Surjective endomorphism of an $R$-module is injective.

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! :)

• No. Don't confuse isomorphism with equality! – darij grinberg Feb 27 '14 at 4:17
• Of course, only structure is preserved, correct? – Zermie Feb 27 '14 at 4:24
• Yes, but you cannot get an equality $I = J$ (or even an isomorphism $I \cong J$) from an isomorphism of quotients $M/I \cong M/J$. – darij grinberg Feb 27 '14 at 4:26
• 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? – KCd Feb 27 '14 at 4:28
• 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? – 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.