Prologue

0.1 References

0.2 Notation

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

0.4 Useful Examples

0.4.1 Varieties

0.4.2 Presheaves / Sheaves

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 \(I\).

An elliptic curve.
A non-manifold curve.

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:

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.

Choosing a square root of a polynomial.

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

After gluing along strips, obtain:

Glusing along strips to obtain a Riemann surface.

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*}

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*}

Effect, a hyperbolic tube around V(J), so both can’t vanish

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.

Intersecting V(y-x^2) and V(y)

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.

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 \(X= X_1 {\coprod}X_2\):

Union of a plane and a parallel line.

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 \(C_i\):