Math 40510, Algebraic Geometry

Problem Set 3 Solutions, Spring 2016

As always, in this problem set k will represent some field. Occasionally we will give a more specific field.

2 1. In §8.1, Definition 1, of CLO, the authors define the projective plane over R, P (R), in a very intuitive way, leaving the more careful definitions to come. They say 2 2 P (R) = R ∪ {one point at infinity for each equivalence class of parallel lines} 2 Given a line, L in R , they denote by [L]∞ the corresponding point at infinity, and they define the associated projective line to be L = L ∪ [L]∞. The collection of all points at infinity (one for each class of parallel lines) is called the line at infinity and is also a projective line. 2 a) Prove that any two distinct points in P (R) determine a unique projective line. [Hint: there are three cases, depending on how many of the points are at infinity.] Solution: 2 Let P1,P2 be two points in P (R). 2 Case 1: If P1 and P2 both lie in R then it is well known that there is a unique affine line, `P1,P2 , joining them. Then adding the point at infinity corresponding to the family of lines parallel to

`P1,P2 gives the unique projective line. 2 Case 2: Assume that P1 ∈ R and P2 lies on the line at inﬁnity. P2 determines a family of parallel lines, and there is exactly one that passes through P1.

Case 3: Assume that both P1 and P2 lie at infinity. They certainly both lie on the line at infinity. Suppose that there were some other projective line, `, that also contains P1 and P2. If ` is not 2 the line at infinity then it passes through some point P ∈ R . But then P1 and P2 represent two different families of parallel lines, and ` can’t be in both. So ` does not exist and there is only one line, namely the line at infinity. 2 b) Prove that any two distinct projective lines in P (R) meet at a unique point. [Hint: you have to divide this into cases too.] Solution: Let `1 and `2 be projective lines. 2 Case 1: Assume that neither line is the line at infinity. If `1 and `2 meet at a point P ∈ R then they are not parallel, so they cross the line at infinity at different points. Hence they meet at exactly P and no other point.

Case 2: Assume that neither line is the line at inﬁnity. If `1 and `2 are parallel then they do not 2 meet in R , and being parallel they meet at a unique point at inﬁnity (corresponding to that family of parallel lines).

Case 3: Assume that `1 is the line at inﬁnity. Then `2 is not the line at inﬁnity, so it meets `1 only at the point corresponding the the family of parallel lines containing `2.

2. Let k be a ﬁeld. In class we proved that if f ∈ k[x0, . . . , xn] is homogeneous of degree d, and if P = n [a0, . . . , an] is a point of P such that f(a0, . . . , an) = 0 then it’s also true that f(λa0, . . . , λan) = 0; n this means that the vanishing of a homogeneous polynomial at a point of P is well-deﬁned, and we say for short that f vanishes at P . We now want to see how far we can push this idea beyond homogeneous polynomials.

1 2

a) If f ∈ k[x0, . . . , xn] is any homogeneous polynomial, we can write it as the sum of its homogeneous parts:

f = f0 + f1 + ··· + fd

where each fi is homogeneous of degree i. To get things started, let f = 5x4yz2 + 6x3y3 + 7xyz4 + 8y3z + 9xyz + 10y3 + 11y + 8.

Tell me what d is, and what each fi is. Solution:

d = 7 4 2 f7 = 5x yz 3 3 4 f6 = 6x y + 7xyz f5 = 0 3 f4 = 8y z 3 f3 = 9xyz + 10y f2 = 0 f1 = 11y f0 = 8 n b) Let f = f0 + f1 + ··· + fd ∈ k[x0, . . . , xn] and let P = [a0, a1, . . . , an] ∈ P . Assume that n each fi ∈ P vanishes at P in the sense of what I said in the ﬁrst paragraph of this problem. Prove that then the vanishing of f is well-deﬁned. [There should be some λ’s somewhere in your answer.] Solution: We know that P = [a0, a1, . . . , an] = [λa0, λa1, . . . , λan] for any λ 6= 0. Then

d X j f(λa0, λa1, . . . , λan) = λ fj(a0, a1, . . . , an) = 0. j=0

c) Now assume that k is an infinite field. Let f = f0 + f1 + ··· + fd ∈ k[x0, . . . , xn] and let n P = [a0, a1, . . . , an] ∈ P . Assume that f vanishes at P . Prove that then each fi vanishes at P . [Again there should be λ’s in your answer, and you should explain what k being infinite has to do with it.] Solution: We have assumed that f(λa0, λa1, . . . , λan) = 0 for all λ 6= 0. So d d−1 λ fd(a0, . . . , an) + λ fd−1(a0, . . . , an) + ··· + λf1(a0, . . . , an) + f0(a0, . . . , an) = 0

for all λ 6= 0. But for each j, fj(a0, . . . , an) is just an element of k, the ﬁeld. This means that every nonzero λ ∈ k is a root of the polynomial

d d−1 fd(a0, . . . , an)x + fd−1(a0, . . . , an)x + ··· + f1(a0, . . . , an)x + f0(a0, . . . , an). Since k is infinite, this is a polynomial of degree d in one variable that has infinitely many roots, which is impossible over a field. 3

3. Homogeneous polynomials satisfy an important relation known as Euler’s Theorem. It says the following. For convenience assume that our ﬁeld is R. Let f ∈ R[x0, . . . , xn] be a homogeneous polynomial of degree d. Then n X ∂f x = d · f. i ∂x i=0 i a) Illustrate Euler’s theorem by cooking up a homogeneous polynomial, f, having three terms and showing that the theorem is true for your example. Solution: Let f(x, y, z) = x3yz + 4x2yz2 + 5xyz3. Notice that d = 5. Then ∂f = 3x2yz + 8xyz2 + 5yz3 ∂x ∂f = x3z + 4x2z2 + 5xz3 ∂y

∂f = x3y + 8x2yz + 15xyz2 ∂z Then ∂f ∂f ∂f x + y + z ∂x ∂y ∂z

= x(3x2yz + 8xyz2 + 5yz3) + y(x3z + 4x2z2 + 5xz3) + z(x3y + 8x2yz + 15xyz2)

= (3x3yz + 8x2yz2 + 5xyz3) + (x3yz + 4x2yz2 + 5xyz3) + (x3yz + 8x2yz2 + 15xyz3)

= 5(x3yz + 4x2yz2 + 5xyz3).

b) Prove Euler’s theorem by considering f(λx0, . . . , λxn) as a function of λ, and diﬀerentiating with respect to λ using the chain rule. Solution: We know that d (1) f(λx0, . . . , λxn) = λ f(x0, . . . , xn). Then diﬀerentiate on both sides with respect to λ. ∂ (2) f(λx , . . . , λx ) = dλd−1f(x , . . . , x ). ∂λ 0 n 0 n

Let’s look at the left-hand side. For 0 ≤ i ≤ n let ui = λxi. n ∂ X ∂f ∂ui f(λx , . . . , λx ) = ∂λ 0 n ∂u ∂λ i=0 i

n ! X ∂f (3) = · xi ∂xi i=0 xi=ui

n X ∂f = λd−1x i ∂x i=0 i 4

where we have used the fact that ∂f is homogeneous of degree d − 1 and applied (1) to the ∂xi partials. Now substitute the result of (3) into (2) and divide by λd−1 (which is non-zero) to obtain the result. c) Let R = [x, y, z] and let f = xyz. In 2 describe (f), (f , f , f ), and the relation between R PR V V x y z these two varieties. (Here fx, fy, fz are the partials with respect to x, y, z respectively.) How is Euler’s theorem relevant to this last part?

Solution: First note that ∂f ∂f ∂f f = = yz, f = = xz, f = = xy. x ∂x y ∂y z ∂z

Now, V(f) = V(xyz) is the union of the three lines deﬁned by x = 0, y = 0 and z = 0. On the other hand, V(fx, fy, fz) is the locus deﬁned by yz = 0 xz = 0 xy = 0. A quick calculation reveals

V(fx, fy, fz) = V(x, y) ∪ V(x, z) ∪ V(y, z).

This is precisely the union of the three points of pairwise intersection of the three lines in V(xyz), that is, the points {[1, 0, 0], [0, 1, 0], [0, 0, 1]}. In particular, V(fx, fy, fz) is a subset of V(xyz). And indeed, since Euler’s theorem gives, in this case, that

x · fx + y · fy + z · fz = 3 · f,

if P ∈ V(fx, fy, fz) then fx, fy, fz all vanish at P , so Euler’s theorem implies that f vanishes at P , so in particular P ∈ V(f). d) Let R = [x, y, z] and let f = xyz(x+y +z). In 2 describe (f), (f , f , f ), and the relation R PR V V x y z between the two. Again, how is Euler’s theorem relevant to this last part?

Solution: First note that f = x2yz + xy2z + xyz2, so

∂f 2 2 fx = ∂x = 2xyz + y z + yz = yz(2x + y + z)

∂f 2 2 fy = ∂y = x z + 2xyz + xz = xz(x + 2y + z)

∂f 2 2 fz = ∂z = x y + xy + 2xyz = xy(x + y + 2z). Now, V(f) = V(xyz(x + y + z)) is the union of the four lines defined by x = 0, y = 0, z = 0 and x + y + z = 0. On the other hand, V(fx, fy, fz) is the locus defined by yz(2x + y + z) = 0 xz(x + 2y + z) = 0 xy(x + y + 2z) = 0. Since each of these is a product of three linear forms, each equation is satisfied exactly when one (or more) of the factors is zero. Then for a point to be in the solution set V(fx, fy, fz) we need 5

one of the following lines to hold: x = 0 ⇒ yz(y + z) = 0 ⇒ y = 0 OR z = 0 OR y = −z

y = 0 ⇒ xz(x + z) = 0 ⇒ x = 0 OR z = 0 OR x = −z

z = 0 ⇒ xy(x + y) = 0 ⇒ x = 0 OR y = 0 OR x = −y

So the solutions are (after eliminating repetitions) {[0, 0, 1], [0, 1, 0], [0, 1, −1], [1, 0, 0], [1, 0, −1], [1, −1, 0].} 4 These points are the pairwise intersections of the four lines. (Note 2 = 6.) In particular, V(fx, fy, fz) is a subset of V(xyz(x + y + z)). And indeed, since Euler’s theorem gives, in this case, that

x · fx + y · fy + z · fz = 4 · f,

if P ∈ V(fx, fy, fz) then fx, fy, fz all vanish at P , so Euler’s theorem implies that f vanishes at P , so in particular P ∈ V(f).

2 FYI: The vanishing locus in P of a polynomial f that is a product of homogeneous linear polynomials, where none is a scalar multiple of another, is called a line arrangement and is an object of interest in current research. The vanishing locus of the partial derivatives, and the ideal that the partial derivatives generate, is an important part of that.

4. Let I and J be homogeneous ideals in k[x0, . . . , xn]. a) Prove that I + J is homogeneous. Solution: Since I and J are homogeneous ideals, we can ﬁnd generators for each that are homogeneous. Say I = hf1, . . . , fsi and J = hg1, . . . , gti. Then

I + J = hf1, . . . , fs, g1, . . . , gti

is generated by homogeneous polynomials, hence is a homogeneous ideal. b) Prove that I ∩ J is homogeneous. Solution: We’ll use the other condition for an ideal to be homogeneous. Let f ∈ I ∩ J. Write f as a sum of homogeneous polynomials, f = fd + fd−1 + ··· + f1 + f0. Since f ∈ I and I is homogeneous, each fi ∈ I. Similarly for J. Thus each fi ∈ I ∩ J, so I ∩ J is homogeneous.

5. Let f1, f2, f3, f4 ∈ k[x0, . . . , xn]. Let I = hf1, f2, f3, f4i. Let di = deg fi for 1 ≤ i ≤ 4. Assume d1 < d2 < d3 < d4.

a) Let g ∈ I be a homogeneous polynomial of degree d1. Prove that g = λf1 for some λ ∈ k. Solution: Since g ∈ I, we have g = a1f1 + a2f2 + a3f3 + a4f4, where a1, a2, a3, a4 ∈ R. Since f2, f3, f4 are homogenous of degree > d1, it’s impossible for a2f2 + a3f3 + a4f4 to contribute any terms of degree d1. Hence since g is homogeneous of degree d1, we must have g = a1f1, and since g and f1 are homogeneous of the same degree, we must have a1 is a constant. 6

b) Let h ∈ I be a homogeneous polynomial of degree d2. Prove that there exist a1, a2 ∈ R such that both a1 and a2 are homogeneous, and a1f1 + a2f2 = h. In particular, what can you say about the degrees of a1 and a2? Solution: Since h ∈ I, we have h = a1f1 + a2f2 + a3f3 + a4f4, where a1, a2, a3, a4 ∈ R. Since f3, f4 are homogenous of degree > d2, it’s impossible for a3f3 + a4f4 to contribute any terms of degree d2. Hence since h is homogeneous of degree d2, we must have g = a1f1 + a2f2.

It remains to show that we can choose a1 homogeneous of degree d2 − d1 and a2 a scalar (i.e. of degree 0). Any terms of a1 of degree not equal to d2 − d1 contribute terms in a1f1 that cannot possible be part of h (which is homogeneous of degree d2), so they have to be cancelled out by terms of a2f2. We can thus disregard these terms and choose as a1 only terms of degree d2 − d1, and as a2 only terms of degree 0, thus making both a1 and a2 homogeneous.

6. Recall that for an ideal I ⊂ k[x0, . . . , xn], a set of polynomials f1, . . . , fr are minimal generators for I if I = hf1, . . . , fri, and if the removal of any of the fi changes the ideal. We also say that {f1, . . . , fr} form a minimal generating set for I. For example, for I = hx2, y2, (x + y)(x − y)i ⊂ k[x, y, z], the generators are not minimal since (x + y)(x − y) = x2 − y2, so removing (x + y)(x − y) does not change the ideal.

a) Give an example of an ideal I ⊂ C[x, y, z] such that • I has a minimal generating set consisting of ﬁve homogeneous polynomials;

• V(I) = ∅; • The ﬁve generators of I all have diﬀerent degrees. Solution:

I = hx4, y5, z6, x2y2z3, x3yz4i.

b) In the statement of the Projective Weak Nullstellensatz (Theorem 8 of the book), the authors mention integers mi (in part (iii)) and r (in part (iv)). For your answer to part (a), what are the values of m1, m2, m3 and r? Be sure to justify your answer. Solution: 4 5 6 m1 = 4 since x ∈ I. m2 = 5 since y ∈ I. m3 = 6 since z ∈ I. Thanks to the proof in class, we can take r = 4 + 5 + 6 = 15. But in fact r = 13 works. c) Find a counterexample to the following statement: If I is a homogeneous ideal and J is an ideal such that J ⊂ I then J is homogeneous. Solution:

I = hx2, y4i and J = hx2 + y4i. √ d) Let I be a homogeneous ideal. Let J be a homogeneous ideal such that J ⊆ I. Prove that J k ⊆ I for some k ≥ 1. [Hint: start with a minimal generating set for J.] Solution: 7 √ Suppose J = hf1, . . . , fti. Since each generator is in J ⊂ I, we have positive integers m1, . . . , mt m1 mt k such that f1 ∈ I, . . . , ft ∈ I. Now let k = m1 + ··· + mt. We claim J ⊆ I. Indeed, for each k f ∈ J we have to show f ∈ I. But since f1, . . . , ft generate J we have

f = a1f1 + ··· + atft for some elements a1, . . . , at ∈ R. We now claim that f k ∈ I. Just as in the proof of the Nullstellensatz in class, this is because in k m k any term of f we have to have at least one fi where m ≥ mi. If this weren’t true, f would have some term i1 i2 it i1 i2 it c(a1f1) (a2f2) ··· (atft) = df1 f2 ··· ft (for suitable c and d) with

i1 ≤ m1 − 1, i2 ≤ m2 − 1, . . . , it ≤ mt − 1 and i1 + i2 + ··· + it = k = m1 + m2 + ··· + mt. This is clearly impossible.