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Math Help - 3 exercises-groups,rings,ideals

  1. #1
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    3 exercises-groups,rings,ideals

    1)prove that every finite subgroup of Q/Z is cyclic.
    Last edited by tom007; January 12th 2010 at 03:33 AM.
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  2. #2
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    Quote Originally Posted by tom007 View Post
    1)prove that every finite subgroup of Q/Z is cyclic.
    2)let R={a/b; a,b ∈ Z and b is not divisable by 11} Show that R is factorization domain. How many prime elements are in R up to associates (Elements a,b are said to be associates if a | b and b | a. ) ? How do ideals look like in R?
    3)a)Find all invertible elements in Z/10Z. Describe a group of invertible elements up to isomorphism. Find all classes associates in Z/10Z.
    b)Find all maximal ideals of Z/10Z
    c)Find all prime ideals of Z/10Z
    d)Find all irreducible elements in Z/10Z
    e)Find all prime elements in Z/10Z

    pls help

    If you want help then first show what've you done so far in each exercise, otherwise it's like asking others to give you the whole answer.

    Tonio
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    1) G={m1/n1 + Z,..., mk/nk + Z} finite simple of Q/Z where gcd(mi,ni)=1, set n=lcm(n1,...nk)
    i don't know how to show that G is <1/n + Z>

    2) invertible :b/a 11 doesnt divide 1 irred: 11k/b

    3) Z/10Z ={(1),...(9)}
    (2)=(4)
    inver: (1)^(-1)=1, (2)^(-1)=8, (3)^(-1)=7, (4)^(-1)=4,(5)^(-1)=5,(6)^(-1)=6,(7)^(-1)=3,(8)^(-1)=2,(9)^(-1)=9
    Last edited by tom007; January 11th 2010 at 10:05 AM.
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    Tom007's solution is incorect.
    you need to understand what operation is invovled in Z/10Z here.
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    Last edited by tom007; January 11th 2010 at 08:33 AM.
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  7. #7
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    is there any1 who could help me with #3
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    Quote Originally Posted by tom007 View Post
    is there any1 who could help me with #3

    An element a\in\mathbb{Z}\slash n\mathbb{Z} is invertible iff (a,n)=1

    Tonio
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    tnx alot Tonio..i'll try now sth with this...(sun)
    do u know #1?

    sorry for bothering u.. i don't have much time, and have to do many things.. so, any help is welcome
    i really appreciate ur help Tonio!
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    Quote Originally Posted by tom007 View Post
    tnx alot Tonio..i'll try now sth with this...(sun)
    do u know #1?

    sorry for bothering u.. i don't have much time, and have to do many things.. so, any help is welcome
    i really appreciate ur help Tonio!


    Let G\le\mathbb{Q}\slash\mathbb{Z} be a finite group. Let us agree to write elements in \mathbb{Q}\slash\mathbb{Z} as \frac{\alpha}{\beta}+\mathbb{Z}\,,\,\,\alpha\,,\,\  beta\in\mathbb{Z}\,,\,\,\beta\neq 0\,,\,\,(\alpha,\beta)=1 .

    \mbox{Let }\frac{a}{n}+\mathbb{Z}\in G\mbox{ be such that }\forall\,\,\frac{\alpha}{\beta}+\mathbb{Z}\,,\,\,  n\geq \beta. Now prove the following:

    1) \exists\,\,x\,,\,y\in\mathbb{Z}\,\,\,s.t.\,\,\,ax+  ny=1

    2) From the above it follows that \frac{1}{n}+\mathbb{Z}\in G

    3) If there's an element \frac{r}{s}+\mathbb{Z}\in G s.t. \frac{r}{s}+\mathbb{Z}\notin \left<\frac{1}{n}+\mathbb{Z}\right> , then we get a contradiction to the maximality of n

    4) It follows that G=\left<\frac{1}{n}+\mathbb{Z}\right>

    Tonio

    Ps. BTW, from the above it almost follows, adding some minor details, that \mathbb{Q}\slash\mathbb{Z} has a unique (cyclic, of course) subgroup of order n for every n\in\mathbb{N}
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  11. #11
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    thank you very much Tonio!!!!
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