# Residue classes

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• Sep 22nd 2009, 03:16 AM
Carl
Residue classes
I have this exercise, where I'm a little stuck. Any help would be nice.

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

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

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

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

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

And (3) I would say yes because $\displaystyle 2^{n_1}3^{m_1}*2^{n_2}3^{m_2}=2^{n_1+n_2}3^{m_1+m_ 2}=2^{n}3^{m}\in S$?
• Sep 22nd 2009, 01:13 PM
NonCommAlg
Quote:

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

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

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

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

Quote:

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

Quote:

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

$\displaystyle S^{-1}(\mathbb{Z}/2\mathbb{Z})=0$ because $\displaystyle 2 \in S$ and thus $\displaystyle 1/s = 2/2s = 0.$ similarly since $\displaystyle 12 \in S,$ we have $\displaystyle S^{-1}(\mathbb{Z}/12\mathbb{Z})=0.$ now since for any $\displaystyle r \in \mathbb{Z}/{5\mathbb{Z}}$ there exists $\displaystyle s \in S$ such that

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

Quote:

And (3) I would say yes because $\displaystyle 2^{n_1}3^{m_1}*2^{n_2}3^{m_2}=2^{n_1+n_2}3^{m_1+m_ 2}=2^{n}3^{m}\in S$?
thst's right.
• Sep 22nd 2009, 02:25 PM
Carl
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 $\displaystyle \mathbb{Z}/2\mathbb{Z}$ and $\displaystyle \mathbb{Z}/5\mathbb{Z}$ like the general case $\displaystyle I_1 \cap I_2$, 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?
• Sep 22nd 2009, 05:25 PM
NonCommAlg
Quote:

Originally Posted by Carl
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 $\displaystyle \mathbb{Z}/2\mathbb{Z}$ and $\displaystyle \mathbb{Z}/5\mathbb{Z}$ like the general case $\displaystyle I_1 \cap I_2$, 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): $\displaystyle R=\mathbb{Z}, \ I_1=2\mathbb{Z}, \ I_2=5\mathbb{Z}$ and so $\displaystyle Ann_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/5\mathbb{Z})=2\mathbb{Z} \cap 5\mathbb{Z}=10\mathbb{Z}.$

for (4): let $\displaystyle M_1=\mathbb{Z}/2\mathbb{Z}, \ M_2=\mathbb{Z}/12\mathbb{Z}, \ M_3=\mathbb{Z}/5\mathbb{Z}.$ then, as i explained in my previous post: $\displaystyle S^{-1}(M_1 \oplus M_2 \oplus M_3) \cong S^{-1}M_1 \oplus S^{-1}M_2 \oplus S^{-1}M_3 \cong S^{-1}M_3$ since $\displaystyle S^{-1}M_1=0, \ S^{-1}M_2=0.$
• Sep 22nd 2009, 09:29 PM
Carl
Ah, ok, I got it know (Happy) thanks.