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I know this is a duplicate question. However, I haven't seen anything that invokes the isomorphism theorem. Here's my idea:

By the isomorphism theorem we have that $M/\ker\varphi \cong \operatorname{im}\varphi = M$ (as $\varphi$ is surjective). Does this imply $\ker\varphi = (0)$?

While typing this I'm beginning suspect the implication does not follow. Can anyone explain why or why not?

Thank you! :)

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    $\begingroup$ No. Don't confuse isomorphism with equality! $\endgroup$ – darij grinberg Feb 27 '14 at 4:17
  • $\begingroup$ Of course, only structure is preserved, correct? $\endgroup$ – Zermie Feb 27 '14 at 4:24
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    $\begingroup$ Yes, but you cannot get an equality $I = J$ (or even an isomorphism $I \cong J$) from an isomorphism of quotients $M/I \cong M/J$. $\endgroup$ – darij grinberg Feb 27 '14 at 4:26
  • $\begingroup$ When $R = \mathbf Z$ this is a question about abelian groups. Consider the $n$th power mapping from $\mathbf C^\times$ to itself when $n > 1$. This is surjective with kernel the $n$th roots of unity $\mu_n$. Thus $\mathbf C^\times/\mu_n \cong \mathbf C^\times$. Does that mean $\mu_n$ is trivial? $\endgroup$ – KCd Feb 27 '14 at 4:28
  • $\begingroup$ I see. According to Zev's answer my proposition is not true. The only reason I even asked this was from an answer here (the second answer). Is it true only for commutative rings then? $\endgroup$ – Zermie Feb 27 '14 at 4:31
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Let $R=\mathbb{Z}$, and consider the $\mathbb{Z}$-module $M=\prod_{i=1}^\infty\mathbb{Z}$ (the direct product of infinitely many copies of $\mathbb{Z}$). Let $\varphi:M\to M$ be the left-shift homomorphism,defined by $$\varphi(a_1,a_2,a_3,\ldots)=(a_2,a_3,a_4,\ldots)$$ Then $\varphi$ is surjective (given any element $a=(a_1,a_2,\ldots)\in M$, we have $\varphi(0,a_1,a_2,\ldots)=a$) but not injective (we have $\varphi(n,0,0,\ldots)=(0,0,0,\ldots)$ for any $n\in\mathbb{Z}$).

However, if $M$ is noetherian $R$-module, we can conclude that $\varphi$ is injective. See my answer here and this thread for more information.

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  • $\begingroup$ So then for the initial proposition to hold we must require the R-module M to be Noetherian? $\endgroup$ – Zermie Feb 27 '14 at 4:21

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