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Hoang
Hoang
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Does exterior product commute functor Hom(-,M)?

Let $N,M$$M$ be an module over the commutative ring $R$. I'd like to ask Is this isopmrhism true:do we have the following isomorphism?

$$Hom_R(\wedge^n_RN,M)\simeq \wedge^n_R Hom_R(N,M)$$$$Hom_R(\wedge^n_RM,R)\simeq \wedge^n_R Hom_R(M,R)$$

We can obviously see it's true for the case $M=R^m$ is a free $R$-module. But I don't know in general.

Does exterior product commute functor Hom(-,M)?

Let $N,M$ be an module over the commutative ring $R$. I'd like to ask Is this isopmrhism true:

$$Hom_R(\wedge^n_RN,M)\simeq \wedge^n_R Hom_R(N,M)$$

Does exterior product commute functor Hom?

Let $M$ be an module over the commutative ring $R$. I'd like to ask do we have the following isomorphism?

$$Hom_R(\wedge^n_RM,R)\simeq \wedge^n_R Hom_R(M,R)$$

We can obviously see it's true for the case $M=R^m$ is a free $R$-module. But I don't know in general.

Source Link
Hoang
Hoang
  • 71
  • 3

Does exterior product commute functor Hom(-,M)?

Let $N,M$ be an module over the commutative ring $R$. I'd like to ask Is this isopmrhism true:

$$Hom_R(\wedge^n_RN,M)\simeq \wedge^n_R Hom_R(N,M)$$