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Let $R$ be a commutative ring. I would like to prove the following two assertions :

(1) If $0 \longrightarrow X \overset{\alpha}{\longrightarrow} Z \overset{\beta}{\longrightarrow} Y$ is an exact sequence of $R$-modules, then $ker(\beta) \simeq X$.

(2) If $X \overset{\alpha}{\longrightarrow} Z \overset{\beta}{\longrightarrow} Y \longrightarrow 0$ is an exact sequence of $R$-modules, then $coker(\alpha) \simeq Y.$

At the moment I only see that by exactness we have $im(\alpha) = Z = ker(\beta)$.
Can someone explain me how to proceed ?

Thanks for your help.

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2 Answers 2

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For (1):

The exact sequence

$0 \longrightarrow X \overset{\alpha}{\longrightarrow} Z \overset{\beta}{\longrightarrow} Y \tag 1$

implies that

$\ker \alpha = \text{Im}(0) = 0, \tag 2$

thus $\alpha:X \to Z$ is injective, meaning

$\alpha:X \to \text{Im} \; \alpha \subset Z \tag 3$

is an isomorphism of $X$ to $\alpha(X) = \text{Im} \; \alpha $; now exactness of (1) yields

$\text{Im} \; \alpha = \ker \beta; \tag 4$

thus $\alpha$ is an isomorphism

$\alpha:X \simeq \ker \beta. \tag 5$

As for (2), the exact sequence

$X \overset{\alpha}{\longrightarrow} Z \overset{\beta}{\longrightarrow} Y \longrightarrow 0 \tag 6$

yields

$\text{Im} \; \beta = \ker (Y \to 0) = Y, \tag 7$

and

$\ker \beta = \text{Im} \; \alpha; \tag 8$

(7) means $\beta$ is surjective; thus

$Z / \ker \beta \simeq Y, \tag 9$

so by (8),

$Z / \text{Im} \; \alpha \simeq Y, \tag{10}$

which is precisely the statement

$\text{coker} \; \alpha \simeq Y; \tag{11}$

see here.

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    $\begingroup$ Thank you very much for this great answer ! I really appreciate your help and your knowledge. $\endgroup$
    – Crystal
    Apr 4, 2018 at 9:13
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Hint:

  1. What is $\ker(\alpha)$? What is the relation between $\text{im}(\alpha)$ and $X$?
  2. $\text{coker}(\alpha)=Z/\text{im}(\beta)$ by definiton. If you can show that $\beta$ is surjective, you can use the 1st isomorphism theorem.
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