M953 Homework 1

Due Friday, January 25, 2002

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 = {f1g1 + ... + frgr : fi is in I for all i, and gi is in J for all i}. Solution: (a) Elements of IJ are sums of products of an element of I with an element of J. Clearly, IJ is not empty (it contains 0, for sure). A sum of two such sums is just another such sum and hence is in IJ. Thus IJ is closed under addition. Also, if b is in IJ then b = f1g1 + ... + frgr for some fi in I and gi in J. So for any h in R, we have hb = (hf1)g1 + ... + (hfr)gr, and since I is an ideal we know hfi is in I for all i, and hence hb is in IJ. Thus IJ is an ideal of R if it is a subgroup of R under addition. But taking h to be -1 shows that IJ is closed under taking additive inverses, so it is a subgroup.

(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 = f1g1 + ... + frgr for some fi in I and gi in J, and hence h(p) = 0 for every h in IJ. Thus p is in Z(IJ). Similarly, if p is in Z(J), then p is in Z(IJ). Thus Z(IJ) contains the union of Z(I) and Z(J). For the reverse inclusion, it is enough to show that if p is in neither Z(I) nor Z(J), then p is not in Z(IJ). But if p is not in Z(I), then there is an f in I such that f(p) is not 0, and if p is not in Z(J) then there is a g in J such that g(p) is not 0. But fg is in IJ and (fg)(p) = f(p)g(p) is not 0 so p is not in Z(IJ), as we wanted to show.

[2] Problem 4 on p. 6: if k is an infinite field and f is a nontrivial element of k[x1, ... ,xn], then f(c) is nonzero for some c = (c1, ... ,cn) in kn.

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[x1, ... ,xn]. If n = 1, f can have at most d roots, where d is the degree of f. Thus f(c) is nonzero for some c in k. So assume n > 1. We can take k[x1, ... ,xn] = R[xn], where R = k[x1, ... ,xn-1]; i.e., f is a polynomial in the single variable xn with coefficients in k[x1, ... ,xn-1]. Since f is not trivial, one of these coefficients is nontrivial, so by induction there is a c = (c1, ... ,cn-1) such that at this c the coefficient is nonzero. Thus f(c1, ... ,cn-1,xn) is not the zero polynomial, so by the n = 1 case we know there is some value cn such that f(c1, ... ,cn-1,cn) is nonzero, as we wanted to show.

[3] Let f be an element of the polynomial ring k[x0, ... ,xn] over an algebraically closed field k. Solution: (a) Consider the ring R = k[x0, ... ,xn][t]. We get an element of R by considering h(t) = f(tx0, ... ,txn). Thus h is a polynomial in the variable t with coefficients in k[x0, ... ,xn]. Since f is not homogeneous, we know the highest and lowest powers of t in h are different. Let gm and gM be the coefficients of these least and largest powers of t (resp.) in h. (For future reference, let dm be the degree of gm and let dM be the degree of gM.) Thus gmgM is a nonzero polynomial in k[x0, ... ,xn], so by Problem 2 there is a b = (b0, ... ,bn) in kn+1 such that (gmgM)(b) is nonzero. Thus f(tb0, ... ,tbn) is a polynomial in t with at least two terms (a term of degree dm and a term of degree dM), which we can factor as f(tb0, ... ,tbn) = (tdm)q(t). Note that q is a nonconstant polynomial with a nonzero constant term. Thus q(0) is not 0, but since k is algebraically closed, we know q has a root so q(r) is zero for some (nonzero) value r of t. On the other hand, q has only a finite number of roots, so since k (being algebraically closed) is infinite, we can pick some nonzero s in k such that q(s) is not zero. Now take a = rb and c = s/r. Then f(a) = f(rb) = (rdm)q(r) = 0, but f(ca) = f(sb) = (sdm)q(s) is nonzero.
(b) This is obvious if f is the 0 polynomial. Otherwise, if f is homogeneous, then f(ca) = cdf(a), where d is the degree of f, so f(ca) = 0 for all c if f(a) = 0.

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

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 = arxr + ... + a0, for some a's in R where ar isn't 0, and we can write G = bsxs + ... + b0, for some b's in R, where bs isn't 0. Now FG = arbsxr+s + ... + a0b0, but R is a domain so arbs is not 0. Since all other terms of FG have degree less than r+s, nothing can cancel with this leading term, so FG has at least one term, hence FG is not 0, so R[x] is a domain.
(b) Suppose F and G are not both forms but FG is a nonzero form. We can write F = Fr + ... + Fr+s and G = Gu + ... + Gu+v, where each Fi and Gi is a form of degree i. (Thus Fi is just the sum of all terms of F of degree i, so what this says is that F has terms of degrees r through r+s.) Since at least one of F and G is not a form, we know that either s > 0 or v > 0 (or both). But the sum of all terms of FG of greatest degree is just Fr+sGu+v, which is nonzero by (a). And the sum of all terms of FG of least degree is FrGu, which is nonzero by (a). Thus FG is not homogeneous, since it has terms of degree r+s and r+s+u+v, and r+s < r+s+u+v.

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

Solution: Note that if I is the ideal in k[x1, ... ,xn] of all polynomials vanishing at a point p = (p1, ... , pn) in kn, it is easy to see that Z(I) = {p} (since xj - pj is in I for each j). Now, if k is finite, then so is kn, so every subset of kn is finite. But given any finite set p1, ..., pr of points, if Ij is the ideal of all polynomials vanishing at pj, by Problem 1 we see that {p1, ..., pr} = Z(I1 ... Ir). Thus {p1, ..., pr}, and hence every finite subset, is algebraic.

[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 Z in the reals k = R. No polynomial vanishes on all of Z except 0, but 0 vanishes everywhere. Thus Z is not the simultaneous set of solutions of any set of polynomials, and hence Z is not algebraic. Since Z is a countable union of single points and any single point is algebraic, we see countable unions of algebraic sets need not be algebraic.

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

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, t2, t3) with t in k is just the set of solutions to y - x2 = 0, z - x3 = 0. Thus it is algebraic.
(b) This set is just the unit circle in the plane; it is the solution set of x2 + y2 - 1 = 0, so it too is algebraic.

[8] Problem 16 on p. 12: Let V and W be algebraic sets in An(k). Show that V = W if and only if I(V) = I(W). Also, show that the statement of this problem can be false if V is not an algebraic set.

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 Z in the reals W = A1(R). Then I(V) = I(W) = (0), but Z is not equal to R.