# commutative diagram

• Feb 12th 2009, 08:38 AM
peteryellow
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.
• Feb 12th 2009, 05:32 PM
NonCommAlg
Quote:

Originally Posted by peteryellow
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: $c_2=u_2(b_2)=-u_2 \beta(b_1)=-\gamma u_1(b_1)=\gamma(-u_1(b_1)) \in \text{im} \gamma. \ \Box$
• Feb 13th 2009, 10:58 AM
peteryellow
whiy is it true that if $

gf=0,
$
then

$

\text{im}f \subseteq \ker g.
$
• Feb 14th 2009, 04:11 AM
kalagota
Quote:

Originally Posted by peteryellow
whiy is it true that if $

gf=0,
$
then

$

\text{im}f \subseteq \ker g.
$

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 $f(\alpha) \in \mbox{Im } f$.
• Feb 16th 2009, 09:53 AM
peteryellow
How can I use this exercise to show the following:

There are given two short exact sequences

$

0 \longrightarrow K {\longrightarrow} P \longrightarrow M\longrightarrow 0,
$

and $

0 \longrightarrow K' {\longrightarrow} P' \longrightarrow M\longrightarrow 0,
$

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

P + K' is isomorphic to P' + K, here + denotes direct sum.
• Feb 16th 2009, 09:03 PM
NonCommAlg
Quote:

Originally Posted by peteryellow

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,
$
and $0 \longrightarrow K' \overset{u'}{\longrightarrow} P' \overset{v'} \longrightarrow M \longrightarrow 0,
$
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!
• Feb 16th 2009, 11:29 PM
peteryellow
Thanksssssss alot for your help. :-)