Algebraic geometry · The Lumenary

MotivationH

Equations are shapes

Draw the line through the point $(-1, 0)$ of the unit circle $x^2 + y^2 = 1$ with slope $t$ . Substituting $y = t(x + 1)$ gives a quadratic equation in $x$ with the known root $x = -1$ , so the other root is forced to be rational in $t$ :

(x, y) = \left( \frac{1 - t^2}{1 + t^2},\; \frac{2t}{1 + t^2} \right).

Every rational slope gives a rational point and every rational point other than $(-1, 0)$ comes from a rational slope: for instance $t = 2/3$ gives $(\tfrac{5}{13}, \tfrac{12}{13})$ and, clearing denominators, the Pythagorean triple $5^2 + 12^2 = 13^2$ . Diophantus used a version of this trick in the third century. It rests on a geometric fact — a line meets a conic in two points — which is false as stated (a line can miss a circle, or touch it) and becomes exactly true only after three repairs: allow complex points, add points at infinity, and count tangency twice. The result is Bézout's theorem: two plane curves of degrees $d$ and $e$ without common component meet in exactly $de$ points.

Algebraic geometry studies the sets of solutions of systems of polynomial equations — varieties — and the polynomial maps between them. Its central idea is a dictionary: a variety $X$ is controlled by its ring of polynomial functions $k[X]$ , points correspond to maximal ideals, subvarieties to prime ideals, and maps to ring homomorphisms in the opposite direction. Grothendieck turned the dictionary into a definition: every commutative ring $R$ is a space, $\operatorname{Spec} R$ , and gluing such spaces gives schemes. The same language then describes curves over $\C$ , curves over finite fields and the integers themselves.

This lesson assumes commutative algebra (prime ideals, localisation, Noetherian rings and the Nullstellensatz) and point-set topology. Throughout, $k$ is an algebraically closed field (such as $\C$ ) unless said otherwise, and $\mathbb{A}^n = k^n$ is affine $n$ -space.

IntuitionH

The dictionary

geometry	algebra
affine algebraic set $X \subseteq \mathbb{A}^n$	radical ideal $I(X) \subseteq k[x_1, \dots, x_n]$
irreducible $X$ (an affine variety)	prime ideal
point $a \in \mathbb{A}^n$	maximal ideal $(x_1 - a_1, \dots, x_n - a_n)$
polynomial functions on $X$	coordinate ring $k[X] = k[x_1, \dots, x_n]/I(X)$
polynomial map $X \to Y$	$k$ -algebra homomorphism $k[Y] \to k[X]$
dimension of $X$	Krull dimension of $k[X]$
$X \cup Y$ , $X \cap Y$	$I(X) \cap I(Y)$ , $\sqrt{I(X) + I(Y)}$

Three warnings shape everything that follows.

The topology is coarse. Closed sets are zero sets of polynomials, so in the line $\mathbb{A}^1$ they are just the finite sets and the whole line. Two non-empty open sets always meet.
Points are not enough. The cusp $y^2 = x^3$ and the line $\mathbb{A}^1$ are in bijection by a polynomial map, but they are not isomorphic: the rings of functions differ. The ideals $(x)$ and $(x^2)$ have the same zero set, but $(x^2)$ remembers a "double point". Rings, not point sets, carry the structure.
Affine space is incomplete. Parallel lines do not meet, and a line may miss a circle. Projective space adds the missing points at infinity; complex numbers add the missing complex points.

DefinitionsH

Definitions

Definition— algebraic sets and their ideals

For $S \subseteq k[x_1, \dots, x_n]$ the affine algebraic set of $S$ is $V(S) = \{a \in \mathbb{A}^n : f(a) = 0 \text{ for all } f \in S\}$ . For $X \subseteq \mathbb{A}^n$ its ideal is $I(X) = \{f : f(a) = 0 \text{ for all } a \in X\}$ , always a radical ideal. $V(S) = V(\langle S \rangle)$ , and by Hilbert's basis theorem every algebraic set is cut out by finitely many polynomials.

Definition— Zariski topology; irreducibility; variety

The Zariski topology on $\mathbb{A}^n$ has the algebraic sets as its closed sets; a subset of $\mathbb{A}^n$ gets the subspace topology. A non-empty topological space is irreducible if it is not the union of two proper closed subsets; equivalently, any two non-empty open subsets meet. An affine variety is an irreducible affine algebraic set. (Some authors, e.g. in scheme theory, allow reducible "varieties"; we follow Hartshorne.)

Definition— coordinate ring, regular functions, morphisms

The coordinate ring of an algebraic set $X \subseteq \mathbb{A}^n$ is $k[X] = k[x_1, \dots, x_n]/I(X)$ , the ring of polynomial functions on $X$ ; it is a finitely generated, reduced (no non-zero nilpotents) $k$ -algebra, and a domain exactly when $X$ is a variety. A morphism $\varphi : X \to Y$ of algebraic sets $X \subseteq \mathbb{A}^n$ , $Y \subseteq \mathbb{A}^m$ is a map $\varphi = (\varphi_1, \dots, \varphi_m)$ with $\varphi_i \in k[X]$ and $\varphi(X) \subseteq Y$ . It induces the comorphism $\varphi^* : k[Y] \to k[X]$ , $g \mapsto g \circ \varphi$ . An isomorphism is a morphism with a morphism inverse.

Definition— function field and dimension

For a variety $X$ the function field $k(X)$ is the fraction field of $k[X]$ ; its elements are rational functions, defined on a dense open subset. Varieties with isomorphic function fields are birational. The dimension of $X$ is the supremum of lengths $n$ of chains $Z_0 \subsetneq Z_1 \subsetneq \cdots \subsetneq Z_n$ of irreducible closed subsets; it equals the Krull dimension of $k[X]$ and the transcendence degree of $k(X)$ over $k$ . Curves have dimension $1$ , surfaces $2$ .

Definition— projective space and projective algebraic sets

Projective $n$ -space $\mathbb{P}^n$ is the set of lines through the origin in $k^{n+1}$ : points are written $[a_0 : \cdots : a_n]$ , not all $a_i = 0$ , with $[a_0 : \cdots : a_n] = [\lambda a_0 : \cdots : \lambda a_n]$ for $\lambda \in k^\times$ . A polynomial $F$ does not have values on $\mathbb{P}^n$ , but if $F$ is homogeneous the condition $F(a) = 0$ is well defined. Projective algebraic sets are common zero sets of homogeneous polynomials (equivalently of a homogeneous ideal), and they are the closed sets of the Zariski topology on $\mathbb{P}^n$ . The standard charts $U_i = \{a_i \neq 0\} \cong \mathbb{A}^n$ , $[a_0 : \cdots : a_n] \mapsto (a_0/a_i, \dots, \widehat{a_i/a_i}, \dots, a_n/a_i)$ , cover $\mathbb{P}^n$ . The projective closure of an affine algebraic set is its closure in $\mathbb{P}^n \supseteq U_0$ , obtained by homogenising the equations.

Definition— plane curves and intersection multiplicity

A projective plane curve of degree $d$ is a non-zero homogeneous $F \in k[x, y, z]$ of degree $d$ , up to scalar multiples (it may have repeated factors, which count as multiple components). For curves $F, G$ and a point $P$ in the chart $z \neq 0$ , with $f, g$ their dehomogenisations, the intersection multiplicity is

I_P(F, G) = \dim_k \mathcal{O}_P / (f, g),

where $\mathcal{O}_P = k[x, y]_{\mathfrak{m}_P}$ is the local ring of the plane at $P$ (rational functions defined at $P$ ); points in other charts are treated in the same way. $I_P$ is finite exactly when $F$ and $G$ have no common component through $P$ , is $0$ when $P$ is not on both curves, is $1$ exactly when $P$ is a smooth point of both curves with different tangent lines, and for a line $L$ equals the order of vanishing at $P$ of $F$ restricted to $L$ (Fulton, Algebraic Curves, §3.3).

Definition— presheaves and sheaves

A presheaf $\mathcal{F}$ of rings on a space $X$ assigns a ring $\mathcal{F}(U)$ to each open $U$ and restriction maps $\mathcal{F}(U) \to \mathcal{F}(V)$ , $s \mapsto s|_V$ , for $V \subseteq U$ , compatibly with composition. It is a sheaf if sections are determined locally and compatible local sections glue: for every open cover $U = \bigcup U_i$ , the sequence

0 \to \mathcal{F}(U) \to \prod_i \mathcal{F}(U_i) \rightrightarrows \prod_{i, j} \mathcal{F}(U_i \cap U_j)

is exact (an equaliser). On a variety, the structure sheaf $\mathcal{O}_X$ assigns to $U$ the ring of regular functions on $U$ : functions that are locally quotients $g/h$ of polynomials with $h \neq 0$ .

Definition— Spec R, the structure sheaf, schemes

For a commutative ring $R$ , $\operatorname{Spec} R$ is the set of prime ideals with the Zariski topology (closed sets $V(I) = \{\mathfrak{p} \supseteq I\}$ ), with basic open sets $D(f) = \{\mathfrak{p} : f \notin \mathfrak{p}\}$ . Its structure sheaf $\mathcal{O}$ is the unique sheaf of rings with $\mathcal{O}(D(f)) = R_f = R[1/f]$ and restriction maps the localisation maps; its stalk at $\mathfrak{p}$ is the local ring $R_\mathfrak{p}$ . The pair $(\operatorname{Spec} R, \mathcal{O})$ is an affine scheme. A scheme is a space with a sheaf of rings (a locally ringed space) that is locally isomorphic to affine schemes; morphisms of schemes are morphisms of locally ringed spaces.

Worked examplesH

Examples and non-examples

Example— The affine line and the cofinite topology

A non-zero polynomial in one variable has finitely many roots, so the algebraic subsets of $\mathbb{A}^1$ are $\mathbb{A}^1$ itself and the finite sets: the Zariski topology on $\mathbb{A}^1$ is the cofinite topology. It is $T_1$ (points are closed) but not Hausdorff, because $k$ is infinite and any two cofinite sets meet. In $\mathbb{A}^2$ the closed sets are finite unions of points and curves, and the Zariski topology on $\mathbb{A}^2 = \mathbb{A}^1 \times \mathbb{A}^1$ is not the product topology: the diagonal $x = y$ is closed in $\mathbb{A}^2$ but not in the product of two cofinite topologies.

Example— A parabola is a line; a hyperbola is not

The parabola $P = V(y - x^2)$ is isomorphic to $\mathbb{A}^1$ : the morphisms $t \mapsto (t, t^2)$ and $(x, y) \mapsto x$ are mutually inverse, and correspondingly $k[P] = k[x, y]/(y - x^2) \cong k[t]$ .

The hyperbola $H = V(xy - 1)$ is not: $k[H] = k[x, y]/(xy - 1) \cong k[t, t^{-1}]$ , in which $t$ is a non-constant unit, whereas the units of $k[t]$ are the non-zero constants. An isomorphism $H \cong \mathbb{A}^1$ would give a ring isomorphism $k[t] \cong k[t, t^{-1}]$ , preserving units. (Projection $(x, y) \mapsto x$ does identify $H$ with the open subset $\mathbb{A}^1 \setminus \{0\}$ — so an open subset of an affine variety can itself be affine.) Note that $H$ and $\mathbb{A}^1$ are homeomorphic in their Zariski topologies — both carry the cofinite topology on a set of cardinality $|k|$ , so any bijection is a homeomorphism: topology alone cannot tell them apart.

Example— The cusp: a bijection that is not an isomorphism

Let $C = V(y^2 - x^3)$ . The morphism $\varphi : \mathbb{A}^1 \to C$ , $t \mapsto (t^2, t^3)$ , is bijective: its inverse is $(x, y) \mapsto y/x$ for $x \neq 0$ and $(0, 0) \mapsto 0$ . But this inverse is not a polynomial map, and no inverse morphism exists: the comorphism $\varphi^* : k[C] \to k[t]$ sends $x \mapsto t^2$ , $y \mapsto t^3$ with image $k[t^2, t^3]$ , which does not contain $t$ (formalised below). So $\varphi^*$ is not an isomorphism, and neither is $\varphi$ . Geometrically, $C$ has a singular point at the origin (both partial derivatives of $y^2 - x^3$ vanish there), while $\mathbb{A}^1$ is smooth. Algebraically, $k[C]$ is not integrally closed, and $\varphi$ is its normalisation.

Example— Reducible sets and non-radical ideals

$V(xy) \subseteq \mathbb{A}^2$ is the union of the two axes $V(x) \cup V(y)$ ; its ideal $(xy)$ is radical but not prime. The ideals $(x^2y, xy^2)$ , $(xy)$ and $(x) \cap (y)$ all have this zero set, and the Nullstellensatz gives $I(V(x^2y, xy^2)) = \sqrt{(x^2y, xy^2)} = (xy)$ . Likewise $V(x^2) = V(x)$ in $\mathbb{A}^1$ : the varieties cannot see the difference between $(x)$ and $(x^2)$ , but schemes can — $\operatorname{Spec} k[x]/(x^2)$ is a point with a tangent direction attached.

Example— Points at infinity

Homogenising with a new variable $z$ embeds $\mathbb{A}^2$ as the chart $z = 1$ of $\mathbb{P}^2$ ; the complement is the line at infinity $z = 0$ .

The parallel lines $y = 0$ and $y = 1$ become $y = 0$ and $y = z$ , which meet at $[1 : 0 : 0]$ : parallel lines meet at infinity, at the point determined by their common direction.
The hyperbola $xy = 1$ becomes $xy = z^2$ and meets the line at infinity at $[1 : 0 : 0]$ and $[0 : 1 : 0]$ — the directions of its asymptotes.
The parabola $y = x^2$ becomes $yz = x^2$ and meets $z = 0$ only at $[0 : 1 : 0]$ , where $x^2 = 0$ gives a double root: the parabola is tangent to the line at infinity.
Every circle $x^2 + y^2 + axz + byz + cz^2 = 0$ passes through the two circular points $[1 : \pm i : 0]$ . So two circles, which meet in at most $2$ affine points, have their remaining $2$ intersections there.

Over $\R$ , ellipse, parabola and hyperbola are projectively equivalent, distinguished only by how they meet the line at infinity: in no real point, in one double point, or in two real points.

Example— Spec of the integers

The points of $\operatorname{Spec}\Z$ are $(2), (3), (5), (7), \dots$ — one closed point for each prime — together with the generic point $(0)$ , whose closure is the whole space. The function $n \in \Z$ has the "value" $n \bmod p$ in $\mathbb{F}_p$ at the point $(p)$ , and vanishes exactly at the primes dividing $n$ . $\operatorname{Spec}\Z$ looks like a curve: it has dimension $1$ , its non-empty open sets are cofinite and contain the generic point, and the rings of integers of number fields give "curves" lying over it.

Spec ℤ drawn as a curve (after Mumford): one closed point for each prime, and a generic point (0) spread along the whole line, lying in every non-empty open set.

ProofH

The main theorems

Theorem— Proposition 1 (the Zariski topology)

For subsets and ideals of $k[x_1, \dots, x_n]$ (any field $k$ ): $V(0) = \mathbb{A}^n$ , $V(1) = \varnothing$ , $V\big(\bigcup_\alpha S_\alpha\big) = \bigcap_\alpha V(S_\alpha)$ , and $V(IJ) = V(I \cap J) = V(I) \cup V(J)$ . Hence the algebraic sets are the closed sets of a topology. Moreover $V$ and $I$ reverse inclusions, $X \subseteq V(I(X))$ with equality exactly when $X$ is algebraic, and $V(I(X)) = \overline{X}$ is the Zariski closure.

Proof

The first three identities are immediate from the definition. Since $IJ \subseteq I \cap J \subseteq I$ , we have $V(I) \subseteq V(I \cap J) \subseteq V(IJ)$ , and similarly for $J$ . Conversely, let $a \in V(IJ)$ and suppose $a \notin V(I)$ : choose $f \in I$ with $f(a) \neq 0$ . For every $g \in J$ , $fg \in IJ$ , so $f(a) g(a) = 0$ and hence $g(a) = 0$ ; thus $a \in V(J)$ . So $V(IJ) \subseteq V(I) \cup V(J)$ . If $X = V(J)$ then $J \subseteq I(X)$ , so $V(I(X)) \subseteq V(J) = X$ ; and $V(I(X))$ is the smallest algebraic set containing $X$ , since any algebraic $V(J) \supseteq X$ has $J \subseteq I(X)$ .

∎

Theorem— Proposition 2 (irreducible sets are prime ideals; components)

An algebraic set $X$ is irreducible if and only if $I(X)$ is a prime ideal. Every algebraic set is a finite union $X = X_1 \cup \cdots \cup X_r$ of irreducible algebraic sets, uniquely if no $X_i$ contains another (the irreducible components).

Proof

Suppose $X$ is irreducible and $fg \in I(X)$ . Then $X = (X \cap V(f)) \cup (X \cap V(g))$ is a union of two closed subsets, so one of them, say $X \cap V(f)$ , is all of $X$ ; that is, $f \in I(X)$ . And $I(X) \neq (1)$ because $X \neq \varnothing$ . Conversely, if $I(X)$ is prime and $X = X_1 \cup X_2$ with $X_i$ closed and proper, then $I(X_i) \supsetneq I(X)$ (Proposition 1: $X_i = V(I(X_i))$ ), so we may pick $f_i \in I(X_i) \setminus I(X)$ ; but $f_1 f_2$ vanishes on $X_1 \cup X_2 = X$ , contradicting primality.

For the decomposition: if some algebraic set had no such decomposition, by Hilbert's basis theorem (ascending chains of ideals stabilise, so descending chains of algebraic sets stabilise) there would be a minimal such $X$ . It is not irreducible, so $X = X_1 \cup X_2$ with proper closed $X_i$ , each of which decomposes by minimality — contradiction. For uniqueness, let $X = \bigcup X_i = \bigcup Y_j$ be irredundant decompositions. Since $X_1 = \bigcup_j (X_1 \cap Y_j)$ is irreducible, $X_1 \subseteq Y_j$ for some $j$ ; symmetrically $Y_j \subseteq X_i$ for some $i$ , and then $X_1 \subseteq X_i$ forces $i = 1$ , so $X_1 = Y_j$ . The same argument applies to every $X_i$ and every $Y_j$ , so the two decompositions have the same members.

∎

Theorem— Theorem 3 (the Nullstellensatz dictionary)

Let $k$ be algebraically closed. For every ideal $J \subseteq k[x_1, \dots, x_n]$ , $I(V(J)) = \sqrt{J}$ . Hence $V$ and $I$ are mutually inverse, inclusion-reversing bijections

\{\text{radical ideals}\} \leftrightarrow \{\text{algebraic sets}\}, \qquad \{\text{prime ideals}\} \leftrightarrow \{\text{affine varieties}\}, \qquad \{\text{maximal ideals}\} \leftrightarrow \{\text{points}\}.

Proof

The equation $I(V(J)) = \sqrt{J}$ is Hilbert's Nullstellensatz, proved in the lesson on commutative algebra (in full for uncountable $k$ such as $\C$ , via Rabinowitsch's trick). It shows $I \circ V$ is the identity on radical ideals, and Proposition 1 shows $V \circ I$ is the identity on algebraic sets. Prime ideals correspond to irreducible sets by Proposition 2, and maximal ideals — minimal non-empty algebraic sets — to points.

∎

Algebraic closure is essential: over $\R$ , $V(x^2 + y^2 + 1) = \varnothing = V(1)$ , although $(x^2 + y^2 + 1)$ is a prime ideal.

Theorem— Proposition 4 (the Zariski topology is not Hausdorff)

In an irreducible space, every non-empty open set is dense, and any two non-empty open sets meet. Consequently an affine variety with more than one point — for instance $\mathbb{A}^n$ for $n \geq 1$ — is not Hausdorff.

Proof

If $U, V$ are non-empty open and $U \cap V = \varnothing$ , then $X = (X \setminus U) \cup (X \setminus V)$ is a union of two proper closed subsets, so $X$ is reducible. In a Hausdorff space with two distinct points $a \neq b$ there are disjoint open neighbourhoods of $a$ and $b$ , which is impossible here. Finally $\mathbb{A}^n$ is irreducible because $I(\mathbb{A}^n) = 0$ (a polynomial vanishing on all of $k^n$ , for infinite $k$ , is zero, by induction on $n$ ) is prime.

∎

For $\mathbb{A}^1$ the proof is concrete: basic open sets $D(f) = \{f \neq 0\}$ and $D(g)$ with $f, g \neq 0$ always meet, because $fg$ has finitely many roots and $k$ is infinite — this is the statement formalised below. The same is true of $\operatorname{Spec}\Z$ , where it is even worse: the generic point lies in every non-empty open set, so $\operatorname{Spec}\Z$ is not even $T_1$ .

Theorem— Theorem 5 (affine varieties are the opposite of algebras)

Let $X \subseteq \mathbb{A}^n$ and $Y \subseteq \mathbb{A}^m$ be affine algebraic sets over an algebraically closed field $k$ .

$\varphi \mapsto \varphi^*$ is a bijection from morphisms $X \to Y$ to $k$ -algebra homomorphisms $k[Y] \to k[X]$ , compatible with composition: $(\psi \circ \varphi)^* = \varphi^* \circ \psi^*$ .
Every finitely generated reduced $k$ -algebra is isomorphic to $k[X]$ for some $X$ .

Hence $X \mapsto k[X]$ is an equivalence between the category of affine algebraic sets and the opposite of the category of finitely generated reduced $k$ -algebras; under it, affine varieties correspond to finitely generated $k$ -algebras that are domains.

Proof

(1) Write $\bar y_1, \dots, \bar y_m$ for the coordinate functions on $Y$ , which generate $k[Y]$ . Given a homomorphism $\alpha : k[Y] \to k[X]$ , set $\varphi = (\alpha(\bar y_1), \dots, \alpha(\bar y_m)) : X \to \mathbb{A}^m$ . For $g \in I(Y)$ and $a \in X$ , $g(\varphi(a)) = \alpha(\bar g)(a) = \alpha(0)(a) = 0$ , because $\alpha$ is a $k$ -algebra map and so commutes with polynomial expressions; hence $\varphi(a) \in V(I(Y)) = Y$ by Proposition 1. So $\varphi$ is a morphism, and $\varphi^*(\bar y_i) = \alpha(\bar y_i)$ , so $\varphi^* = \alpha$ since both are algebra maps agreeing on generators. Conversely a morphism is determined by its coordinates $\varphi_i = \varphi^*(\bar y_i)$ . Compatibility with composition is $g \circ (\psi \circ \varphi) = (g \circ \psi) \circ \varphi$ .

(2) A finitely generated $k$ -algebra is $A \cong k[x_1, \dots, x_n]/J$ for some ideal $J$ ; it is reduced exactly when $J$ is radical. Then $X = V(J)$ has $I(X) = \sqrt{J} = J$ by the Nullstellensatz, so $k[X] = A$ . A functor that is fully faithful (by (1)) and essentially surjective (by (2)) is an equivalence; domains correspond to prime $I(X)$ , i.e. to varieties by Proposition 2.

∎

So isomorphism of varieties is isomorphism of coordinate rings — which is how the parabola, hyperbola and cusp examples were decided. A point $a \in X$ is the same as a homomorphism $k[X] \to k$ (evaluation at $a$ ), which is a first glimpse of the functor of points: $X(k) = \operatorname{Hom}_{k\text{-alg}}(k[X], k)$ .

Theorem— Proposition 6 (projective space over a finite field)

$\mathbb{P}^n$ is covered by the $n + 1$ charts $U_i \cong \mathbb{A}^n$ , and $\mathbb{P}^n = \mathbb{A}^n \sqcup \mathbb{P}^{n-1}$ (the chart $U_0$ and the hyperplane at infinity $a_0 = 0$ ). Over the finite field $\mathbb{F}_q$ ,

\#\mathbb{P}^n(\mathbb{F}_q) = \frac{q^{n+1} - 1}{q - 1} = q^n + q^{n-1} + \cdots + q + 1.

Proof

The chart maps are inverse to $(b_1, \dots, b_n) \mapsto [b_1 : \cdots : 1 : \cdots : b_n]$ ( $1$ in position $i$ ), and every point has some non-zero coordinate. Points with $a_0 \neq 0$ are $[1 : b_1 : \cdots : b_n]$ , a copy of $\mathbb{A}^n$ ; points with $a_0 = 0$ are $[0 : a_1 : \cdots : a_n]$ , a copy of $\mathbb{P}^{n-1}$ . Over $\mathbb{F}_q$ the non-zero vectors of $\mathbb{F}_q^{n+1}$ , $q^{n+1} - 1$ of them, fall into lines of $q - 1$ non-zero vectors each; alternatively induct with $\#\mathbb{P}^n = q^n + \#\mathbb{P}^{n-1}$ .

∎

Theorem— Theorem 7 (Bézout for a line)

Let $k$ be algebraically closed, $L \subseteq \mathbb{P}^2$ a line and $F$ a curve of degree $d$ not containing $L$ . Then $L$ meets $F$ in exactly $d$ points counted with multiplicity: $\sum_{P \in L} I_P(L, F) = d$ .

Proof

Choose two distinct points $A, B$ spanning $L$ and parametrise $L$ by $[s : t] \mapsto sA + tB$ , a bijection $\mathbb{P}^1 \to L$ . Then $h(s, t) = F(sA + tB)$ is a homogeneous polynomial of degree $d$ in $s, t$ (each monomial of $F$ has degree $d$ ). It is not identically zero, since $L \not\subseteq V(F)$ . Over an algebraically closed field a non-zero binary form of degree $d$ factors as a product of $d$ linear forms: $h = c \prod_{j=1}^{d} (\beta_j s - \alpha_j t)$ — dehomogenise at $t = 1$ to get a polynomial of degree $d - e$ in $s$ (where $t^{e} \parallel h$ ), which splits into linear factors, and the $e$ missing roots are at $[1 : 0]$ . The zeros of $h$ are the parameters of the points of $L \cap V(F)$ , and the multiplicity of the root $[\alpha_j : \beta_j]$ is the order of vanishing of $F$ along $L$ at that point, which is $I_P(L, F)$ . Summing the multiplicities gives $d$ .

∎

For example, a line meets a smooth cubic in $3$ points; if two are rational, so is the third — which is the source of the group law on an elliptic curve (see arithmetic geometry).

Theorem— Theorem 8 (Bézout for a smooth conic)

Let $Q \subseteq \mathbb{P}^2$ be a smooth conic and $F$ a curve of degree $d$ not containing $Q$ . Then $\sum_P I_P(Q, F) = 2d$ . In particular two conics without common component meet in $4$ points counted with multiplicity.

Proof

Over an algebraically closed field of characteristic $\neq 2$ , a smooth conic is a non-degenerate quadratic form in three variables, and after a linear change of coordinates it becomes $Q = V(xz - y^2)$ (diagonalise the form to $x'^2 + y'^2 + z'^2$ , then substitute). The map $\nu : \mathbb{P}^1 \to Q$ , $[s : t] \mapsto [s^2 : st : t^2]$ , is a bijection onto $Q$ : a point $[x : y : z]$ of $Q$ with $x \neq 0$ is $\nu[x : y]$ , and the only point with $x = 0$ is $[0 : 0 : 1] = \nu[0 : 1]$ . Now $h(s, t) = F(s^2, st, t^2)$ is a binary form of degree $2d$ , non-zero because $Q \not\subseteq V(F)$ , so it has $2d$ roots counted with multiplicity, as in Theorem 7. The multiplicity of a root equals $I_P(Q, F)$ because $\nu$ restricts, near each point, to an isomorphism of a neighbourhood in $\mathbb{P}^1$ with a neighbourhood in the smooth curve $Q$ , and the local ring of $Q$ at $P$ is a discrete valuation ring in which $I_P(Q, F)$ is the order of vanishing of $F$ (Fulton, §3.3).

∎

The same argument works for any curve with a parametrisation $\mathbb{P}^1 \to C$ by forms of degree $m$ that is an isomorphism onto a smooth curve: it meets $F$ in $md$ points. The general case needs a different tool.

Theorem— Theorem 9 (Bézout's theorem; proof sketched)

Let $k$ be algebraically closed and $F, G \subseteq \mathbb{P}^2$ curves of degrees $d$ and $e$ with no common component. Then

\sum_{P \in \mathbb{P}^2} I_P(F, G) = de.

In particular $F$ and $G$ meet in at least one and at most $de$ points.

Proof sketch. Change coordinates so that $[0 : 0 : 1]$ lies on neither curve and on no line joining two of the (finitely many) intersection points. Write $F$ and $G$ as polynomials in $z$ with coefficients in $k[x, y]$ ; their leading coefficients are non-zero constants. The resultant $R(x, y) = \operatorname{Res}_z(F, G)$ , the determinant of the Sylvester matrix, is a homogeneous polynomial of degree exactly $de$ in $x, y$ — this is a degree count in the determinant — and it vanishes at $(a, b)$ exactly when $F(a, b, z)$ and $G(a, b, z)$ have a common root $z$ , that is, when the line through $[0 : 0 : 1]$ and $[a : b : 0]$ contains an intersection point. Since $R \neq 0$ (no common component) it has $de$ roots in $\mathbb{P}^1$ counted with multiplicity, one for each intersection point by the choice of coordinates; the remaining work is to show that the multiplicity of each root equals $I_P(F, G)$ , using the axiomatic characterisation of $I_P$ . Complete proofs: Fulton, Algebraic Curves, Chapter 5; Kirwan, Complex Algebraic Curves, Chapter 3; Shafarevich, Basic Algebraic Geometry 1, §IV.2 (via intersection theory). The first experiment below carries out this resultant computation. $\square$

Each hypothesis is needed: over $\R$ a line can miss a circle; in $\mathbb{A}^2$ parallel lines do not meet; without multiplicities a tangent line meets a conic once; and curves with a common component share infinitely many points.

Theorem— Theorem 10 (the structure sheaf of Spec R)

Let $R$ be a commutative ring and $f_1, \dots, f_r \in R$ with $D(f_1) \cup \cdots \cup D(f_r) = \operatorname{Spec} R$ . Then the sequence

0 \longrightarrow R \longrightarrow \prod_{i} R_{f_i} \rightrightarrows \prod_{i, j} R_{f_i f_j}

is exact: an element of $R$ that vanishes in every $R_{f_i}$ is $0$ , and elements $s_i \in R_{f_i}$ that agree in every $R_{f_i f_j}$ come from a unique $a \in R$ . Consequently $D(f) \mapsto R_f$ satisfies the sheaf condition on covers by basic open sets, extends to a sheaf $\mathcal{O}$ on $\operatorname{Spec} R$ , and $\mathcal{O}(\operatorname{Spec} R) = R$ .

Proof

The $f_i$ generate the unit ideal, and so do their powers. If $(f_1, \dots, f_r) \neq R$ , it lies in a maximal ideal $\mathfrak{m}$ , which is then in no $D(f_i)$ . For any $N$ , the $f_i^N$ also satisfy the hypothesis ( $D(f^N) = D(f)$ ), so $1 = \sum_i c_i f_i^N$ for some $c_i \in R$ .

Injectivity. If $a/1 = 0$ in each $R_{f_i}$ , then $f_i^{N} a = 0$ for all $i$ and some common $N$ . Writing $1 = \sum c_i f_i^N$ gives $a = \sum c_i f_i^N a = 0$ .

Gluing. Write $s_i = a_i / f_i^n$ with a common $n$ . Agreement in $R_{f_i f_j}$ means $(f_i f_j)^m (a_i f_j^n - a_j f_i^n) = 0$ for some $m$ , which we may take independent of $i, j$ . Put $b_i = a_i f_i^m$ and $g_i = f_i^{n + m}$ ; then $s_i = b_i / g_i$ in $R_{f_i}$ and $b_i g_j = b_j g_i$ for all $i, j$ . Write $1 = \sum_j c_j g_j$ and set $a = \sum_j c_j b_j$ . Then

a\, g_i = \sum_j c_j b_j g_i = \sum_j c_j b_i g_j = b_i,

so $a/1 = b_i/g_i = s_i$ in $R_{f_i}$ . Uniqueness is injectivity. Applying this to $R_f$ in place of $R$ handles covers of $D(f) = \operatorname{Spec} R_f$ by smaller basic open sets; a sheaf on a basis extends uniquely to a sheaf (Hartshorne, II.1 and II.2.2).

∎

R

is the equaliser of the two restriction maps: local data that agree on overlaps glue uniquely to a global element.

Theorem 10 is why Grothendieck's definition works: an affine scheme remembers its ring ( $\Gamma(\operatorname{Spec} R, \mathcal{O}) = R$ ), Theorem 5 becomes the equivalence of affine schemes with the opposite of all commutative rings, and gluing affine schemes along open subsets gives schemes such as $\mathbb{P}^n_\Z$ , defined over every ring at once. A variety over $k$ is (the set of closed points of) a reduced, irreducible scheme of finite type over $k$ . Theorem 10 is stated here for the ring $R$ itself; the same proof for modules gives the quasi-coherent sheaves $\widetilde{M}$ that underlie modern cohomology.

Explore

Explore: chords, conics and cubics

The chord trick. Move the slope $m$ of the line through $(-1, 0)$ . The line meets the circle in exactly one more point $P = \big(\tfrac{1 - m^2}{1 + m^2}, \tfrac{2m}{1 + m^2}\big)$ — Bézout's count $2 = 1 + 1$ . As $m \to \pm\infty$ the line becomes vertical and $P \to (-1, 0)$ : the tangent line meets the circle twice at the same point. Rational $m$ gives rational points, which is the whole parametrisation of Pythagorean triples.

A line through a point of a conic meets it once more

Rational slope m ⇔ rational point P: the circle is isomorphic to the projective line.

x² + y² = 1
line through (−1, 0) with slope m

slope mm = 0.5

Move along tt = 0.79

(0.707, 0.707)
(−1.125, −0.063)

Cubics and their singularities. The real points of $y^2 = x^3 + ax + b$ are drawn as the two graphs $y = \pm\sqrt{x^3 + ax + b}$ . The curve is singular exactly when $4a^3 + 27b^2 = 0$ : try $a = -3$ , $b = 2$ , where $x^3 - 3x + 2 = (x - 1)^2(x + 2)$ and the curve has a node at $(1, 0)$ (two branches crossing), and $a = b = 0$ , the cusp $y^2 = x^3$ . For other values the cubic is smooth — an elliptic curve — with one or two real components; over $\C$ it is a torus, of genus $1$ .

The family y² = x³ + ax + b

Smooth unless 4a³ + 27b² = 0: try (a, b) = (−3, 2) for a node and (0, 0) for a cusp.

y = +√(x³ + ax + b)
y = −√(x³ + ax + b)

aa = −1bb = 0

Trace along xx = 1

y = +√(x³ + ax + b): y = 0
y = −√(x³ + ax + b): y = 0

Formal mathematicsF

Algebraic geometry in Lean and Mathlib

Mathlib has both classical algebraic sets (MvPolynomial.zeroLocus, MvPolynomial.vanishingIdeal) and the full theory of schemes (AlgebraicGeometry.Scheme). We begin with Proposition 1, proved here exactly as in the text:

zeroLocus_mulLean 4Proved here from basic factsFormally verified

/-- **Finite unions of algebraic sets are algebraic**: `V(IJ) = V(I) ∪ V(J)`. Human proof: if
`x` lies in `V(I)` (or `V(J)`), every element of `IJ ⊆ I` (or `⊆ J`) vanishes at `x`.
Conversely, if `x` lies in neither, choose `f ∈ I` and `g ∈ J` with `f(x) ≠ 0 ≠ g(x)`; then
`fg ∈ IJ` and `(fg)(x) = f(x) g(x) ≠ 0` because `K` is a field. Together with
`V(∑ Iᵢ) = ⋂ V(Iᵢ)`, this makes the algebraic sets the closed sets of a topology. -/
theorem zeroLocus_mul {σ k K : Type*} [Field k] [Field K] [Algebra k K]
    (I J : Ideal (MvPolynomial σ k)) :
    MvPolynomial.zeroLocus K (I * J) = MvPolynomial.zeroLocus K I ∪ MvPolynomial.zeroLocus K J := by
  ext x
  simp only [Set.mem_union, MvPolynomial.mem_zeroLocus_iff]
  constructor
  · intro hx
    by_contra h
    simp only [not_or, not_forall] at h
    obtain ⟨⟨f, hfI, hf⟩, ⟨g, hgJ, hg⟩⟩ := h
    have hmem : f * g ∈ I * J := Ideal.mul_mem_mul hfI hgJ
    have hfg : MvPolynomial.aeval x (f * g) = 0 := hx (f * g) hmem
    rw [map_mul] at hfg
    exact mul_ne_zero hf hg hfg
  · rintro (hx | hx) p hp
    · have hpI : p ∈ I := Ideal.mul_le_left (I := I) (J := J) hp
      exact hx p hpI
    · have hpJ : p ∈ J := Ideal.mul_le_right (I := I) (J := J) hp
      exact hx p hpJ