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I think this is true if the module is finite, using Lagrange's Theorem. However, is this the case for infinite modules (or rings, or groups)? This comes from me trying to prove that tensoring is right-exact; at some point in my proof, I am trying to show that if $0\to M\xrightarrow[]{f} N\xrightarrow[]{g} P\to 0$ is exact, and $\text{im}(f\otimes \text{id}_C) \subseteq \text{ker}(g\otimes \text{id}_C)$, then $(N\otimes C)/\text{im}(f\otimes \text{id}_C) \cong (N\otimes C)/\text{ker}(g\otimes \text{id}_C)$ implies that $\text{im}(f\otimes \text{id}_C)= \text{ker}(g\otimes \text{id}_C)$.

IAAW
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  • For the ring case, do you mean $B, C$ to be ideals of $A$. Anyway, for an example with vector spaces, consider $A = \bigoplus_{n=1}^\infty \mathbb{R}$, $B = \langle e_1 \rangle$, $C = { 0 }$. (On the other hand, if you have that $A/B \simeq A/C$ specifically via the induced map $A/B \to A/C$, then of course that does imply $B = C$.) – Daniel Schepler Feb 04 '23 at 00:40
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    For the question in the title: No, for example $\Bbb Q/2\Bbb Z \cong \Bbb Q/\Bbb Z$ as $\Bbb Z$-modules (= abelian groups). For the problem in the body: How did you get $(N\otimes C)/\text{im}(f\otimes \text{id}_C) \cong (N\otimes C)/\text{ker}(g\otimes \text{id}_C)$? – azif00 Feb 04 '23 at 04:07
  • @azif00 first I showed that $g\otimes \text{id}_C$ is surjective. By the First Isomorphism Theorem, $(N\otimes C)/\text{ker}(g\otimes \text{id}_C) \cong P\otimes C$. Then I defined a map $h:(N\otimes C)/\text{im}(f\otimes \text{id}_C) \to P\otimes C$ by $\overline{n\otimes c} \mapsto g(n)\otimes c \in P\otimes C$, and in fact, $h$ is an isomorphism. Since isomorphism is transitive... well, you get the rest. Thank you for the $\mathbb{Q}/2\mathbb{Z}$, $\mathbb{Q}/\mathbb{Z}$ example. It was really elucidating! – IAAW Feb 04 '23 at 05:34
  • I'm also noticing that this question has been answered here. Perhaps I should delete this post? – IAAW Feb 04 '23 at 06:01
  • @iateawalrus You’re almost done it! Note that $\ker h = \ker(g \otimes \text{id}_C)/\text{im}(f \otimes \text{id}_C)$, and since $h$ is injective we get… – azif00 Feb 04 '23 at 08:13

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You can fit those modules into the following diagram, enter image description here

So now your want to construct a morphism from $B$ to $C$, which seems is doable by diagram-chasing (though, I am lazy and did not check this, I will just let other people, or you, tell me if my guess is correct).

Nicolas Bourbaki
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    This only works if the right square commutes. And, even if $B \cong C$, we don’t necessarily have $B=C$. – azif00 Feb 04 '23 at 04:08