Due Friday, January 25, 2002:
Notes: In Problem 1, I wrote IJ down wrong on the problem handout. It's corrected below. Also, you can assume that an algebraically closed field is always infinite. (The proof is similar to the fact that there are an infinite number of primes in the integers.) Finally, Fulton uses V(I) for the set of simultaneous solutions of a set I of polynomial equations (V for variety), whereas I've been using Z (for zero). You can see either approach depending on what book you look at, although some people prefer to use the word variety only when an algebraic set is irreducible (i.e., not empty and not the union of two proper algebraic sets, neither of which contains the other).

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

Read pp. 1-12 of the text and read each of the problems on those pages. Write down any questions you have to discuss on Monday. Feel free to discuss the problems with each other and with me, or to ask about them in class.

[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}.
• (a) Show that IJ is an ideal.
• (b) If R = k[x1, ... ,xn] is a polynomial ring over a field k, show that Z(IJ) is the union of Z(I) and Z(J).
[2] Problem 4 on p. 6.

[3] Let f be an element of the polynomial ring k[x0, ... ,xn] over an algebraically closed field k.
• (a) If f is not homogeneous, show that there is a point a = (a0, ... ,an) and a nonzero constant c of k such that f(a) = 0 but f(ca) is not 0. [Hint: look at the polynomial f(tx0, ... , txn) in the ring k[x0, ... ,xn][t]. Keeping in mind Problem 2, try to plug in for the x's so you get a nonhomogeneous polynomial in t.]
• (b) If f is homogeneous, show that if f(a) = 0, then f(ca) = 0 for all c in k.
[4] Problem 1 on p. 6.

[5] Problem 9 on p. 9.

[6] Problem 10 on p. 9.

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

[8] Problem 16 on p. 12. Also, show that the statement of this problem can be false if V is not an algebraic set.