"1. Let X be a set in C^n. Show V(I(X)) contains X. "
If x is in X, then every function vanishing on X vanishes on x, so x is in V(the functions vanishing on X), which is V(I(X)).
"2. Give an example of X where they're equal, and an example where they're not. "
If X is empty, then I(X) = C[x_1..x_n], and V(I(X)) is also empty (because for any point, there's a function not vanishing there).
If X is an infinite set in C^1, then I(X) = {0} because a polynomial p vanishes at z iff p is divisible by x-z, so it would have to be divisible by all x-z for all z in X. And then it would be infinite degree, if it weren't 0. Then, V(I(X)) = C^1. If X is infinite but not all of C^1, we have our example.
"3. Let I be an ideal in C[x_1,...,x_n]. Show I(V(I)) contains I. "
If i is in I, then i vanishes on V(I), so i is in the ideal of functions vanishing on V(I), i.e. i is in I(V(I)).
"4. If I,J are ideals, let I+J := {i+j : i in I, j in J}. Show I+J is an ideal. "
0's in there, and it's obviously closed under addition.
If p is in there, then p is of the form i+j. Then for any r, rp = r(i+j) = ri+rj, where ri is in I and rj is in J. Hence ri+rj is in I+J, so rp is in I+J, which was the condition we needed for an ideal.
"5. Show I+I = I. "
I+I is the ring elements of the form i+j, where i,j are in I.
(It is NOT the ring elements of the form i+i. For example, if the ring is Z, and I = Z, then I+I = I not the even integers.)
Since I is assumed to be closed under addition, each i+j is in I. That's one containment.
For the other, note that if j=0 then i+j = i, so everything in I is in I+I.
"6. Let I = < g_1, ..., g_m >. Show that V(I) = the set of x in C^n where every g_i vanishes. (Make sure you understand why that's different from the definition!)"
V(I) is the set of x where every p in I vanishes. p is in I if it's of the form sum_j r_j g_j.
If g_j vanishes at x, then so does r_j g_j, and so if each g_j vanishes there then so does p.
Hence, if every g_i vanishes at x, then every p in I vanishes at x, i.e. x is in V(I).
Conversely, if x is in V(I), then every p in I vanishes at x, so in particular every g_i vanishes at X.
"7. Assume g_1, ..., g_m are homogeneous linear polynomials, and let I be the ideal generated by them. Let p be another homogeneous linear polynomial. How would you test whether p is in I? (Describe an algorithm, perhaps, that correctly answers "yes" or "no" after finite time.)"
We're trying to write p = sum_i q_i g_i. Break q_i into its constant term c_i plus the rest, r_i.
Then p = sum_i c_i g_i + sum_i r_i g_i. The first of these two sums has only linear terms, and the second has no linear terms (just higher degree), so there can be no cancelation between them. Moreover p has only linear terms by assumption, so the sum_i r_i g_i part must vanish. The upshot is that it's no easier or harder to do this with general q_i than it is with constant coefficients. So the question becomes, can we write p as sum_i c_i g_i?
Write out the n coefficients of g_i as a row vector, and put them together into an mxn matrix.
Now we're asking whether p's row vector is in the row span of that matrix.
The algorithm that tests this is usually called Gaussian elimination (even though the Chinese invented it many centuries earlier).
"8. What if the (g_i) and p in #7 are all homogeneous of the same degree, but that degree isn't necessarily 1? "
The same degree argument as in #7 says that we only need to consider constant coefficients c_i. Now again, make each g_i and p into a row vector, except that the columns are now indexed by the monomials of that degree, instead of x_1..x_n (the monomials of degree 1). Then Gaussian elimination does the job once more.
