1. ## Short exact sequence

How can I show image of \lambda is equal to kernel $\mu$?

$0\rightarrow {\frak m} \cap {\frak a} \stackrel{\lambda}{\rightarrow} {\frak m} \oplus {\frak a} \stackrel{\mu}{\rightarrow} {\frak m} + {\frak a} \rightarrow 0,$
where
$\lambda$ is defined as $x \mapsto (x,x)$ and $\mu$ is given by $(x,ya)\mapsto (x-ya).$

2. Originally Posted by peteryellow
How can I show image of \lambda is equal to kernel $\mu$?

$0\rightarrow {\frak m} \cap {\frak a} \stackrel{\lambda}{\rightarrow} {\frak m} \oplus {\frak a} \stackrel{\mu}{\rightarrow} {\frak m} + {\frak a} \rightarrow 0,$
where
$\lambda$ is defined as $x \mapsto (x,x)$ and $\mu$ is given by $(x,ya)\mapsto (x-ya).$
I have no idea what the second line is supposed to mean... anyway:

Assuming $\lambda :X \to X\times X, \ \mu :X \times X \to Z$, we get:
$Im(\lambda) = \{(x,x):x \in X\} = span\{(1,1)\}$
$Ker(\mu) = \{(x,y\alpha) \in X \times X : x = y\alpha \} = \{(x,x):x \in X\} = Im(\lambda)$

3. Originally Posted by peteryellow
How can I show image of \lambda is equal to kernel $\mu$?

$0\rightarrow {\frak m} \cap {\frak a} \stackrel{\lambda}{\rightarrow} {\frak m} \oplus {\frak a} \stackrel{\mu}{\rightarrow} {\frak m} + {\frak a} \rightarrow 0,$
where
$\lambda$ is defined as $x \mapsto (x,x)$ and $\mu$ is given by $(x,ya)\mapsto (x-ya).$

I think it'd be nice to have the slightest idea what are we talking here: are ${\frak m}\,,\,{\frak a}$ an ideal in some ring and a module over that ring? Anyway, both opposite containments are pretty obvious, me thinks...unless I'm missing something:

$(x,x)\in Im(\lambda)\Longrightarrow \mu (x,x)=x-x=0\Longrightarrow Im(\lambda)\subset Ker(\mu)\,;\,\,(x,ya)\in Ker(\mu)\Longrightarrow$ $x=ya\in {\frak m}\cap {\frak a}\Longrightarrow (x,ya=x)=\lambda(x)\Longrightarrow (x,ya)\in Im(\lambda)$

Tonio