Each worksheet prints cleanly on A4 or Letter. Answer keys are on a separate page.

The Lumenary · First-year graduate

Algebraic geometry

Name: ______________________

Date: ______________________

You will learn to

Pass between ideals and affine algebraic sets with the Nullstellensatz, matching irreducible sets with prime ideals and morphisms with homomorphisms of coordinate rings
Work with the Zariski topology and explain why it is not Hausdorff
Use projective space and homogeneous coordinates, and state Bézout's theorem for plane curves, proving it when one curve is a line or a smooth conic
Distinguish varieties that are in bijection but not isomorphic, using coordinate rings
Explain why schemes generalise varieties, using $\operatorname{Spec} R$ and its structure sheaf

1.
Which of these subsets of the complex line $\mathbb{A}^1 = \mathbb{C}$ are closed in the Zariski topology? (Choose all that apply.)
- A. $\{0, 1, 2\}$
- B. $\mathbb{Z}$
- C. $\mathbb{C}$
- D. $\varnothing$
- E.the closed unit disc $\{z : |z| \le 1\}$
2.
A smooth cubic curve and a conic in the complex projective plane have no common component. In how many points do they meet, counted with multiplicity?
3.
Over $\mathbb{C}$ , let $X = V(x^2 y,\, x y^2) \subseteq \mathbb{A}^2$ . Which ideal is $I(X)$ ?
- A. $(x^2 y, x y^2)$
- B. $(xy)$
- C. $(x, y)$
- D. $(x + y)$
4.
Which argument correctly proves that the hyperbola $H = V(xy - 1)$ is not isomorphic to the affine line $\mathbb{A}^1$ (over $\mathbb{C}$ )?
- A. $\mathbb{C}[H] \cong \mathbb{C}[t, t^{-1}]$ has a non-constant unit, while the units of $\mathbb{C}[t]$ are constants; an isomorphism of varieties would give an isomorphism of coordinate rings.
- B. $H$ and $\mathbb{A}^1$ are not homeomorphic in the Zariski topology.
- C. $H$ has dimension $2$ because it lives in $\mathbb{A}^2$ .
- D. $H$ misses the point $0$ of the line, so it has fewer points.
5.
How many points does the projective plane $\mathbb{P}^2(\mathbb{F}_5)$ have?
6.
The concentric circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$ have no common point in $\mathbb{C}^2$ . Bézout's theorem promises $4$ intersection points in $\mathbb{P}^2(\mathbb{C})$ . Where are they?
- A.At the circular points $[1 : i : 0]$ and $[1 : -i : 0]$ , each with multiplicity $2$ .
- B.Nowhere: Bézout's theorem does not apply to concentric circles.
- C.At four complex points of $\mathbb{C}^2$ that we have overlooked.
- D.At $[1 : 0 : 0]$ and $[0 : 1 : 0]$ , each with multiplicity $2$ .
7.
The morphism $\varphi : \mathbb{A}^1 \to C = V(y^2 - x^3)$ , $t \mapsto (t^2, t^3)$ , is a bijection. Which statement explains why it is not an isomorphism?
- A.Its comorphism $k[C] \to k[t]$ has image $k[t^2, t^3]$ , which does not contain $t$ ; so $\varphi^*$ is not surjective and no inverse morphism exists.
- B. $\varphi$ is not continuous in the Zariski topology.
- C.Its inverse $(x, y) \mapsto y/x$ is a morphism, but it is not defined at the origin.
- D. $\varphi$ is not injective: $t$ and $-t$ have the same image.
8.
Which Lean statement formalises Proposition 1's identity "the zero set of a product of ideals is the union of the zero sets"? (zeroLocus K I is the set of points of $K^\sigma$ where every polynomial in I vanishes.)
- A.zeroLocus K (I * J) = zeroLocus K I ∪ zeroLocus K J
- B.zeroLocus K (I * J) = zeroLocus K I ∩ zeroLocus K J
- C.zeroLocus K (I + J) = zeroLocus K I ∪ zeroLocus K J
- D.zeroLocus K I ⊆ zeroLocus K (I * J) only
9.
Write projective_count(f, p) returning the number of points of $\mathbb{P}^2(\mathbb{F}_p)$ at which the homogeneous polynomial f(x, y, z) (a Python function with integer values) vanishes modulo $p$ . Use one representative per point: $(x, y, 1)$ , $(x, 1, 0)$ and $(1, 0, 0)$ .
Computational — Practice

Answer key — Algebraic geometry

1.
The closed sets of $\mathbb{A}^1$ are $V(f)$ for $f \in \mathbb{C}[x]$ (every ideal of $\mathbb{C}[x]$ is principal): for $f \ne 0$ a finite set, for $f = 0$ the whole line. So $\{0, 1, 2\}$ , $\mathbb{C}$ and $\varnothing$ are closed, while $\mathbb{Z}$ and the unit disc are not. The Zariski topology on $\mathbb{A}^1$ is the cofinite topology.
2.
By Bézout's theorem, $\sum_P I_P(F, G) = de = 3 \cdot 2 = 6$ . Some of these points may be complex, at infinity, or counted more than once if the curves are tangent.
3.
$x^2y = xy^2 = 0$ exactly when $x = 0$ or $y = 0$ , so $X = V(x) \cup V(y)$ , the union of the axes. By the Nullstellensatz $I(X) = \sqrt{(x^2y, xy^2)}$ , which is $(xy)$ : indeed $(xy)^2 = y \cdot x^2y \in (x^2y, xy^2)$ , and $(xy) = (x) \cap (y)$ is radical, being an intersection of primes. Equivalently, $I(X) = I(V(x)) \cap I(V(y)) = (x) \cap (y) = (xy)$ .
4.
$\mathbb{C}[H] = \mathbb{C}[x, y]/(xy - 1) \cong \mathbb{C}[t, t^{-1}]$ via $x \mapsto t$ , $y \mapsto t^{-1}$ , and $t$ is a non-constant unit. In $\mathbb{C}[t]$ the units are the non-zero constants (degrees add under multiplication). An isomorphism $H \cong \mathbb{A}^1$ would induce an isomorphism $\mathbb{C}[t] \cong \mathbb{C}[t, t^{-1}]$ , which must map units to units — impossible. The topological and cardinality arguments fail because the Zariski topology and the point sets are the same; only the ring of functions sees the difference.
5.
$\#\mathbb{P}^2(\mathbb{F}_q) = (q^3 - 1)/(q - 1) = q^2 + q + 1$ , so for $q = 5$ it is $25 + 5 + 1 = 31$ : the $25$ points of the affine plane and the $6$ points of the line at infinity.
6.
Homogenised: $x^2 + y^2 - z^2 = 0$ and $x^2 + y^2 - 4z^2 = 0$ . Subtracting, $3z^2 = 0$ , so every common point has $z = 0$ , and then $x^2 + y^2 = (x + iy)(x - iy) = 0$ gives the two circular points $[1 : \pm i : 0]$ . Each has intersection multiplicity $2$ (the factor $z^2$ is a double root; the resultant computation in the first experiment confirms the pattern '2 points ×2'). Total: $2 \cdot 2 = 4 = 2 \cdot 2$ , as Bézout says. Every circle passes through the circular points, which is why two circles meet in at most two affine points.
7.
$\varphi^*(x) = t^2$ and $\varphi^*(y) = t^3$ , so the image of $\varphi^*$ is the subalgebra $k[t^2, t^3]$ of polynomials with no $t$ -term (the $t$ -coefficient of a product $pq$ is $p_0 q_1 + p_1 q_0$ ). Since $t$ is not in it, $\varphi^*$ is not surjective, hence not an isomorphism, and by Theorem 5 neither is $\varphi$ . Geometrically $C$ is singular at the origin, while $\mathbb{A}^1$ is smooth.
8.
Since $IJ \subseteq I$ and $IJ \subseteq J$ , $V(IJ) \supseteq V(I) \cup V(J)$ ; conversely if $f(a) \ne 0$ and $g(a) \ne 0$ for some $f \in I$ , $g \in J$ , then $(fg)(a) \ne 0$ with $fg \in IJ$ . So zeroLocus K (I * J) = zeroLocus K I ∪ zeroLocus K J. The sum of ideals corresponds to the intersection of zero sets.
9.
```
def projective_count(f, p):
    points = [(x, y, 1) for x in range(p) for y in range(p)]
    points += [(x, 1, 0) for x in range(p)]
    points.append((1, 0, 0))
    return sum(1 for x, y, z in points if f(x, y, z) % p == 0)
```
Every point $[a : b : c]$ has a unique representative of one of the three forms: scale so that the last non-zero coordinate is $1$ . The starter missed the $p + 1$ points at infinity: for the line $z = 0$ it found nothing, although the line at infinity is a whole $\mathbb{P}^1$ . The conic $x^2 + y^2 = z^2$ has $p + 1$ points, as a copy of $\mathbb{P}^1$ should; the Fermat cubic has $6 = 5 + 1$ points over $\mathbb{F}_5$ and $9$ over $\mathbb{F}_7$ , both within Hasse's bound $|\#E - (p + 1)| \le 2\sqrt{p}$ .