Endℝ[x](M)&nbsp;where&nbsp;M=R[x](x2+1)&nbsp;is a module over the ring&nbsp;R[x]

Solved: "Endℝ[x](M)whereM=R[x](x2+1)is a module over the ringR[x]"

Octavio Chen

2022-03-31

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

Tristatex9tw

Beginner2022-04-01Added 18 answers

Step 1
For any R-module N, it can be shown that

{End}_{R} (N)

is a R-module, hence the result is certainly not gonna be a group like

G L_{2}

. In fact,

{End}_{R [x]} \frac{R [x]}{(x^{2} + 1)} \tilde{=} {End}_{\frac{R [x]}{(x^{2} + 1)}} \frac{R [x]}{(x^{2} + 1)} \tilde{=} \frac{R [x]}{(x^{2} + 1)} \tilde{=} C

The first isomorphism follows from the general fact

{Hom}_{R} (\frac{M}{I M}, \frac{N}{I N}) \tilde{=} {Hom}_{\frac{R}{I}} (\frac{M}{I M}, \frac{N}{I N})

It's clear that any

\frac{R}{I}

-hom is also R-hom. On the other hand, for any R-homomorphism

ρ : \frac{M}{I M} \to \frac{N}{I N}

, we have

ρ (a m) = a ρ (m) = 0

for any

a \in I, m \in \frac{M}{I M}

. Therefore

ρ

factors through

\frac{R}{I}

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