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Math Help - ring isomorphism theorem

  1. #1
    Senior Member sfspitfire23's Avatar
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    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?
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  2. #2
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    Quote Originally Posted by sfspitfire23 View Post
    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
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  3. #3
    Senior Member sfspitfire23's Avatar
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    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
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