# 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 $\displaystyle 0 \longrightarrow A_j \overset{k_j}{\longrightarrow} B_j \overset{u_j} \longrightarrow C_j\longrightarrow 0, \ \ j=1,2.$ we also have the sequence $\displaystyle 0 \longrightarrow A_1 \overset{f}{\longrightarrow} A_2 \oplus B_1 \overset{g} \longrightarrow B_2 \longrightarrow 0$ defined by:

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

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

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

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

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

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

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

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

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

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

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

that: $\displaystyle k_2(a_2)-\beta(b_1)=g(a_2,b_1)=b_2.$ thus: $\displaystyle c_2=u_2(b_2)=-u_2 \beta(b_1)=-\gamma u_1(b_1)=\gamma(-u_1(b_1)) \in \text{im} \gamma. \ \Box$
• Feb 13th 2009, 10:58 AM
peteryellow
whiy is it true that if $\displaystyle gf=0,$ then

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

Originally Posted by peteryellow
whiy is it true that if $\displaystyle gf=0,$ then

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

if $\displaystyle gf=0,$ then for all $\displaystyle \alpha$, $\displaystyle gf(\alpha)= g(f(\alpha))=0$. this means that $\displaystyle f(\alpha) \in \ker g$ for all $\displaystyle \alpha$.. but $\displaystyle f(\alpha) \in \mbox{Im } f$.
• Feb 16th 2009, 09:53 AM
peteryellow
How can I use this exercise to show the following:

There are given two short exact sequences

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

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 $\displaystyle 0 \longrightarrow K \overset{u}{\longrightarrow} P \overset{v}\longrightarrow M\longrightarrow 0,$ and $\displaystyle 0 \longrightarrow K' \overset{u'}{\longrightarrow} P' \overset{v'} \longrightarrow M \longrightarrow 0,$ whereP and P' are projective R-modules. I want to show that $\displaystyle P \oplus K'$

is isomorphic to $\displaystyle P' \oplus K.$

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

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

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

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

and hence it splits, because $\displaystyle P'$ is projective. thus $\displaystyle P' \oplus K \simeq P \oplus K',$ and you're happily done!
• Feb 16th 2009, 11:29 PM
peteryellow
Thanksssssss alot for your help. :-)