0

If I have $A$ and $B$ two modules, is the following reasoning true? $(A+B)/B=\{a+b+B : a\in A, b \in B \}=\{a+B : a \in A \}=A/B$

I am just doubting it since it is weirdly similar to the second isomorphism theorem but still not the same thing.

roi_saumon
  • 4,196

1 Answers1

0

No, consider abelian groups as $\Bbb Z$-modules:

By the second isomorphism theorem your proposition gives $A / B = B / (A \cap B)$

Let $A=\Bbb Z_{12}$, $B = \Bbb Z_3$. Then $A/B = \Bbb Z_4$ but $B/(A \cap B)$ is a quotient group of $\Bbb Z_3$ so it is not equal to $\Bbb Z_6$.

bsbb4
  • 3,636
  • But what does $\mathbb{Z_{12}} \cap \mathbb{Z_3}$ mean? – roi_saumon Nov 22 '18 at 12:58
  • You can identify $\Bbb Z_3$ with its corresponding subgroup in $\Bbb Z_{12}$. Then $\cap$ gives just $\Bbb Z_3$, so the quotient is trivial. – bsbb4 Nov 22 '18 at 15:32