1. ## First Isomorphism Theorem

Let S,T be subspaces of the vector spaces V ,U (respectively), and let

$T : V × U -> (V/S)× (U/T)$ be the linear map defined by $T(x, y) = (x +S, y + T).$ Find the kernel of T and prove that

$(V× U)/(S× T) (~ =) (V/S) × (W/T).$

Plx help...

2. First, do you mean to define $T(x, y) = (x +S, y + T)$?

I think that your first step is to determine the kernel of $T$. If $T(x,y)=(0 +S, 0 +T)$, what can be said about $x$ and $y$?

3. Originally Posted by roninpro
First, do you mean to define $T(x, y) = (x +S, y + T)$?

I think that your first step is to determine the kernel of $T$. If $T(x,y)=(0 +S, 0 +T)$, what can be said about $x$ and $y$?
is it Ker T= T(0,0)=(0+S, 0+T) =(S,T)
therefor Ker T = (S,T)?

but I still don't know how to prove the following....

4. Originally Posted by mathbeginner
is it Ker T= T(0,0)=(0+S, 0+T) =(S,T)
therefor Ker T = (S,T)?

but I still don't know how to prove the following....
You have that $F:V\boxplus U\to (V/S)\boxplus (U/T)x,y)\mapsto \left(x+S,y+T)" alt="F:V\boxplus U\to (V/S)\boxplus (U/T)x,y)\mapsto \left(x+S,y+T)" /> (I use $\boxplus$ to differentiate between external and internal direct sum), this is evidently a surjective linear homomorphism. Then, we are trying to look for what values $(x,y)$ is $T(x,y)=...$...equals what? We're trying to look for the origin in the vector spaces $V/S$ and $U/T$ but these are precisely $S$ and $T$ respectively. So, for what $x$ is $x+S=S$ and for what $y$ is $y+T=T$? It is fairly clear that these statements are true if and only if $x\in S$and $y\in T$. Thus, a quick proof shows that $\ker F=S\boxplus T$ and so by the F.I.T. $\left(V\boxplus U\right)/\left(S\boxplus T\right)\cong \left(V/S\right)\boxplus\left(U/T\right)$

5. I don't reli get why Ker F = S X T and how to prove that???

6. Originally Posted by mathbeginner
I don't reli get why Ker F = S X T and how to prove that???
Come on friend. Review your definitions. You're trying to show that $\ker F=S\boxplus T$ which is equivalent to showing $F(x,y)=(S,T)\Leftrightarrow (x,y)\in S\boxplus T$. I give the solution below "hidden". Try your damnedest, and only look at it as a last result.
[spoil]Clearly if $(x,y)\in S\boxplus T$ then $F(x,y)=(x+S,y+T)=(S,T)$ since $x+S=\left\{x+s:s\in S\right\}=S$ since $S\leqslant V$ and similarly $y+T=T$. Conversely, suppose that \$latex F(x,y)=(S,T)[/tex] then $x+S=S$. But, note that $x\in x+S$ since $0\in S$ and so $x\in S$. Similarly $y+T=T\implies y\in T$ and so $F(x,y)=(S,T)\implies (x,y)\in (S,T)$. Combining these we get $F(x,y)=(S,T)\Leftrightarrow (x,y)\in S\boxplus T$[/spoil]

EDIT: I can't figure out how to hide it...so work on the honor code haha

7. Originally Posted by Drexel28
EDIT: I can't figure out how to hide it...so work on the honor code haha
[Spoiler] [/Spoiler] You have been gone for so long that you forgot the tags!

8. Originally Posted by TheCoffeeMachine
[Spoiler] [/Spoiler] You have been gone for so long that you forgot the tags!
Haha, I guess so!