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Math Help - Short exact sequence

  1. #1
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    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).
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  2. #2
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    Quote Originally Posted by peteryellow View Post
    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)
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  3. #3
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    Quote Originally Posted by peteryellow View Post
    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
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