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! :)

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.