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Math Help - Quotient Rings and Homomorphisms

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    Quotient Rings and Homomorphisms

    Let R and S be rings. Show that pi: RxS--> R given by pi(r,s)=r is a surjective homomorphism whose kernel is isomorphic to S.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by iwonde View Post
    Let R and S be rings. Show that \pi:R\times S\mapsto R given by (r,s)\mapsto r is a surjective homomorphism whose kernel is isomorphic to S
    Clearly since 0_R\in R,0_S\in S we know that R,S\ne\varnothing. Obviously every projection is surjective since given any r\in R then for an arbitrary s\in S we have that (r,s)\in R\times S and (r,s)\mapsto r.

    It is clearly a homomorphism too since \pi\left((r,s)+(r',s')\right)=\pi\left(\left(r+r',  s+s'\right)\right)=r+r'=\pi\left((r,s)\right)+\pi\  left((r',s')\right). You can figure out the multiplicative part.

    Also, r,s)\mapsto 0_R\right\}" alt="\ker\pi=\left\{(r,s)r,s)\mapsto 0_R\right\}" />. Clearly then \ker \pi=\{0_R\}\times S. So, simply define \phi:\ker \pi\mapsto S by (0_R,s)\mapsto s. This is readily verified to be an isomorphism
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