# Ideals in an integral domain and Galois theory S_5

1. Jun 12, 2007

### bham10246

Hi, I've attempted these problems since Sunday morning but I'm still stuck! Please help... These are not homework problems but these are some problems that I'm working on over the summer by myself.

1. The problem statement, all variables and given/known data

1. Let I_1, I_2,... be ideals in an integral domain.
a. If I_1 intersect I_2 intersect ... intersect I_m = 0, show at least one I_i must equal zero.
b. Assume that I_1 intersect I_2 intersect ... = 0 for infinitely many non-zero I_i's. Prove that I_k intersect I_{k+1} intersect .... = 0 for all positive integers k.

2.
a. Let E be a Galois extension of a field F with characteristic 0. Prove that there is a unique smallest subfield K such that F subset of K subset of E, K is normal over F and E is subradical over K. Use this fact: Let A and B be solvable subgroups of a group G and suppose that A is normal in G. Then AB is solvable.
b. Let f be an irreducible polynomial over the rationals Q which has degree 5 and at least two complex roots. Prove that Gal(f) has order 10, 20, 60 or 120.

2. The attempt at a solution
My attempts at these problems:
1. a. Well, let x be an element of the intersection I_1 intersect I_2 intersect ... I_m. Let r be a nonzero element of the ring R. Then rx is in the intersection. Since the intersection equals 0, rx = 0. Since R is an integral domain and r is not equal to zero, x = 0. But this doesn't show that I_i =0 for some i.... THIS METHOD IS WRONG. BUT I SOLVED IT USING INDUCTION!
b. Suppose J := I_k intersect I_{k+1} intersect .... is NOT equal to 0 for some k >0. Then consider I_1 intersect I_2 intersect ... intersect J. Since this equals zero and J is not equal to zero, I_i equals zero for some i. But I don't see my contradiction... THIS METHOD IS WRONG AS WELL. THIS PROBLEM IS WRONG, I THINK. DO OTHERS AGREE WITH ME?
Counterexample 1: Let I_1 = 0, I_j =Z (the integers) for all j >1.
Counterexample 2: Let I_1 = 0, I_2 contains I_1, I_3 contains I_2, etc...

But I STILL need HELP with this....
2.a. ?? See Hungerford?
b. I see that G(f) is isomorphic to a subgroup of S_5. So |G(f)| divides 120. Maybe we can take these into cases:
Case 1. We have only two complex roots.
Case 2. We have only two complex roots and three real roots that permute.
Case 3. We have four complex roots.

What I don't understand is why do we have the possibility that |Gal(f)| = 10?

Thank you so much. I tried talking about these with a friend but we made little progress.

Last edited: Jun 12, 2007
2. Jun 12, 2007

### Hurkyl

Staff Emeritus
These proposed counterexamples do not satisfy the hypothesis that the I_i's are nonzero.

3. Jun 12, 2007

### Hurkyl

Staff Emeritus
I don't follow the part in green: you assumed x was in the zero ideal, and proved that it was zero. I'm not sure why you say you solved it. And I don't see induction used anywhere in this solution...

4. Jun 13, 2007

### bham10246

Hi Hurkyl,

My Counterexample 1: Let I_1 = 0, I_j =Z (the integers) for all j >1. It didn't say all the I_i's must be non-zero. Here, all except I_1 is equal to the integers. The I_j's are not distinct.

Counterexample 2: Let I_1 = 0, I_2 contains I_1, I_3 contains I_2, etc...
Now suppose the I_j's must be distinct. Then take I_1 = 0, I_2 = 2Z, I_3 = 4Z, I_4 = 8Z, I_5 = 16Z....

Another counterexample with all I_j's distinct: Take I_1 = 0, I_2 = 2Z, I_3 = 3Z, I_4 = 4Z, .... The intersection is still 0 but the conclusion doesn't hold.

Last edited: Jun 13, 2007
5. Jun 13, 2007

### bham10246

As for the induction, the contrapositive of "If I_1 intersect I_2 intersect ... intersect I_m = 0, show at least one I_i must equal zero" is "if none of the I_i equal 0, then I_1 intersect I_2 intersect ... intersect I_m does not equal zero."

Proof: Take m=2 as my base case. Suppose I_1 and I_2 are not the zero ideal. Then there exists nonzero elements x_1 in I_1 and x_2 in I_2. Then x_1*x_2 is not equal to 0 because R is an integral domain. x_1*x_2 is in I_1 and x_1*x_2 is in I_2. Thus x_1*x_2 is in I_1 intersect I_2. Since I_1 intersect I_2 contains a nonzero element, I_1 intersect I_2 does not equal zero.

Now suppose I_1, ... I_{m-1} do not equal 0. Our induction step says that I_1 intersect ... I_{m -1} is not 0. Consider a nonzero idea I_m. Then I_1 intersect ... I_m = ( I_1 intersect ... I_{m -1} ) intersect I_m. Since R is an integral domain, this doesn't equal zero. Thus we're done.

6. Jun 13, 2007

### Hurkyl

Staff Emeritus
Yes, it did. I would certainly agree that the wording is quite unfortunate, but the problem does state that all of the I_i's are nonzero.

Since the point of disagreement is whether or not any of the I_i's are allowed to be zero, I don't see why you would proceed from here by stating more examples with I_1 = 0. But, it's good you did, because I can point out some other mistakes.

For positive integers k, both

$$\bigcap_{n = k}^{+\infty} 2^n \mathbb{Z} = 0$$

and

$$\bigcap_{n = k}^{+\infty} (n+1) \mathbb{Z} = 0$$

are true. In fact, we have the following:

Exercise: Let {Ii} be an infinite collection of distinct ideals of Z (or of any Dedekind domain). Then, $\bigcap_i I_i = 0$.

Last edited: Jun 13, 2007
7. Jun 19, 2007

### bham10246

Thanks Hurkyl. I don't see an immediate way to do your exercises, but I did go back and worked on 1b.