Answers to HW #3

"1. Let phi: R->S be a ring homomorphism, and I an ideal of S. Let J = phi^{-1}(I) := {r in R : phi(r) in I}. Show J is an ideal of R."

The slow way: if j is in J and r is in R, then phi(rj) = phi(r) phi(j) = phi(r) * something in I, and I is an ideal, so phi(r) phi(j) is in I, hence phi(rj) is, hence rj is in J.

The quick way: J is the kernel of the composite ring homomorphism R -> S -> S/I.

"2. Let R = C[x_1..x_n]. For each i in 1..n, and t a nonzero complex number, define the ring homomorphism phi_{i,t} : R -> R by phi_{i,t}(x_j) = x_j for j not equal to i, and phi_{i,t}(x_i) = tx_i. Also phi_{i,t}(c) = c for c in C. (From here, you should be able to figure out what phi_{i,t} does to an arbitrary element of R.)

Let I be a monomial ideal. Show that phi_{i,t}^{-1}(I) = I for every i,t."

Let m_1,..,m_k be a bunch of monomials that together generate I.
Let m'_j = phi_{i,t}^{-1}(m_j), hence m'_j = t^{-e} m_j, where e is the exponent on x_i in m_j.
phi_{i,t}^{-1}(I) is generated by m'_1..m'_j, but the numbers t^{-e} are invertible so don't change the ideal generated. Hence it's again I.

"3. Harder: assume that phi_{i,t}^{-1}(I) = I for every i,t. Show I is a monomial ideal.
Hint: if g is a generator of I, then phi_{i,t}(g) in I for every i,t. Use this to show that g's monomials are also in I."

Let g be in I, and break g = sum_e x_i^e g_i, where g_i doesn't use the variable x_i.
By the assumption, sum_e (tx_i)^e g_e is in I also, for all nonzero t.
Let E be the lowest power of x_i occurring in g; then sum_e t^{e-E} x_i^e g_e is in I for all nonzero t.
Take the limit as t goes to zero, and we get just x_i^E g_E.
Since that's in the ideal, we can subtract it off of g, and start the process over with the new E.

(You might think it's weird to take limits, since this is an algebra problem. But if you think about dealing with ideals in (Z/<2>)[x_1..x_n] instead of C[x_1..x_n], you notice that the only nonzero t is t=1, and now every ideal I satisfies the stated condition!)

"4. Fix i, and call an ideal I of R[x_1..x_n]  x_i-homogeneous if phi_{i,t}^{-1}(I) = I for every t. Show that if I is x_i-homogeneous, then radical(I) is also x_i-homogeneous."

Let p be in radical(I), so p^m is in I for some m.
Let the lowest x_i-term of p be x_i^E p_E (as in question #3).
Then the lowest x_i-term of p^m is x_i^{mE} p_E^m.
Since I is x_i-homogeneous, we learn that x_i^{mE} p_E^m is in I.
Hence its mth root x_i^E p_E is in radical(I).
Subtract that off p and continue (as in question #3).

"5. Let I be a monomial ideal. Show that radical(I) is also a monomial ideal."

Put #3 and #4 together.

"6. Describe an algorithm that, given a system of generators of a monomial ideal, computes the radical (by giving a system of generators)."

Obviously we want the generators to be monomials; separate them into monomials if they're not.
Then for each monomial, if there's an exponent >1 replace it with 1, so now we have a list of squarefree monomials. That's the system of generators of rad(I).

"7. Prove your algorithm works."

First we need to be sure that the squarefree monomials produced this way are indeed in rad(I).
If m = \prod_{i=1}^n x_i^{e_i}, with E = max({e_i}), then multiply it by \prod_{i: e_i not 0} x_i^{E-e_i} to get a new monomial m' in I that uses the same variables, but now all with exponent E.
Its Eth root \prod_{i: e_i not 0} x^i is then the squarefree monomial in rad(I) described in #6.

Now we need to be sure that these generate rad(I).
By #5, we know that rad(I) is a monomial ideal. So we only want to know which monomials are in rad(I). By the argument in the last paragraph, any square-ful monomial in rad(I) is generated by a squarefree monomial also in rad(I), so we only need to know which squarefree monomials are in rad(I).  If g is a squarefree monomial, it's in rad(I)
iff some power of it is in I
iff that power is a multiple of some single generator m (we did this last week)
iff the variables used in g are a superset of the variables used in m
iff the variables used in g are a superset of the variables used in the m' constructed above.
So those m' generate the squarefree monomials in rad(I), which generate rad(I).