Galous group, modules

  • Thread starter tohauz
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  • #1
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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)[tex]\cong[/tex]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 [tex]\phi[/tex]:M->N be an R-module hom. Prove that if induced map [tex]\bar{\phi}[/tex]:M/IM->N/IN is surjective, then [tex]\phi[/tex] is surjective.
4. show that 2[tex]\otimes[/tex]1 [tex]\neq[/tex]0 in 2Z[tex]\otimes[/tex]Z/2Z.
 

Answers and Replies

  • #2
matt grime
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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).
 
  • #3
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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
 
  • #4
matt grime
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Have you tried working to find an inverse? The second part of my hint wasn't just decoration.
 
  • #5
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Have you tried working to find an inverse? The second part of my hint wasn't just decoration.
yeah, i got it. thanks
 
  • #6
139
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#3 is somewhat tricky. The key is to prove that [tex] \varphi(IM) = IN [/tex]. This can be done by induction on [tex] k [/tex], but first you'll need a lemma to the effect that if [tex] \bar{\varphi} [/tex] is surjective, then the induced map [tex] \psi_r : I^r M/I^{r+1} \to I^r N/I^{r+1} [/tex] is surjective for all [tex] r [/tex].

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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