Please just explain how the sequence are exact?

Asked Nov 08 '19 at 06:51

Active Nov 08 '19 at 07:55

Viewed 59 times

$\text{Ramification Lemma}:$

Let $A$ be a ring, $\alpha$ an ideal of $A$, and $M$ an $A$-module. Then prove $M/ \alpha M \cong (A/\alpha) \otimes_A M $.

Proof: We have the exact sequence: $$0 \to \alpha \to A \to A/\alpha \to 0.$$ (Why ?)

Tensoring with $M$ gives another exact sequence: $$ 0 \to \alpha \otimes A \to A \otimes M \to A/ \alpha \otimes M \to 0.$$ i.e, $$0 \to \alpha \otimes A \to M \to A/ \alpha \otimes M \to 0, $$ because $A \otimes M=M$.

The image of $\alpha \otimes_A M$ in $M$ are sums of elements of the form $a_i \cdot m_i$,

where $a_i \in \alpha$ and $m_i \in M$. This implies $M/ \alpha M \cong A / \alpha \otimes_A M$.

Please just explain how the sequence are exact?

what are the functions like $\alpha \to A$, $A \to A/\alpha$ etc?

edited Nov 08 '19 at 07:55

asked Nov 08 '19 at 06:51

MAS

10,898

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The second sequence is not exact in general. – Mohammad Bagheri Nov 08 '19 at 07:01
Why can't you use $Im$ and $ker$? – Arctic Char Nov 08 '19 at 07:04
@ArcticChar, I mean I can not but it is has to be used. Actually I can not find the required functions like $ \alpha \to A$ and then $A \to A/\alpha$. What are those? – MAS Nov 08 '19 at 07:50
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Ok, the maps are the inclusion $\alpha \to A$ and the projection $A\to A/\alpha$, where $f\mapsto f+\alpha$. – Arctic Char Nov 08 '19 at 07:53
@ArcticChar, Why $A \otimes M=M$ ? Is it by absorption property of tensor product? – MAS Nov 08 '19 at 07:58
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@M.A.SARKAR Observe that the map $x\otimes m \to xm$ yields an isomorphism between $A \otimes M $ and $M$. – juan diego rojas Nov 08 '19 at 23:31
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related: https://math.stackexchange.com/questions/2714188/a-mathfraka-otimes-a-m-simeq-m-mathfrakam-by-tensoring-the-canonic?rq=1 – juan diego rojas Nov 09 '19 at 18:38

Please just explain how the sequence are exact?

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