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Math Help - Residue classes

  1. #1
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    Residue classes

    I have this exercise, where I'm a little stuck. Any help would be nice.

    1) Find Ann_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z})

    2) For ideals I_1 and I_2 in a ring R find Ann_R(R/I_{1}\oplus R/I_{2})

    3) Define S=\{2^n3^m | n,m \in \mathbb{N}_0\}. Is S multiplicatively closed?
    Prove that \overline{5/1}=\overline{0} as an element in S^{-1}(\mathbb{Z}/10\mathbb{Z}) and that \overline{2/1}\neq \overline{0} as an element in S^{-1}(\mathbb{Z}/10\mathbb{Z}).

    4) With S as above find S^{-1}M for each of following modules;
    \mathbb{Z}/2\mathbb{Z}
    \mathbb{Z}/12\mathbb{Z}
    \mathbb{Z}/5\mathbb{Z} and
    \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/12\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z}


    Well, regarding (1) I know that the sum module \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z} is just rows  \left(<br />
\begin{array}{c}<br />
x  \\<br />
y \\<br />
\end{array}\right), right? So I would say that the Annulator would be  \left(<br />
\begin{array}{c}<br />
0 \\<br />
0 \\<br />
\end{array}\right)

    And (3) I would say yes because 2^{n_1}3^{m_1}*2^{n_2}3^{m_2}=2^{n_1+n_2}3^{m_1+m_  2}=2^{n}3^{m}\in S?
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  2. #2
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    Quote Originally Posted by Carl View Post
    I have this exercise, where I'm a little stuck. Any help would be nice.

    1) Find Ann_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z})

    2) For ideals I_1 and I_2 in a ring R find Ann_R(R/I_{1}\oplus R/I_{2})

    if R is a (commutative) ring with identity, we have r \in Ann_R(R/I_1 \oplus R/I_2) \Longleftrightarrow r(R/I_1)=0, \ r(R/I_2)=0 \Longleftrightarrow r \in I_1 \cap I_2. so Ann_R(R/I_1 \oplus R/I_2)=I_1 \cap I_2.



    3) Define S=\{2^n3^m | n,m \in \mathbb{N}_0\}. Is S multiplicatively closed?
    Prove that \overline{5/1}=\overline{0} as an element in S^{-1}(\mathbb{Z}/10\mathbb{Z}) and that \overline{2/1}\neq \overline{0} as an element in S^{-1}(\mathbb{Z}/10\mathbb{Z}).
    in S^{-1}(\mathbb{Z}/10\mathbb{Z}) we have 5/1 = 10/2 = 0/2 = 0. now suppose 2/1 = 0. so there exists s=2^a3^b \in S such that 2s =0, i.e. 10 \mid 2s, which is impossible.


    4) With S as above find S^{-1}M for each of following modules;
    \mathbb{Z}/2\mathbb{Z}
    \mathbb{Z}/12\mathbb{Z}
    \mathbb{Z}/5\mathbb{Z} and
    \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/12\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z}
    S^{-1}(\mathbb{Z}/2\mathbb{Z})=0 because 2 \in S and thus 1/s = 2/2s = 0. similarly since 12 \in S, we have S^{-1}(\mathbb{Z}/12\mathbb{Z})=0. now since for any r \in \mathbb{Z}/{5\mathbb{Z}} there exists s \in S such that

    rs = 1, we have S^{-1}(\mathbb{Z}/5\mathbb{Z})=\{1/s : \ s \in S \} \cup \{0 \}. for the last part use this fact that S^{-1}(M \oplus N) \cong S^{-1}M \oplus S^{-1}N.


    And (3) I would say yes because 2^{n_1}3^{m_1}*2^{n_2}3^{m_2}=2^{n_1+n_2}3^{m_1+m_  2}=2^{n}3^{m}\in S?
    thst's right.
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  3. #3
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    Thanks a lot, you have been very helpful, I appreciate it!

    Ok, I must have misunderstood in (1). Apparently I thought one was just supposed to take the Annihilator of every one of them. So you're saying I should just take the intersection between \mathbb{Z}/2\mathbb{Z} and \mathbb{Z}/5\mathbb{Z} like the general case I_1 \cap I_2, but they have only [0] and [1] in common, correct?

    So in (4) the last one is isomorphic to the sum of each of the fraction sets? And the only thing they all three have in common is zero, therefore zero is the set I'm after?
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  4. #4
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    Quote Originally Posted by Carl View Post
    Thanks a lot, you have been very helpful, I appreciate it!

    Ok, I must have misunderstood in (1). Apparently I thought one was just supposed to take the Annihilator of every one of them. So you're saying I should just take the intersection between \mathbb{Z}/2\mathbb{Z} and \mathbb{Z}/5\mathbb{Z} like the general case I_1 \cap I_2, but they have only [0] and [1] in common, correct?

    So in (4) the last one is isomorphic to the sum of each of the fraction sets? And the only thing they all three have in common is zero, therefore zero is the set I'm after?
    for (1): R=\mathbb{Z}, \ I_1=2\mathbb{Z}, \ I_2=5\mathbb{Z} and so Ann_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/5\mathbb{Z})=2\mathbb{Z} \cap 5\mathbb{Z}=10\mathbb{Z}.

    for (4): let M_1=\mathbb{Z}/2\mathbb{Z}, \ M_2=\mathbb{Z}/12\mathbb{Z}, \ M_3=\mathbb{Z}/5\mathbb{Z}. then, as i explained in my previous post: S^{-1}(M_1 \oplus M_2 \oplus M_3) \cong S^{-1}M_1 \oplus S^{-1}M_2 \oplus S^{-1}M_3 \cong S^{-1}M_3 since S^{-1}M_1=0, \ S^{-1}M_2=0.
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  5. #5
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    Ah, ok, I got it know thanks.
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