# Galous group, modules

tohauz
I was doing some self study and have questions:
1. p(x)=x^{7}+11 over Q(a), R.
where a is 7-th root of unity. What are Galouis groups?
For the 1st case I got Z_{7}, second not sure. need hint for that
2. need hint. I know it is easy: M is an R-module. Show that Hom_{R}(R,M)$$\cong$$M.
3. Spse that I is an ideal of R such that I^{k}=0 for some k>0 integer. Let M, N be R-modules and let $$\phi$$:M->N be an R-module hom. Prove that if induced map $$\bar{\phi}$$:M/IM->N/IN is surjective, then $$\phi$$ is surjective.
4. show that 2$$\otimes$$1 $$\neq$$0 in 2Z$$\otimes$$Z/2Z.

Homework Helper
2. You need a map from

{f : R-->M } to M

(or vice versa).

Assuming that R is unital, then there is only one possible map you can write down:

f ---> f(1)

You have to try to show that is an isomorphism. You also may want to think about the other direction;

M --> {f: R --> M}

again, there is only one possible map you can write down - given m in M, then the only candidate in Hom_R(R,M) is translation by m:

f_m(r)= r.m

so you have to show that the map m--->f_m is an isomorphism (note we've dropped the explicit use use R being unital).

tohauz
2. You need a map from

{f : R-->M } to M

(or vice versa).

Assuming that R is unital, then there is only one possible map you can write down:

f ---> f(1)
This is what I tried. g(f)=f_{1}. But I had hard time showing that it is surjective

Homework Helper
Have you tried working to find an inverse? The second part of my hint wasn't just decoration.

tohauz
Have you tried working to find an inverse? The second part of my hint wasn't just decoration.

yeah, i got it. thanks

VKint
#3 is somewhat tricky. The key is to prove that $$\varphi(IM) = IN$$. This can be done by induction on $$k$$, but first you'll need a lemma to the effect that if $$\bar{\varphi}$$ is surjective, then the induced map $$\psi_r : I^r M/I^{r+1} \to I^r N/I^{r+1}$$ is surjective for all $$r$$.

Let me think about the others some more...I'll get back to you in a bit. (Although #4 looks pretty trivial.)

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