1. Let V be a k-dimensional subspace of C^n.
Let I(V) be generated by the linear polynomials { \sum_i v_i x_i : (v_1,...,v_n) in V }.
How would you find a finite generating set for I(V), in ten words or less?
"Pick a basis."
"Pick a basis of V, and use only those."
That sort of thing.
The point being that all the other linear polynomials are already C-linear combinations, much less R-linear.
2. A Gr\"obner basis is reduced if
a) the coefficient on every leading term is 1,
b) no term in any generator can be reduced using another generator.
Show that every ideal in a polynomial ring has a reduced Gr\"obner basis.
We proved already that I has a finite Gr\"obner basis.
Apply the reduction algorithm to each element of the basis, using only the other generators; if you get to zero throw the element away, otherwise replace it with the reduced version.
Obviously this doesn't change the ideal; to see that it's still a Gr\"obner basis, observe that anything that would reduce to 0 before will still reduce to 0.
As we do these reductions, the leading monomials can only decrease. So at some point they stop decreasing.
Reduce all the generators one last time (not changing the leading monomials). For each one, we get stuck when no term can be reduced using another generator. Since we don't change the leading monomials, we will remain stuck, i.e. we will have condition (b).
Now divide each generator by its leading coefficient, to achieve (a). This obviously doesn't change whether we have a Gr\"obner basis, nor what ideal it generates.
3 = 1+2. Let M be an m x n matrix. Let I(M) = I(the span of the row vectors) from Q#1.
Describe an algorithm to fiddle with M, from which one can read off a reduced Gr\"obner basis for I(M).
The usual algorithm to put a matrix in reduced row-echelon form. Then the conditions on that form exactly say that the generating set is a reduced Gr\"obner basis.
4. Let I be an ideal, and define Rad(I) = {p : for some natural number n, p^n is in I}.
a) Show that Rad(I) contains I.
b) Show that Rad(I) is an ideal.
c) If I = I_X for some subset X of C^n, show that Rad(I) = I.
a) For any p, we can take n=1.
b) It contains I, so it contains 0.
Let p,q be in Rad(I), and m,n such that p^m, q^n are in I.
Then for any r, (rp)^m = r^m p^m is in I too. So rp is in Rad(I).
Consider the binomial expansion of (p+q)^{m+n-1}.
The Ath term is a multiple of p^A b^{m+n-1 - A}.
If A is at least m, then p^A = p^m p^{A-m}, so p^A is in I, so p^A b^{m+n-1 - A} is.
If A < m, then m+n-1-A = n + (m-1 - A) is at least n, so b^{m+n-1-A} is in I.
Hence every term in the binomial expansion is in I. So p+q is in Rad(I).
(If you used m+n instead of m+n-1, or something larger, that's totally fine.)
c) p in Rad(I_X)
=> some p^m in I
=> p^m(x) = 0 for all x in X
=> (p(x))^m = 0 for all x in X
=> p(x) = 0 for all x in X
=> p in I_X
Hence Rad(I_X) is contained in I_X. But we already knew the opposite containment.
