For us a ring will always mean a commutative ring with multiplicative identity 1 not equal to 0.

[1] Recall that if I and J are ideals in a ring R, then IJ = {f

- (a) Show that IJ is an ideal.
- (b) If R = k[x
_{1}, ... ,x_{n}] is a polynomial ring over a field k, show that Z(IJ) is the union of Z(I) and Z(J).

(b) If p is a point of Z(I), then f(p) = 0 for every f in I. Thus h(p) = 0 for every h of the form h = f

[2] Problem 4 on p. 6: if k is an infinite field and f is a nontrivial element of k[x

Solution: Note that k must be an infinite field, since we can always find a nontrivial polynomial over a finite field k which has every element of k as a root. So let f be a nontrivial element of k[x

[3] Let f be an element of the polynomial ring k[x

- (a) If f is not homogeneous, show that there is a
point a = (a
_{0}, ... ,a_{n}) and a nonzero constant c of k such that f(a) = 0 but f(ca) is not 0. - (b) If f is homogeneous, show that if f(a) = 0, then f(ca) = 0 for all c in k.

(b) This is obvious if f is the 0 polynomial. Otherwise, if f is homogeneous, then f(ca) = c

[4] Problem 1 on p. 6: Let R be a domain.

- (a) If F and G are forms of degree r and s resp.
in R[x
_{1}, ... ,x_{n-1}], show FG is a form of degree r+s. - (b) Show that any factor of a form in R[x
_{1}, ... ,x_{n-1}] is also a form.

Solution: (a) Every term of F has degree r and every term of G has degree s, so by the distributive property of multiplication, every term of FG has degree r+s, so FG is homogeneous. And unless everything cancels so that FG ends up being 0, we see that FG must have degree r+s; i.e., FG is a form of degree r+s. So to show that FG can't be 0, it's enough by induction to show that R[x] is a domain if R is. But if F and G are nonzero polynomials in R[x] of degrees r and s resp., then we can write F = a

(b) Suppose F and G are not both forms but FG is a nonzero form. We can write F = F

[5] Problem 9 on p. 9: if k is a finite field, show that every subset of

Solution: Note that if I is the ideal in k[x

[6] Problem 10 on p. 9: Give an example of a countable collection of algebraic sets whose union is not algebraic.

Solution: Take the integers

[7] Problem 11(a,b) on p. 9:

- (a) Show that { (t, t
^{2}, t^{3}) : t in k } is an algebraic subset of**A**^{3}(k). - (b) Show that { (cos(t), sin(t)) : t real } is an algebraic subset
of
**A**^{2}(**R**).

Solution: (a) By Problem 6 we may as well assume k is infinite. We will work in k[x,y,z]. The set of points (t, t

(b) This set is just the unit circle in the plane; it is the solution set of x

[8] Problem 16 on p. 12: Let V and W be algebraic sets in

Solution: Clearly, if V = W, then I(V) = I(W). For the other implication, note that if X is an algebraic set then by definition X = Z(J) for some ideal J. In particular, every polynomial f in J vanishes on X, so J is contained in the ideal I(X) of all polynomials that vanish on X. But if you add equations to a set of equations, you can lose solutions but you can't gain solutions. Thus J being in I(X) means that Z(I(X)) is contained in Z(J) = X. But clearly X is contained in Z(I(X)) (since every polynomial in I(X) by definition vanishes on all of X), so X = Z(I(X)) = Z(J) = X.

Thus we see that Z(I(V)) = V and Z(I(W)) = W, since V and W are algebraic. So if I(V) = I(W), then V = Z(I(V)) = Z(I(W)) = W, as we wanted to show.

This can be false if V or W is not algebraic. Take V to be the integers