Show that each g_i is squarefree, i.e. is not divisible by f^2 for any polynomial f of degree > 0.
If g_i = f^2 h, then h g_i = f^2 h^2 is in I too, so fh is in I (by the assumption I = Rad(I)). Hence some init(g_j) divides init(fh). By the assumption deg f > 0, j is not i. Since init(g_j) divides init(fh), it divides init(
2. Let I be an ideal such that for all f, if f^2 is in I, then f is in I. Show that I is radical.
If I is not radical, then there exists g not in I, and a number N, such that g^N is in I. Let k be the least number such that g^k is in I. (So k > 1, and is at most N.) Let f = g^{k-1}. Then f^2 = g^{2(k-1)} = g^k g^{k-2}, which only makes sense because k is at least 2. In particular, f is not in I, but f^2 is in I.
3. Let I be a monomial ideal generated by squarefree monomials. Show that I is radical.
Hint: show that if p is in I, then init(p) and p-init(p) are in I. Use this to show that it is enough to test the condition in question #2 when f is a monomial.
4. Let I be an ideal with a Gr\"obner basis {f_1,...,f_n}, such that each init(f_i) is a squarefree monomial. Show that I is radical.
#3 is actually the special case of #4 where each f_i = init(f_i). So we'll just do #4.
By #2 it's enough to check that f^2 in I => f in I.
So say f is a polynomial such that f^2 is in I. If f=0 we're done; otherwise we can talk about init(f).
Since f^2 is in I and we have a Gr\"obner basis, some init(f_i) | init(f^2) = init(f)^2.
Since init(f_i) is squarefree, it already divides init(f).
So we can replace f by f - m f_i where m = init(f)/init(f_i), canceling the leading term of f. This new guy has (f - mf_i)^2 = f^2 + f_i(-2m + f_i), so again in I, so we can run the reduction algorithm again. Since it only gets stuck when f=0, we see that f reduces to 0, which means f is in I.
5. Let I and J be two radical ideals.
a) Show that I intersect J is radical.
b) Give an example where I+J (which concatenates their generators) is not radical.
a) If f^N is in I intersect J, then f^N is in I and f^N is in J, hence f is in I and J, hence f is in I intersect J.
b) My favorite example is I = < y >, J = < y-x^2 >. (We can see that J is radical by taking a term order for which y is the leading term and applying #4.)
6. Let I = < ac,bc,bd,ae,de > inside C[a,b,c,d,e]. Decompose V(I) as a union of subspaces. Make it minimal, so no subspace in your list contains another.
If a=0, we have bc,bd,de=0 left.
If b=0, we have de=0 left, so either d=0 or e=0. So far {a=b=d=0} union {a=b=e=0}.
If not b=0, we have c=d=0. This adds {a=c=d=0}.
If not a=0, we have c=e=0, leaving bd=0, so either b=0 or d=0: {c=e=b=0} union {c=e=d=0}.
In the end, the solutions are the union of five 2-planes in 5-space.
