# Question on the annihilator $Ann_{\Bbb{Z}} (\Bbb{Z}_6)$ of $\Bbb{Z}_6$ in $\Bbb{Z}$.

by Chris   Last Updated November 18, 2018 18:20 PM

We have the unital commutative ring $$\mathbb{Z}$$ and the $$\Bbb{Z}-$$module $$\Bbb{Z}_6$$. We want to find the (2-sided ideal of $$\mathbb{Z}$$) annihilator

$$Ann_{\Bbb{Z}} (\Bbb{Z}_6):=\{z\in \Bbb{Z}:z\cdot \overline{m} = \overline{0},\ \forall \overline{m} \in \Bbb{Z}_6 \} \subseteq \Bbb{Z}_6$$ of $$\Bbb{Z}_6$$ in $$\Bbb{Z}$$.

The question seems simple, but I m a little bit confused. See:

We want to find $$z\in \Bbb{Z}$$, such that $$z\cdot \overline{m} =\overline{0},\ \forall \bar{m} \in \Bbb{Z}$$. So, \begin{align} z\cdot \overline{m}& =\overline{0} && \iff\\ \overline{zm}=\overline{z}\ \overline{m}& =\overline{0} \end{align}

If $$\overline{m} \in U(\Bbb{Z}_6)=\{\overline{1},\overline{5}\}$$, then $$\overline{z}=\overline{0}\iff 6|z$$.

If $$\overline{m} \in \mathbb{Z}_6 \backslash U(\Bbb{Z}_6)=\{ \overline{0}, \overline{2},\overline{3},\overline{4} \}$$, then

\begin{align} z\cdot \overline{0}=\overline{0},\ z\cdot \overline{2}=\overline{0},\ z\cdot \overline{3}=\overline{0},\ z\cdot \overline{4}=\overline{0} & \iff \\ 6|z,6|2z,6|3z,6|4z& \implies \\ 6|z \end{align} because $$6|z \iff 3|z \wedge 2|z$$ and $$\ 6|4z\implies 3|2z \implies 3|z$$.

So, $$Ann_{\Bbb{Z}} (\Bbb{Z}_6):=\{z\in \Bbb{Z}:6|z,\ z \in \Bbb{Z} \}=6\Bbb{Z} \trianglelefteq \Bbb{Z}.$$

Is this ok? Can we use the relation $$zm\equiv 0 \pmod 6$$ and Number Theory?

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#### Answers 1

Your answer is correct.

Your proof is rather lengthy and too formal.

Please rewrite a shorter proof without missing the essence of the question.

Mohammad Riazi-Kermani
November 18, 2018 17:54 PM

## modul left multiplication

Updated December 22, 2017 22:20 PM