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

ngihlungeqtr
2022-05-02
Answered

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

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Luke Kane

Answered 2022-05-03
Author has **14** answers

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,

$\text{End}}_{\mathbb{R}\left[x\right]}\frac{\mathbb{R}\left[x\right]}{({x}^{2}+1)}\stackrel{\sim}{=}{\text{End}}_{\frac{\mathbb{R}\left[x\right]}{({x}^{2}+1)}}\frac{\mathbb{R}\left[x\right]}{({x}^{2}+1)}\stackrel{\sim}{=}\frac{\mathbb{R}\left[x\right]}{({x}^{2}+1)}\stackrel{\sim}{=}\mathbb{C$

The first isomorphism follows from the general fact

${\text{Hom}}_{R}(\frac{M}{IM},\frac{N}{IN})\stackrel{\sim}{=}{\text{Hom}}_{\frac{R}{I}}(\frac{M}{IM},\frac{N}{IN})$

It's clear that any$\frac{R}{I}$ -hom is also R-hom. On the other hand, for any R-homomorphism $\rho :\frac{M}{IM}\Rightarrow \frac{N}{IN}$ , we have $\rho \left(am\right)=a\rho \left(m\right)=0$ for any $a\in I,\text{}m\in \frac{M}{IM}$ . Therefore $\rho$ factors through $\frac{R}{I}$ .

For any R-module N, it can be shown that

The first isomorphism follows from the general fact

It's clear that any