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# dec02 - Let R be an integral domain and let S be a subring...

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Math 5125 Friday, December 2 December 2, Ungraded Homework Let R be a ring with a 1 and let Q be an injective left R -module. Suppose x R is not a right zero divisor, that is rx 6 = 0 whenever 0 6 = r R . Prove that xQ = Q . Deﬁne ι : R R by ι ( r ) = rx . Then ι is an R -module homomorphism and is one-to-one, because x is not a right zero divisor. Suppose q Q and deﬁne θ : R Q by θ ( 1 ) = q , or more explicitly θ ( r ) = rq . Since Q is injective, there exists an R -map φ : R Q such that φι = θ . Set p = φ ( 1 ) . Then q = θ ( 1 ) = φι ( 1 ) = φ ( x ) = x φ ( 1 ) = xp and the result is proven.
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Unformatted text preview: Let R be an integral domain and let S be a subring of R which is a PID. Let Q be an injective R-module. Prove that Q is also an injective S-module. Note that R and S have the same 1 (or maybe we assume this). Since Q is an injective R-module, we see that rQ = Q for all r ∈ R \ 0, in particular sQ = Q for all s ∈ S \ 0. It follows that Q is an injective S-module, because S is a PID....
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