## Tuesday, May 11, 2010

Here is an additional hint on 3b: Note that with such a substantial hint, I expect cleaner proofs ;)

i) Prove that (I:f) is monomial by induction on the number of terms of f, which we may assume are all *not* in I by part a). To do this, let g be in (I:f). We need to show that every term of g is in (I:f). Consider gf. It is in I, so LT(g)LT(f) is in I.

Now consider (I:LT(g)f). Now (I:f) is a subset of (I:LT(g)f). Also, note that (I:LT(g)f), if prime, is monomial by the inductive hypothesis (Why?!?). Now show that if (I:f) is a proper subset of (I:LT(g)f), then LT(g) \in (I:f), and now repeat the argument for g-LT(g).

The following facts will help you to prove that (I:f) = (I:m) for some monomial m:

ii) Note that since I is monomial, one has that (I:f) = intersection of (I:f_i) for all the terms f_i of f. [You should prove this]

iii) Recall a (jazzed up version of a) homework problem: If P is prime and the product of the ideals I_1, I_2, ..., I_t (denoted I_1I_2I_3...I_t) is in P, then I_j is in P for some j. You may just cite this version without giving the (easy) proof using induction and the previous homework problem.

To show that if I is squarefree, then one can take m to be squarefree, you should look at Theorem 4.4.11 in the book.