Question

Let R be a commutative ring. Prove that Hom_R(R, M) and M are isomorphic R-modules

Commutative Algebra
ANSWERED
asked 2021-01-19
Let R be a commutative ring. Prove that \(\displaystyle{H}{o}{m}_{{R}}{\left({R},{M}\right)}\) and M are isomorphic R-modules

Answers (1)

2021-01-20
Define \(\displaystyle{F}:{H}{o}{m}{\left({R},{M}\right)}\to{M}\) by \(\displaystyle{F}{\left(\phi\right)}=\phi{\left({1}\right)}\in{M}\)
First, we proved that F is an R-module homomorphism.
Let \(\displaystyle\phi,\rho\in{H}{o}{m}{\left({R},{M}\right)}{\quad\text{and}\quad}{r}\in{R}\). Then,
\(\displaystyle{F}{\left({r}\phi-\rho\right)}={\left({r}\phi-\rho\right)}{\left({1}\right)}={r}\phi{\left({1}\right)}-\rho{\left({1}\right)}={r}{F}{\left({r}\right)}-{F}{\left({r}\right)}\)
Hence, F is an R-module homomorphism.
Now, to prove that it is an isomorphic R-module, we must prove that F is bijective.
Suppose \(\displaystyle{F}{\left(\phi\right)}={0}\) for some \(\displaystyle\phi\in{H}{o}{m}{\left({R},{M}\right)}\). Then, \(\displaystyle\phi{\left({1}\right)}={0}.\)
For any \(\displaystyle{r}\in{R},\phi{\left({r}\right)}={r}\phi{\left({1}\right)}={r}\cdot{0}={0}\). Hence, \(\displaystyle\phi={0}\). Thus, F is injective.
Now, suppose \(\displaystyle{m}\in{M}\).
Define a map \(\displaystyle\phi:{R}\to{M}\) by \(\displaystyle\phi{\left({r}\right)}={r}{m}\) for any \(\displaystyle{r}\in{R}.\)
Let \(\displaystyle{r},{s},{t}\in{R}.\)
\(\displaystyle\Rightarrow\phi{\left({r}{s}-{t}\right)}={\left({r}{s}-{t}\right)}{m}={r}{\left({s}{m}\right)}-{t}{m}={r}\phi{\left({s}\right)}-\phi{\left({t}\right)}\)
Hence, \(\displaystyle\phi\in{H}{o}{m}{\left({R},{M}\right)}.\)
Futher, F \(\displaystyle{\left(\phi\right)}=\phi{\left({1}\right)}={m}.\)
Thus, F is surjective.
F is homomorphism and bijective. Therefore, \(\displaystyle{H}{o}{m}{\left({R},{M}\right)}{\quad\text{and}\quad}{M}\) are isomorphic R-module.
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