Given any $R$-module $M$, there is a scheme which corresponds to the 'total space' of $M$, given by $$ Tot(M):=Spec( Sym_R(M^*))$$ where $M^*$ is the dual module $Hom_R(M,R)$ and $Sym_RM^*$ is the symmetric algebra of $M^*$ over $R$. If $M$ happens to a free rank $n$ $R$-module, then $Sym_RM\simeq R[X_1,...,X_n]$. The scheme $Tot(M)$ has a natural map to $Spec(R)$, which is dual to the obvious inclusion $$R\rightarrow Sym_RM$$

If you start with a locally free, finite rank $R$-module $P$, and then consider its total space $Tot(P)$, the corresponding scheme is a vector bundle by your definition. This follows from considering open sets on which $P$ is free, and considering the restriction of $Tot(P)$ over those open sets. Since restriction to an open set is the same as tensoring over the localization, and localization commutes with forming symmetric algebras, the locally freeness becomes your second condition. The first condition is also straightforward.

Then, observe that every vector bundle by your definition arises this way. To see this, follow Mike's comment. Associate to a vector bundle $V$ its sheaf of sections over $Spec(R)$, which is an $R$-module in a natural way. It will be free over the open cover $D(a_i)$, with constant rank $n$.

Edit: As pointed out by roger, the total space construction should use the dual of $M$. As a side note, this means that it is the same if you replace $M$ with $M^{**}$, and so it is not interesting to apply this construction to non-reflexive modules.

Given any $R$-module $M$, there is a scheme which corresponds to the 'total space' of $M$, given by $$ Tot(M):=Spec( Sym_R(M*))$$ where $M*$ is the dual module $Hom_R(M,R)$ and $Sym_RM*$ is the symmetric algebra of $M^*$ over $R$. If $M$ happens to a free rank $n$ $R$-module, then $Sym_RM\simeq R[X_1,...,X_n]$. The scheme $Tot(M)$ has a natural map to $Spec(R)$, which is dual to the obvious inclusion $$R\rightarrow Sym_RM$$

If you start with a locally free, finite rank $R$-module $P$, and then consider its total space $Tot(P)$, the corresponding scheme is a vector bundle by your definition. This follows from considering open sets on which $P$ is free, and considering the restriction of $Tot(P)$ over those open sets. Since restriction to an open set is the same as tensoring over the localization, and localization commutes with forming symmetric algebras, the locally freeness becomes your second condition. The first condition is also straightforward.

Then, observe that every vector bundle by your definition arises this way. To see this, follow Mike's comment. Associate to a vector bundle $V$ its sheaf of sections over $Spec(R)$, which is an $R$-module in a natural way. It will be free over the open cover $D(a_i)$, with constant rank $n$.

Edit: As pointed out by roger, the total space construction should use the dual of $M$. As a side note, this means that it is the same if you replace $M$ with $M^{**}$, and so it is not interesting to apply this construction to non-reflexive modules.

Given any $R$-module $M$, there is a scheme which corresponds to the 'total space' of $M$, given by $$ Tot(M):=Spec( Sym_R(M))$$ where $Sym_RM$ is the symmetric algebra of $M$ over $R$. If $M$ happens to a free rank $n$ $R$-module, then $Sym_RM\simeq R[X_1,...,X_n]$. The scheme $Tot(M)$ has a natural map to $Spec(R)$, which is dual to the obvious inclusion $$R\rightarrow Sym_RM$$

If you start with a locally free, finite rank $R$-module $P$, and then consider its total space $Tot(P)$, the corresponding scheme is a vector bundle by your definition. This follows from considering open sets on which $P$ is free, and considering the restriction of $Tot(P)$ over those open sets. Since restriction to an open set is the same as tensoring over the localization, and localization commutes with forming symmetric algebras, the locally freeness becomes your second condition. The first condition is also straightforward.

Then, observe that every vector bundle by your definition arises this way. To see this, follow Mike's comment. Associate to a vector bundle $V$ its sheaf of sections over $Spec(R)$, which is an $R$-module in a natural way. It will be free over the open cover $D(a_i)$, with constant rank $n$.

Given any $R$-module $M$, there is a scheme which corresponds to the 'total space' of $M$, given by $$ Tot(M):=Spec( Sym_R(M^*))$$ where $M^*$ is the dual module $Hom_R(M,R)$ and $Sym_RM^*$ is the symmetric algebra of $M^*$ over $R$. If $M$ happens to a free rank $n$ $R$-module, then $Sym_RM\simeq R[X_1,...,X_n]$. The scheme $Tot(M)$ has a natural map to $Spec(R)$, which is dual to the obvious inclusion $$R\rightarrow Sym_RM$$

If you start with a locally free, finite rank $R$-module $P$, and then consider its total space $Tot(P)$, the corresponding scheme is a vector bundle by your definition. This follows from considering open sets on which $P$ is free, and considering the restriction of $Tot(P)$ over those open sets. Since restriction to an open set is the same as tensoring over the localization, and localization commutes with forming symmetric algebras, the locally freeness becomes your second condition. The first condition is also straightforward.

Then, observe that every vector bundle by your definition arises this way. To see this, follow Mike's comment. Associate to a vector bundle $V$ its sheaf of sections over $Spec(R)$, which is an $R$-module in a natural way. It will be free over the open cover $D(a_i)$, with constant rank $n$.

Edit: As pointed out by roger, the total space construction should use the dual of $M$. As a side note, this means that it is the same if you replace $M$ with $M^{**}$, and so it is not interesting to apply this construction to non-reflexive modules.