M953 Homework 4

Due Friday, February 22, 2002



[1] Let f : V -> Am be a morphism of algebraic sets. Let W be an algebraic set in Am. Show that f -1(W) is an algebraic subset of V. [Aside: this is analogous to the fact that the inverse image of a closed subset under a continuous map is closed.]

Solution: Since W is an algebraic subset of Am, there are polynomials g1, ..., gr in k[Am] such that W = V(g1, ..., gr). If (since it is is hard to write math symbols on the web) we denote the intersection of a list S1, ..., Sj of sets by Int(S1, ..., Sj), then recall that W = V(g1, ..., gr) = Int(V(g1), ..., V(gr)), so f - 1(W) = f - 1(V(g1, ..., gr)) = f - 1(Int(V(g1), ..., V(gr))). But the inverse image of an intersection is the intersection of the inverse images, so f - 1(Int(V(g1), ..., V(gr))) = Int(f - 1(V(g1)), ..., f - 1(V(gr))). Also, f - 1(V(g1)) = {v in V : f(v) is in V(gi) for all i} = {v in V : gi((f(v)) = 0 for all i} = Int(g1f, ..., grf), and this is an algebraic set in V since each gif (by our alternate definition of morphism) is in k[V].

[2] Again let f : V -> Am be a morphism of algebraic sets.

Solution: (a) Arguing contrapositively, assume V(I(f(V))) is not irreducible. Then there are proper algebraic subsets A and B of V(I(f(V))) whose union is V(I(f(V))); we will write Union(A,B) for the union of A and B. Thus Union(A,B) = V(I(f(V))), so f - 1(V(I(f(V)))) = f - 1(Union(A,B)) = Union(f - 1(A), f - 1(B)). By Problem 1, we know that f - 1(A) and f - 1(B) are algebraic subsets of V, and since Union(A, B) = V(I(f(V))) contains f(V), we know Union(f - 1(A), f - 1(B)) contains V. If we can show neither f - 1(A) nor f - 1(B) is all of V, then V is not irreducible, as we wanted to show. But if, say, f - 1(A) = V, then f(V) is contained in A, so I(f(V)) contains I(A) so W = V(I(f(V))) is contained in A, contrary to assumption.

(b) Say W = f(V) is an algebraic set. Then W = V(I(W)) = V(I(f(V))) is irreducible if V is, by part (a).

[3] Show that projection fi : An -> A1, defined by fi((a1, ... , an)) = ai, is a morphism, and determine the corresponding homomorphism of coordinate rings.

Solution: Let k[A1] = k[y], and let k[An] = k[x1, ... , axsub>n]. Then fi(x1, ... , xn) = xi, so for any g(y) in k[y], we have g(fi(x1, ... , xn)) g(xi). I.e., gf is just g with xi in place of y, so gf is in k[x1, ... , axsub>n] for all g in k[y]. Thus fi is a morphism, and the homomorphism fi~ : k[A1] -> k[An] is the one that sends y to xi (i.e., in any polynomial g(y), substitute xi in for y).

[4] Show that the image of an algebraic set need not be an algebraic set. [Hint: look at the projection of the hyperbola xy = 1.]

Solution: Look at V = V(xy-1) under the morphism f1 : A2 -> A1. Then V is an algebraic set, but its image f1(V) is A1 - {0}. Since the algebraic subsets of A1 are either finite or all of A1, f1(V) is not an algebraic set, in spite of the fact (by the last problem) that f1 is a morphism.

[5] Do Problem 2-8(b) on p. 39; i.e., show that X = V(ideal(x*z-y^2, y*z-x^3, z^2-x^2*y)) is irreducible [N.B.: the book gives a hint]. Also, check to see what Macaulay2 has to say about the irreducible components of X.

Solution: Define a morphism f : A1 -> A3 by f = (t3, t4, t5). [The book's hint, that y^3 - x^4 and z^3 - x^5 are in I(X), suggested to me that the points of X are all of the form (a, a4/3, a5/3). Thus X is the image of the map A1 -> A3 defined as f(t) = (t, t4/3, t5/3). But this is not a morphism, since the component functions are not polynomials: they have fractional powers. If we replace t by t3 in (t, t4/3, t5/3) we get f(t) = (t3, t4, t5), which is a morphism.]

Note that f(t) is in X for every t: the value of x*z-y^2 at f(t) is (t^3)(t^5) - (t^4)^2 = t^8 - t^8 = 0. Likewise, y*z-x^3 and z^2-x^2*y are 0 at f(t), so the image of f is in X. In fact, f(A1) = X. To see this, take any (a,b,c) in X. If a = 0, then x*z-y^2 = 0 means that b = 0, so by z^2-x^2*y = 0, we see that c = 0, too. Thus f(0) = (0,0,0) = (a,b,c). Now say a is not 0. Then from y*z-x^3 = 0 we see that b and c are also not 0. Let t = c^2/a^3. Then f(t) = ((c^2/a^3)^3, (c^2/a^3)^4, (c^2/a^3)^5). But z^2-x^2*y = 0 means (as long as x is not 0) that y = (z/x)^2, and so from x*z-y^2 = 0 we get z = y^2/x = z^4/x^5 or z^3 = x^5 (as long as z is not 0). Thus c^6 = (c^3)^2 = (a^5)^2 = a^10, so (c^2/a^3)^3 = c^6/a^9 = a^10/a^9 = a. Likewise, (c^2/a^3)^5 = c^10/a^15 = c^10/(a^5)^3 = c^10/c^9 = c, and using y = (z/x)^2 we see that (c^2/a^3)^4 = ((c^2/a^3)^5/(c^2/a^3)^3)^2 = b. Thus f(c^2/a^3) = (a,b,c).

Since X = f(A1) and A1 is irreducible, it follows by Problem 2b that X is irreducible.

Finally, we check Macaulay2's answer. We find that Macaulay2 also sees only one irreducible component; i.e., X is irreducible over the rationals: 
i1 : R = QQ[x,y,z]

o1 = R

o1 : PolynomialRing

i2 : J = ideal(x*z-y^2, y*z-x^3, z^2-x^2*y)

               2           3           2     2
o2 = ideal (- y  + x*z, - x  + y*z, - x y + z )

o2 : Ideal of R

i3 : decompose J

              2         3         2     2
o3 = {ideal (y  - x*z, x  - y*z, x y - z )}

o3 : List