\mathrm{End}_{\mathbb R[x]}(M) where M = \frac{\mathbb R[x]}{(x^2 + 1)} is

ngihlungeqtr 2022-05-02 Answered

End[x](M) where M=R[x](x2+1) is a module over the ring R[x]

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Answers (1)

Luke Kane
Answered 2022-05-03 Author has 14 answers
Step 1
For any R-module N, it can be shown that EndR(N) is a R-module, hence the result is certainly not gonna be a group like GL2. In fact,
EndR[x]R[x](x2+1)=EndR[x](x2+1)R[x](x2+1)=R[x](x2+1)=C
The first isomorphism follows from the general fact
HomR(MIM,NIN)=HomRI(MIM,NIN)
It's clear that any RI -hom is also R-hom. On the other hand, for any R-homomorphism ρ:MIMNIN, we have ρ(am)=aρ(m)=0 for any aI, mMIM. Therefore ρ factors through RI.
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