## M953 Homework 4

Due Friday, February 22, 2002

[1] Let f : V -> **A**^{m} be a morphism of algebraic sets.
Let W be an algebraic set in **A**^{m}. 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 **A**^{m}, there are
polynomials g_{1}, ..., g_{r} in k[**A**^{m}]
such that W = V(g_{1}, ..., g_{r}). If (since it is is hard to
write math symbols on the web) we denote the intersection
of a list S_{1}, ..., S_{j} of sets by
Int(S_{1}, ..., S_{j}), then
recall that W = V(g_{1}, ..., g_{r}) =
Int(V(g_{1}), ..., V(g_{r})), so
f^{ - 1}(W) = f^{ - 1}(V(g_{1}, ..., g_{r}))
= f^{ - 1}(Int(V(g_{1}), ..., V(g_{r}))). But the inverse
image of an intersection is the intersection of the inverse images, so
f^{ - 1}(Int(V(g_{1}), ..., V(g_{r}))) =
Int(f^{ - 1}(V(g_{1})), ..., f^{ - 1}(V(g_{r}))).
Also, f^{ - 1}(V(g_{1})) = {v in V : f(v) is in V(g_{i}) for all i}
= {v in V : g_{i}((f(v)) = 0 for all i}
= Int(g_{1}f, ..., g_{r}f), and this is an algebraic set in V
since each g_{i}f (by our alternate definition of morphism) is in k[V].

[2] Again let f : V -> **A**^{m} be a morphism of algebraic sets.
- (a) If V is irreducible, show that V(I(f(V))) is an irreducible
algebraic subset of
**A**^{m}.
- (b) In particular, if f : V -> W is a
surjective morphism of algebraic sets with V irreducible, conclude that
W is irreducible, too.

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 f_{i} : **A**^{n} -> **A**^{1},
defined by f_{i}((a_{1}, ... , a_{n})) =
a_{i}, is a morphism, and determine the corresponding homomorphism
of coordinate rings.

Solution: Let k[**A**^{1}] = k[y], and let
k[**A**^{n}] = k[x_{1}, ... , axsub>n].
Then f_{i}(x_{1}, ... , x_{n}) = x_{i},
so for any g(y) in k[y], we have g(f_{i}(x_{1}, ... , x_{n}))
g(x_{i}). I.e., gf is just g with x_{i} in place of y, so
gf is in k[x_{1}, ... , axsub>n] for all g in k[y]. Thus
f_{i} is a morphism, and the homomorphism
f_{i}^{~} : k[**A**^{1}] -> k[**A**^{n}]
is the one that sends y to x_{i} (i.e., in any polynomial g(y),
substitute x_{i} 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
f_{1} : **A**^{2} -> **A**^{1}. Then V is an algebraic
set, but its image f_{1}(V) is **A**^{1} - {0}. Since the algebraic
subsets of **A**^{1} are either finite or all of **A**^{1},
f_{1}(V) is not an algebraic set, in spite of the fact (by the last problem)
that f_{1} 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 : **A**^{1} -> **A**^{3}
by f = (t^{3}, t^{4}, t^{5}). [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, a^{4/3}, a^{5/3}).
Thus X is the image of the map **A**^{1} -> **A**^{3}
defined as f(t) = (t, t^{4/3}, t^{5/3}). But this is not a
morphism, since the component functions are not polynomials: they have fractional powers.
If we replace t by t^{3} in (t, t^{4/3}, t^{5/3}) we get
f(t) = (t^{3}, t^{4}, t^{5}), 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(**A**^{1}) = 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(**A**^{1}) and **A**^{1} 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