1. ## ring isomorphism theorem

I have the matrices R=[s,s;0,s] and A=[0,s;0,0]. Now, A is the ideal and it is also supposed to be the Kernel, as I am trying to show that R/ker f is isomorphic to SxS. So, when trying to show A is the kernel, can I just show it takes everything and brings it to the additive identity or is this wrong b/c there is multiplication and addition in rings?

2. Originally Posted by sfspitfire23
I have the matrices R=[s,s;0,s] and A=[0,s;0,0]. Now, A is the ideal and it is also supposed to be the Kernel, as I am trying to show that R/ker f is isomorphic to SxS. So, when trying to show A is the kernel, can I just show it takes everything and brings it to the additive identity or is this wrong b/c there is multiplication and addition in rings?

Please do read carefully what you wrote: what "ideal" is A? In what ring? (Of 2 x 2 matrices, of course...but over what field or ring?)Is it a matrix or a set of a matrices of a definite form? Kernel of what? What is f? What is its definition set(ring, field...whatever) and what its image...??
Really...

Tonio

3. In the first isomorphism theorem for rings, $R/A\cong S\times S$ (which is what my problem is) is equivalent to $R/ker f\cong image f$, yes? So I would have to prove that $A$ from the factor ring $R/A$ which is the ideal because multiplication must be well defined is infact the kernel. But my matrix A looks to be the additive identity, not the multiplicative. Can I just say then that since $im(f)=S\times S$ and we know that the identity is (0,0), A will be the kernel? I defined my mapping by $f[s_1,s_2;0,s_3]=(s_1,s_3)$. Would I somehow have to include the multiplicative identity as well? Does this make sense? haha