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Let $M = \frac{ \mathbb{Z}[X] }{(X^2,2X)} $ et $A = \mathbb{Z} $

$M$ is an $A$-module that is finitely generated because $M$ is generated by $\overline{2}$ and $\overline{X}$. In addition, the torsion of $2$ is $\left\{ 0 \right\} $

I want to show that the torsion of $X$ is not $\left\{ 0 \right\} $.

Let $a \in A$, then $aX = 0 \Leftrightarrow a = 0 $ or $X = 0$ i.e $a = 0$ or $a \neq 0$. It means

$$ \mathrm{Ann}_A{X} : = \left\{ a \in A \mid aX = 0 \right\} = \left\{ 0 \right\} \cup A^* = A \neq \left\{ 0 \right\} $$

Then $\mathrm{Ann}_A{X} \neq \left\{ 0 \right\}$.

Is it correct? Thank you

Tohiea
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    What do you mean by the torsion of an element? – Randall Dec 02 '20 at 15:45
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    Do you perhaps mean the annihilator of $X?$ If so, it would help to notice that $2X$ is in the ideal $(X^2,2X),$ so $2\bar{X}=0$ in $M.$ – Chris Leary Dec 02 '20 at 15:52
  • @Randall Yes sorry, I believe that I did not express myself well. I would mean that $\mathrm{Ann}_A(X) = \left{ a \in A \mid aX = 0 \right} \neq \left{ 0 \right}$ where $\mathrm{Ann}_A(X) $ is the annihilator of $X$ – Tohiea Dec 02 '20 at 15:52
  • @ChrisLeary Yes, okay, since $2 \overline{X} = 0$ in $M$, so the annihilator of $X$ is not $\left{ 0 \right}$ right? – Tohiea Dec 02 '20 at 16:02
  • @Toheia - Right. You should note that there are other integers that also annihilate $X.$ Remember that the annihilator of an element is an ideal in $A.$ Can you see what that ideal must be? – Chris Leary Dec 02 '20 at 17:02
  • @ChrisLeary I am a beginner in abstract algebra so I don’t really know... $A = \mathbb{Z}$ and the only ideals of $A$ is $(n \mathbb{Z})$. Here it may be $( n \mathbb{Z})$ with $n \neq 0$ it is correct ? – Tohiea Dec 02 '20 at 17:34
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    @Toheia - Yes, specifically $2\mathbb{Z}$, what we more usually write as $(2).$ – Chris Leary Dec 02 '20 at 18:56

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