$\Bbb Z$ as a $\Bbb Z$-Module is not artinian but noetherian

Question

I' ve read that $\Bbb Z$ as a $\Bbb Z$-Module is not artinian, but noetherian.

I' ve thought a lot about this, but I can' t find a proof of that. It might be easy, but I just can' t imagine how to proof that.

Does somebody can help me? I really want to understand this.

score 0 · Answer 1 · edited Jul 12 '16 at 20:25

0

$Z$ is not artinian since the sequence $(2)\supset\dots\supset(2^{n})\supset (2^{n+1})\supset\dotsb$ does not stabilize.

It is Notherian since every subring is generated by a positive integer, $(n)\subset (m)$ implies that $m|n$ thus for a sequence $(a_i)\subset (a_{i+1})$, $a_{i+1}\leq a_i$ the sequence of positive integers $a_i$ has a minimal element $a_{i_0}$ and $(a_i)$ stabilizes at $(a_{i_0})$

edited Jul 12 '16 at 20:25

egreg

238,574

answered Jul 12 '16 at 20:11

Tsemo Aristide

87,475

It would be better to write is as $;(2)\supset \dots\supset(2^n)\supset(2^{n+1}\supset\dots$. – Bernard Jul 12 '16 at 20:14

$\Bbb Z$ as a $\Bbb Z$-Module is not artinian but noetherian

1 Answers1