(X_1 \cup X_2) &= I(X_1) \cap I(X_2) \tag{a} \\ I(X_1) \cap I(X_2) &= \sqrt{I(X_1) + I(X_2)} \tag{b} .\end{align*}

We proved (a) on the variety side. For (b), by the Nullstellensatz we have $$X_i = V(I(X_i))$$, so \begin{align*} I(X_1\cap X_2) &= I\qty{ VI(X_1) \cap VI(X_2)} \\ &= IV\qty{I(X_1) + I(X_2)} \\ &= \sqrt{I(X_1) + I(X_2)} .\end{align*}

Example of property (b):

Take $$X_1 = V(y-x^2)$$ and $$X_2 = V(y)$$, a parabola and the $$x{\hbox{-}}$$axis.

Then $$X_1 \cap X_2 = \left\{{(0, 0)}\right\}$$, and $$I(X_1) + I(X_2) = \left\langle{y-x^2, y}\right\rangle = \left\langle{x^2, y}\right\rangle$$, but \begin{align*} I(X_1 \cap X_2) = \left\langle{x, y}\right\rangle = \sqrt{\left\langle{x^2, y}\right\rangle} \end{align*}

If $$f, g\in k[x_1, \cdots, x_n]$$, and suppose $$f(x) = g(x)$$ for all $$x\in {\mathbb{A}}^n$$. Then $$f = g$$.

Since $$f-g$$ vanishes everywhere, \begin{align*} f-g \in I({\mathbb{A}}^n) = I(V(0)) = \sqrt{0} = 0 \end{align*}

More generally suppose $$f(x) = g(x)$$ for all $$x\in X$$, where $$X$$ is some affine variety. Then by definition, $$f-g \in I(X)$$, so a “natural” space of functions on $$X$$ is $$k[x_1,\cdots, x_n]/I(X)$$.

For an affine variety $$X$$, the coordinate ring of $$X$$ is \begin{align*} A(X) \mathrel{\vcenter{:}}= k[x_1, \cdots, x_n]/ I(X) .\end{align*}

Elements $$f\in A(X)$$ are called polynomial or regular functions on $$X$$.

The constructions $$V({\,\cdot\,}), I({\,\cdot\,})$$ work just as well with $$A(X)$$ instead of $$k[x_1, \cdots, x_{n}]$$ and $$X$$ instead of $${\mathbb{A}}^n$$.

Given any $$S\subset A(Y)$$ for $$Y$$ an affine variety, \begin{align*} V(S) = V_Y(S) \mathrel{\vcenter{:}}=\left\{{x\in Y {~\mathrel{\Big|}~}f(x) = 0\,\,\forall f\in S}\right\} .\end{align*}

Given $$X\subset Y$$ a subset, \begin{align*} I(X) = I_Y(X) \mathrel{\vcenter{:}}=\left\{{f\in A(Y) {~\mathrel{\Big|}~}f(x) = 0\,\,\forall x\in X}\right\} \subseteq A(Y) .\end{align*}

For $$X\subset Y \subset {\mathbb{A}}^n$$, we have $$I(X) \supset I(Y) \supset I({\mathbb{A}}^n)$$, so we have maps

Let $$X\subset Y$$ be an affine subvariety, then

1. $$A(X) = A(Y) / I_Y(X)$$

2. There is a correspondence \begin{align*} \left\{{\substack{\text{Affine subvarieties of }Y}}\right\} &\iff \left\{{\substack{\text{Radical ideals in }A(Y)}}\right\} \\ X &\mapsto I_Y(X) \\ V_Y(J) &\mapsfrom J .\end{align*}

Properties are inherited from the case of $${\mathbb{A}}^n$$, see exercise in Gathmann.

Let $$Y = V(y-x^2) \subset {\mathbb{A}}^2/{\mathbb{C}}$$ and $$X = \left\{{(1, 1)}\right\} = V(x-1, y-1)\subset {\mathbb{A}}^2/{\mathbb{C}}$$.

Then there is an inclusion $$\left\langle{y-x^2}\right\rangle \subset \left\langle{x-1, y-1}\right\rangle$$, e.g. by Taylor expanding about the point $$(1, 1)$$. and thus there is a map

4 Zariski Topology (Tuesday, September 01)

4.1 The Zariski Topology

Last time: \begin{align*} V(I) &= \left\{{x\in {\mathbb{A}}^n {~\mathrel{\Big|}~}f(x) = 0 \, \forall x\in I}\right\} \\ I(X) &= \left\{{f\in k[x_1, \cdots, x_n] {~\mathrel{\Big|}~}f(x) = 0\, \forall x\in X}\right\} .\end{align*}

We proved the Nullstellensatz $$I(V(J)) = \sqrt{J}$$, defined the coordinate ring of an affine variety $$X$$ as \begin{align*} A(X) \mathrel{\vcenter{:}}= k[x_1, \cdots, x_n] / I(X) \end{align*} the ring of regular (polynomial) functions on $$X$$. Recall that a topology on $$X$$ can be defined as a collection of closed subsets of $$X$$ that are closed under arbitrary intersections and finite unions. A subset $$Y\subset X$$ inherits a subspace topology with closed sets of the form $$Z\cap Y$$ for $$Z\subset X$$ closed in $$X$$.

Let $$X$$ be an affine variety. The closed sets are affine subvarieties $$Y\subset X$$.

This satisfies the axioms for a topological space.

• We have $$\emptyset, X$$ closed, since

1. $$V_X(1) = \emptyset$$,
2. $$V_X(0) = X$$
• Closure under finite unions:

Let $$V_X(I), V_X(J)$$ be closed in $$X$$ with $$I, J \subset A(X)$$ ideals. Then $$V_X(IJ) = V_X(I) \cup V_X(J)$$.

• Closure under intersections:

We have $$\bigcap_{i\in \sigma} V_X(J) = V_X\qty{ \sum_{i\in \sigma} J_i}$$.

There are few closed sets, so this is a “weak” topology.

Compare the classical topology on $${\mathbb{A}}^1_{/{\mathbb{C}}}$$ to the Zariski topology. Consider the set \begin{align*} A\mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{A}}^1_{/{\mathbb{C}}} {~\mathrel{\Big|}~}{\left\lVert {x} \right\rVert} \leq 1}\right\} \end{align*} which is closed in the classical topology. However, $$A$$ is not closed in the Zariski topology, since the closed subsets are finite sets or the whole space. In fact, the topology here is the cofinite topology.

Let $$f: {\mathbb{A}}^1_{/k} \to {\mathbb{A}}^1_{/k}$$ be any injective map. Then $$f$$ is necessarily continuous wrt the Zariski topology. Thus the notion of continuity is too weak in this situation.

Consider $$X\times Y$$ a product of affine varieties. Then there is a product topology where open sets are of the form $$\bigcup_{i=1}^n U_i \times V_i$$ with $$U_i, V_i$$ open in $$X, Y$$ respectively. This is the wrong topology! On $${\mathbb{A}}^1 \times{\mathbb{A}}^1 = {\mathbb{A}}^2$$, the diagonal $$\Delta \mathrel{\vcenter{:}}= V(x-y)$$ is closed in the Zariski topology on $${\mathbb{A}}^2$$ but not in the product topology.

4.2 Irreducibility and Connectedness

Consider $${\mathbb{A}}^2_{/{\mathbb{C}}}$$, so the closed sets are curves and points. Observation: $$V(x_1 x_2 ) \subset {\mathbb{A}}^2_{/{\mathbb{C}}}$$ decomposed into the union of the coordinate axes $$X_1 \mathrel{\vcenter{:}}= V(x_1)$$ and $$X_2 \mathrel{\vcenter{:}}= V(x_2)$$. The Zariski topology can detect these decompositions.

Let $$X$$ be a topological space.

1. $$X$$ is reducible iff there exist nonempty proper closed subsets $$X_1 ,X_2 \subset X$$ such that $$X = X_1 \cup X_2$$. Otherwise, $$X$$ is said to be irreducible.

2. $$X$$ is disconnected if there exist $$X_1, X_2 \subset X$$ such that $$X = X_1 {\coprod}X_2$$. Otherwise, $$X$$ is said to be connected.

$$V(x_1 x_2)$$ is reducible but connected.

$${\mathbb{A}}^1_{/{\mathbb{C}}}$$ is not irreducible, since we can write \begin{align*} {\mathbb{A}}^1_{/{\mathbb{C}}} = \left\{{{\left\lVert {x} \right\rVert} \leq 1}\right\} \cup\left\{{{\left\lVert {x} \right\rVert} \geq 1}\right\} \end{align*}

Let $$X$$ be a disconnected affine variety with $$X = X_1 {\coprod}X_2$$. Then $$A(X) \cong A(X_1) \times A(X_2)$$.

We have \begin{align*} X_1 \cup X_2 = X \implies I(X_1) \cap I(X_2) = I(X) = (0) \in A(X) ,\end{align*} recalling that the coordinate ring $$A(X)$$ is a quotient by $$I(X)$$. Since $$X_1 \cap X_1 = \emptyset$$, we have \begin{align*} I(X_1 \cap X_2) = \sqrt{I(X_1) + I(X_2) } = I(\emptyset) = \left\langle{1}\right\rangle .\end{align*}

Thus $$I(X_1) + I(X_2) = \left\langle{1}\right\rangle$$, and by the Chinese Remainder Theorem, the following map is an isomorphism: \begin{align*} A(X) \to A(X)/I(X_1) \times A(X) / I(X_2) .\end{align*}

However, the codomain is precisely $$A(X_1) \times A(X_2)$$.

4.2.1 Irreducibility on the Algebra Side

An affine variety $$X$$ is irreducible $$\iff$$ $$A(X)$$ is an integral domain.

$$\implies$$: By contrapositive, suppose $$f_1, f_2 \in A(X)$$ are nonzero with $$f_1 f_2 = 0$$. Let $$X_i \mathrel{\vcenter{:}}= V(f_i)$$, then \begin{align*} X= V(0) = V(f_1 f_2) = X_1 \cup X_2 \end{align*} which are closed and proper since $$f_i \neq 0$$.

$$\impliedby$$: Suppose $$X$$ is reducible with $$X = X_1 \cup X_2$$ with $$X_i$$ proper and closed. Define $$J_i \mathrel{\vcenter{:}}= I(X_i)$$, then by part (a) of the Nullstellensatz. \begin{align*} V(J_i) = V(I(X_i)) = X_i \implies J_i \neq 0 .\end{align*} So there exists a nonzero $$f_i \in J_i = I(X_i)$$, so $$f_i$$ vanishes on $$X_i$$. But then \begin{align*} V(f_1) \cup V(f_2) \supset X_1 \cup X_2 = X ,\end{align*} so $$X= V(f_1 f_2)$$ and $$f_1 f_2 \in I(X) = \left\langle{0}\right\rangle$$ and $$f_1 f_2 = 0$$, and $$A(X)$$ is thus not a domain.

Let $$X = \left\{{\mathbf{p}^1, \cdots, \mathbf{p}^d}\right\}$$ be a finite set in $${\mathbb{A}}^n$$. The Zariski topology on $$X$$ is the discrete topology, and $$X = {\coprod}_{i=1}^d \left\{{\mathbf{p}^i}\right\}$$. So \begin{align*} A(X) = A\qty{\displaystyle\coprod_{i=1}^d \left\{{\mathbf{p}^i}\right\}} = \prod_{i=1}^d A\qty{\left\{{\mathbf{p}^i}\right\}} = \prod_{i=1}^d \frac{ k[x_1, \cdots, x_n]} {\left\langle{x_1 - p^i_1, \cdots, x_n - p^i_n}\right\rangle } \end{align*} where $$p_j^i$$ is the $$j$$th component of $$\mathbf{p}^i$$.

Set $$V(x_1 x_2) = X$$, then $$A(X) = k[x_1, x_2]/ \left\langle{x_1 x_2}\right\rangle$$. This not being a domain (since $$x_1 x_2 = 0$$) corresponds to $$X = V(x_1) \cup V(x_2)$$ not being irreducible.

Let $$X_1$$ be the $$xy{\hbox{-}}$$plane and $$X_2$$ be the line parallel to the $$y{\hbox{-}}$$axis through $${\left[ {0,0,1} \right]}$$, and let $$X= X_1 {\coprod}X_2$$:

Then $$X_1 = V(z)$$ and $$X_2 = V(x, z-1)$$, and \begin{align*} I(X) = \left\langle{z}\right\rangle \cdot \left\langle{x, z-1}\right\rangle= \left\langle{xz, z^2 - z}\right\rangle \end{align*} The coordinate ring is then given by \begin{align*} A(X) = { {\mathbb{C}}[x, y, z] \over \left\langle{xz, z^2 - z}\right\rangle } = { {\mathbb{C}}[x, y, z] \over \left\langle{z}\right\rangle } \oplus { {\mathbb{C}}[x, y,z] \over \left\langle{x, z-1}\right\rangle } \end{align*}

5 Irreducibility (Thursday, September 03)

5.1 Irreducibility and Prime Ideals

Recall that the Zariski topology is defined on an affine variety $$X = V(J)$$ with $$J {~\trianglelefteq~}k[x_1, \cdots, x_n]$$ by describing the closed sets.

$$X$$ is irreducible if its coordinate ring $$A(X)$$ is a domain.

There is a 1-to-1 correspondence \begin{align*} \left\{{\substack{\text{Irreducible subvarieties} \\ \text{of }X}}\right\} \iff \left\{{\substack{\text{Prime ideals} \\ \text{in }A(X)}}\right\} .\end{align*}

Suppose $$Y\subset X$$ is an affine subvariety. Then \begin{align*} A(X) / I_X(Y) = A(Y) .\end{align*}

By the Nullstellensatz, there is a bijection between subvarieties of $$X$$ and radical ideals of $$A(X)$$ where $$Y\mapsto I_X(Y)$$. A quotient is a domain iff quotienting by a prime ideal, so $$A(Y)$$ is a domain iff $$I_X(Y)$$ is prime.

Recall that $${\mathfrak{p}}{~\trianglelefteq~}R$$ is prime when $$fg\in {\mathfrak{p}}\iff f\in {\mathfrak{p}}$$ or $$g\in {\mathfrak{p}}$$. Thus $$\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu = 0$$ in $$R/{\mathfrak{p}}$$ implies $$\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu = 0$$ or $$\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu = 0$$ in $$R/{\mathfrak{p}}$$, i.e. $$R/{\mathfrak{p}}$$ is a domain. Finally, note that prime ideals are radical (easy proof).

Consider $${\mathbb{A}}^2/{\mathbb{C}}$$ and some subvarieties $$C_i$$: