 # \mathrm{End}_{\mathbb R[x]}(M) where M = \frac{\mathbb R[x]}{(x^2 + 1)} is ngihlungeqtr 2022-05-02 Answered

${\mathrm{End}}_{\mathrm{ℝ}\left[x\right]}\left(M\right)$ where $M=\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}$ is a module over the ring $\mathbb{R}\left[x\right]$

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Step 1
For any R-module N, it can be shown that ${\text{End}}_{R}\left(N\right)$ is a R-module, hence the result is certainly not gonna be a group like $G{L}_{2}$. In fact,
${\text{End}}_{\mathbb{R}\left[x\right]}\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}\stackrel{\sim }{=}{\text{End}}_{\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}}\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}\stackrel{\sim }{=}\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}\stackrel{\sim }{=}\mathbb{C}$
The first isomorphism follows from the general fact
${\text{Hom}}_{R}\left(\frac{M}{IM},\frac{N}{IN}\right)\stackrel{\sim }{=}{\text{Hom}}_{\frac{R}{I}}\left(\frac{M}{IM},\frac{N}{IN}\right)$
It's clear that any $\frac{R}{I}$ -hom is also R-hom. On the other hand, for any R-homomorphism $\rho :\frac{M}{IM}⇒\frac{N}{IN}$, we have $\rho \left(am\right)=a\rho \left(m\right)=0$ for any . Therefore $\rho$ factors through $\frac{R}{I}$.