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Math Help - commutative diagram

  1. #1
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    commutative diagram

    R is a ring and consider commutative diagram of R -modules with exact rows.

    http://images.planetmath.org:8080/ca...8/l2h/img1.png

    then I want to show that Gamma is isomorphism if and only if

    the sequence
    0----->A_1----->A_2+B_1------>B_2----->0

    here + is direct sum and teh first homomorphism sends a to \alpha(a),k_1(a)
    and the second homomorphism sends (a_2,b) to k_2(a_2)-\beta(b)

    here k_1 is homomorphism from A_1 to B_1 and k_2 is homomorphism from
    A_2 to B_2.
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  2. #2
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    Quote Originally Posted by peteryellow View Post
    R is a ring and consider commutative diagram of R -modules with exact rows.

    http://images.planetmath.org:8080/ca...8/l2h/img1.png

    then I want to show that Gamma is isomorphism if and only if

    the sequence
    0----->A_1----->A_2+B_1------>B_2----->0

    here + is direct sum and teh first homomorphism sends a to \alpha(a),k_1(a)
    and the second homomorphism sends (a_2,b) to k_2(a_2)-\beta(b)

    here k_1 is homomorphism from A_1 to B_1 and k_2 is homomorphism from
    A_2 to B_2.
    ok, like other diagram chasing problems the solution is fairly straightforward but quite lengthy and annoying! i'll give the solution this time and that would also be the last time!

    so the rows in your commutative diagram are 0 \longrightarrow A_j \overset{k_j}{\longrightarrow} B_j \overset{u_j} \longrightarrow C_j\longrightarrow 0, \ \ j=1,2. we also have the sequence 0 \longrightarrow A_1 \overset{f}{\longrightarrow} A_2 \oplus B_1 \overset{g} \longrightarrow B_2 \longrightarrow 0 defined by:

    f(a_1)=(\alpha(a_1),k_1(a_1)), \ \ g(a_2,b_1)=k_2(a_2)-\beta(b_1). so suppoose first that \gamma in your diagram is an isomorphism. see that since k_1 is injective, f is injective too.

    also by the definition: gf =k_2 \alpha -\beta k_1. but since the diagram is commutative, we have k_2 \alpha = \beta k_1. thus gf=0, i.e. \text{im}f \subseteq \ker g. next we'll show that \ker g \subseteq \text{im}f:

    let z=(a_2,b_1) \in \ker g, i.e. k_2(a_2)=\beta(b_1). hence: 0=u_2k_2(a_2)=u_2 \beta (b_1)=\gamma u_1(b_1), which gives us u_1(b_1)=0, because \gamma is an isomorphism. so b_1 \in \ker u_1 = \text{im}\ k_1.

    thus b_1=k_1(a_1), for some a_1 \in A_1. now since k_2(a_2)=\beta k_1(a_1)=k_2 \alpha (a_1), we'll have a_2=\alpha(a_1) because k_2 is injective. hence: z=(\alpha(a_1), k_1(a_1))=f(a_1) \in \text{im} f.

    so we've proved that \text{im} f = \ker g. finally we need to show that g is surjective: let b_2 \in B_2. then u_2(b_2) \in C_2=\gamma(C_1), because \gamma is an isomorphism. so \gamma(c_1)=u_2(b_2),

    for some c_1 \in C_1. now since u_1 is surjective, c_1=u_1(b_1), for some b_1 \in B_1. thus: u_2(b_2)=\gamma u_1(b_1)=u_2 \beta (b_1), which gives us: b_2 - \beta(b_1) \in \ker u_2 = \text{im} \ k_2. so there exists

    a_2 \in A_2 such that b_2 - \beta(b_1) = k_2(a_2). hence b_2=k_2(a_2)+\beta(b_1)=g(a_2, - b_1) \in \text{im} g. this completes the proof of the first half of the problem.

    conversely, suppose that the sequence is exact. we want to show that \gamma is an isomorphism. first we show that \gamma is injective: so suppose \gamma(c_1)=0, for some c_1 \in C_1. since

    u_1 is surjective, we have c_1=u_1(b_1), for some b_1 \in B_1. then: 0=\gamma u_1(b_1)=u_2 \beta(b_1) and hence \beta(b_1) \in \ker u_2 = \text{im} \ k_2. so \beta(b_1)=k_2(a_2), for some a_2 \in A_2, which gives us:

    g(a_2,b_1)=0. thus (a_2,b_1) \in \ker g = \text{im} f. so there exists a_1 \in A_1 such that (a_2,b_1)=f(a_1)=(\alpha(a_1), k_1(a_1)). hence b_1=k_1(a_1) and thus c_1=u_1(b_1)=u_1k_1(a_1)=0.

    finally we need to show that \gamma is surjective: let c_2 \in C_2. then \exists \ b_2 \in B_2: \ u_2(b_2)=c_2, because u_2 is surjective. now since g is surjective, there exists (a_2,b_1) \in A_2 \oplus B_1 such

    that: k_2(a_2)-\beta(b_1)=g(a_2,b_1)=b_2. thus: c_2=u_2(b_2)=-u_2 \beta(b_1)=-\gamma u_1(b_1)=\gamma(-u_1(b_1)) \in \text{im} \gamma. \ \Box
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  3. #3
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    whiy is it true that if <br /> <br />
gf=0,<br />
then

    <br /> <br />
\text{im}f \subseteq \ker g.<br />
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  4. #4
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    Quote Originally Posted by peteryellow View Post
    whiy is it true that if <br /> <br />
gf=0,<br />
then

    <br /> <br />
\text{im}f \subseteq \ker g.<br />

    if gf=0, then for all \alpha, gf(\alpha)= g(f(\alpha))=0. this means that f(\alpha) \in \ker g for all \alpha.. but f(\alpha) \in \mbox{Im } f.
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  5. #5
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    How can I use this exercise to show the following:


    There are given two short exact sequences

    <br /> <br />
0 \longrightarrow K {\longrightarrow} P  \longrightarrow M\longrightarrow 0, <br />
    and <br /> <br />
0 \longrightarrow K' {\longrightarrow} P' \longrightarrow M\longrightarrow 0, <br />

    whereP and P' are projective R-modules. I want to show that

    P + K' is isomorphic to P' + K, here + denotes direct sum.
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  6. #6
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    Quote Originally Posted by peteryellow View Post

    How can I use this exercise to show the following:

    There are given two short exact sequences 0 \longrightarrow K \overset{u}{\longrightarrow} P \overset{v}\longrightarrow M\longrightarrow 0, <br />
and  0 \longrightarrow K' \overset{u'}{\longrightarrow} P' \overset{v'} \longrightarrow M \longrightarrow 0, <br />
whereP and P' are projective R-modules. I want to show that P \oplus K'

    is isomorphic to P' \oplus K.
    let \gamma: M \longrightarrow M be the identity map, which is obviously an isomorphism. now since P is projective and v' is surjective, there exists a map \beta: P \longrightarrow P' such that v'\beta=\gamma v=v. next we will

    define a map \alpha: K \longrightarrow K' such that u' \alpha = \beta u. here's how to define \alpha: let x \in K. then v' \beta u(x)=vu(x)=0. thus \beta u(x) \in \ker v' = \text{im} \ u'. so there exists y \in K' such that \beta u(x)=u'(y).

    define \alpha(x)=y. then \alpha is well-defined because if z \in K' is another element with \beta u(x)=u'(z), then u'(y)=u'(z) and so y=z, because u' is injective. see that \alpha is a homomorphism.

    also from the definition it's clear that u' \alpha = \beta u. now we have all conditions needed in your exercise. hence since \gamma is an isomorphism, the sequence 0 \longrightarrow K \longrightarrow P \oplus K' \longrightarrow P' \longrightarrow 0 is exact

    and hence it splits, because P' is projective. thus P' \oplus K \simeq P \oplus K', and you're happily done!
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  7. #7
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    Thanksssssss alot for your help. :-)
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