Assume thatV is irreducible

Duke University – Spring 2017 – MATH 627 January 24, 2017

Problem set 1 Olivier Debarre Due Thursday February 2, 2017

Problem 1. Recall that a topological spaceX isirreducibleif it is non-empty and is not the union of two strict closed subsets. In other words, ifX₁ andX₂are closed subsets ofXandX =X₁∪X₂, thenX =X₁orX =X₂.

a) LetX be a topological space and let V ⊂ X be a subset (endowed with the induced topology).

Prove thatV is irreducible if and only if its closureV is irreducible.

Proof. Assume thatV is irreducible. IfV = W1 ∪W2, whereW1 andW2 are closed in V (hence also inX), we haveV = (V ∩W1)∪(V ∩W2), whereV ∩W1andV ∩W2 are closed inV. Since V is irreducible, we obtain V = V ∩W₁ orV = V ∩W₂, henceV ⊂ W₁ orV ⊂ W₂. SinceW₁ andW₂ are closed inX, this impliesV ⊂ W₁ orV ⊂W₂, henceV =W₁orV =W₂. This proves thatV is irreducible.

For the converse, assume thatV is irreducible. If V = W₁ ∪W₂, whereW₁ andW₂ are closed in V, we can writeW_i = V ∩F_i, where F_i is closed inX (we could take F_i = W_i). We then have V ⊂F₁∪F₂, henceV ⊂F₁∪F₂andV = (V ∩F₁)∪(V ∩F₂). SinceV is irreducible, we obtain V =V ∩F₁ orV =V ∩F₂, henceV ⊂F₁ orV ⊂ F₂ andV =W₁ orV =W₂. This proves that V is irreducible.

b) LetX andY be topological spaces and letu :X →Y be a continuous map. IfX is irreducible, prove thatu(X)is irreducible.

Proof. Assume u(X) = W₁ ∪W₂, where W₁ and W₂ are closed in u(X). We can writeW_i = u(X)∩F_i, whereF_iis closed inY. We have thenu(X)⊂F₁∪F₂, henceX =u⁻¹(F₁)∪u⁻¹(F₂).

SinceX is irreducible andu⁻¹(F_i)is closed in X, we getX = u⁻¹(F₁) orX = u⁻¹(F₂), hence u(X) ⊂ F₁ or u(X) ⊂ F₂. This implies u(X) = W₁ or u(X) = W₂ and proves that u(X) is irreducible.

Problem 2. Let k be an infinite (not necessarily algebraically closed) field. Let C ⊂ k² be the vanishing setV(X²−Y³).

a) Prove that the ideal ofCis the ideal ink[X, Y]generated byX²−Y³ and thatC is irreducible (Hint: use the “parametrization”k→ C given byt 7→ (t³, t²)and express A(C) = k[X, Y]/I(C) as a subring ofk[T]).

Proof. Obviously,X²−Y³is inI(C). Assume thatP ∈k[X, Y]vanishes onC. We haveP(t³, t²) for all t ∈ k. Since the field k is infinite, this implies that the polynomial P(T³, T²) ∈ k[T] vanishes. Modulo the idealI generated byX²−Y³, one can writeP ≡A(Y) +XB(Y). We then haveA(T²) +T³B(T²) = 0ink[T]. Since only even powers ofT appear in A(T²)and only odd powers of T appear in T³B(T²), we obtain A = B = 0 and P ∈ I. This proves the opposite inclusionI(C)⊂I.

So we have A(C) = k[X, Y]/I and the parametrizationk → C induces an isomorphism between A(C) and the subring k[T², T³] of k[T]. The latter is obviously an integral domain, hence so is A(C)andCis irreducible.

b) Prove thatCis not isomorphic tok(Hint: prove thatA(C)is not a principal ideal domain).

Proof. We saw in a) thatA(C)is isomorphic to the subringk[T², T³]ofk[T]. LetJ be the ideal of A(C)generated byT²andT³. If it is generated by one elementP(T)∈J ⊂k[T², T³], we can write T² =A(T)P(T)andT³ =B(T)P(T), hence2 = deg(A) + deg(P)and3 = deg(B) + deg(P), with deg(P) ≥ 2 (because P ∈ J). Since each of these degrees is different from 1, we get a contradiction.

c) How do these these results generalize to the vanishing setV(X^r−Y^s), whererandsare relatively prime positive integers?

Proof. The conclusions are the same but the arguments are slightly more complicated. For a), a polynomialP ∈k[X, Y]that vanishes onC is such thatP(T^s, T^r)vanishes ink[T]. Write

P(X, Y) = P₀(Y) +XP₁(Y) +· · ·+X^r−1Pr−1(Y)

modulo the idealI generated byX^r−Y^s. We then have

0 =P(T^s, T^r) = P₀(T^r) +T^sP₁(T^r) +· · ·+T^s(r−1)Pr−1(T^r)∈k[T].

The powers ofT that appear in the termT^siP_i(T^r)are congruent tosimodulor. Sincerandsare relatively prime, these numbers are all different modulo r for i ∈ {0, . . . , r −1}. It follows that P_i = 0for alliandP ∈I. This provesI(C) =I andA(C)'k[T^s, T^r].

To prove that A(C) is not a principal ideal domain, we may assumer < s. As a k-vector space, A(C) is generated by all monomial T^mr+ns, with m, n non-negative integers. When of degree

< s, these monomials are of the type T^mr. IfT^r = A(T)P(T) andT^s = B(T)P(T), we have r = deg(A) + deg(P) and s = deg(B) + deg(P), with deg(P) ≥ r, hence we may assume P(T) = T^r. This impliesB(T) = T^s−r ∈ k[T^s, T^r]. Sinces−r < s, it must be a multiple ofr, which is absurd.

Problem 3. Letkbe aninfinite(not necessarily algebraically closed) field, letu: P¹_k →P³_kbe the regular map defined byu(s, t) = (s³, s²t, st², t³), and setC :=u(P¹_k).

a) Prove that no 4 distinct points ofC are contained in a hyperplane inP³_k.

Proof. A hyperplane has equation L(x₀, x₁, x₂, x₃) = 0, where L is a non-zero linear form. If L vanishes at 4 points ofC, we haveL(s³, s²t, st², t³) = 0for 4 distinct points(s, t)ofP¹_k. But this is a non-zero homogeneous polynomial of degree 3 in two variables, hence it cannot have 4 distinct zeroes inP¹_k.

b) Prove that any quadric inP³_k(i.e., any subset ofP³_kdefined by a non-zero homogoneous polyno- mial of degree 2) that contains 7 distinct points ofC containsC.

Proof. A quadric has equationQ(x₀, x₁, x₂, x₃) = 0, whereQis a non-zero quadratic form. If Q vanishes at 7 points ofC, we have Q(s³, s²t, st², t³) = 0for 7 distinct points(s, t)ofP¹_k. But this

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is a non-zero homogeneous polynomial of degree 6 in two variables, hence it cannot have 7 distinct zeroes inP¹_k.

c) Prove thatCis the vanishing set inP³_kof the (homogeneous) idealI ink[T0, T1, T2, T3]generated by the homogeneous polynomialsT₀T₂−T₁², T₂²−T₁T₃, T₁T₂−T₀T₃, which can be neatly expressed as the2×2-minors of the matrix

T₀ T₁ T₂ T₁ T₂ T₃

.

Proof. The setCis contained inV(I). Conversely, assumex:= (x0, x1, x2, x3)∈V(I). Ifx0 6= 0, we may takex₀ = 1and we havex₂ =x²₁ andx₃ =x₁x₂ =x³₁, hencex=u(1, x₁)∈C. Ifx₀ = 0, we havex₁ = 0,x₂ = 0, hencex=u(0,1)∈C.

d) Prove that the ideal ofCisI (Hint: prove that any polynomialP ∈k[T0, T1, T2, T3]is congruent moduloI to a polynomial of the typeA(T₀, T₁, T₃) +T₂B(T₃)and that ifP vanishes onC, one has B = 0; then, use a similar method to show thatAis divisible byT₁³−T₀²T₃).

Proof. The inclusionI ⊂I(C)is clear. Using the fact thatI containsT0T2−T₁², T1T2−T0T3, T₂²− T₁T₃, we reduce moduloIany polynomialP to the formA(T₀, T₁, T₃) +T₂B(T₃). IfP vanishes on C, we obtain as in Problem 2 (using the fact thatkis infinite)A(S³, S²T, T³) +ST²B(T³) = 0in k[S, T]. Monomials of the typeST^monly appear inST²B(T³), henceB = 0andA(S³, S²T, T³) = 0. The polynomial T₁³ − T₀²T₃ = −T₁(T₀T₂ − T₁²) + T₀(T₁T₂ − T₀T₃) is in I, hence one can write A(T₀, T₁, T₃) ≡ A₀(T₀, T₃) + T₁A₁(T₀, T₃) + T₁²A₂(T₀, T₃) (mod I), with A₀(S³, T³) + S²T A₁(S³, T³)+S⁴T²A₂(S³, T³) = 0. Looking at the exponents ofT that appear in this polynomial modulo 3, we obtainA₀ =A₁ =A₂, henceA∈I. This proves the opposite inclusionI(C)⊂I.

e)(Extra credit)How do these results generalize to the regular mapu: P¹_k →Pⁿ_k(n ≥ 3) defined byu(s, t) = (sⁿ, sⁿ⁻¹t, . . . , stⁿ⁻¹, tⁿ)?

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