If $\mathcal{A}$ is some abelian category and $S$ is some set of objects in $\mathcal{A}$, then its "abelian closure" $\overline{S}$ should be the smallest abelian full subcategory of $\mathcal{A}$ containing $S$ such that the inclusion functor is exact (in other words, it's a "sub-abelian category" as Qiaochu calls it). Note that it is automatically closed under isomorphisms (consider $0 \to A \cong A' \to 0$).

It always exists: Consider the set of all full subcategories $T$ of $\mathcal{A}$ containing $S$, which also contain $0$ and are closed with respect to kernels and cokernels. By this I mean that if $f$ is some morphism between objects in $T$, then every kernel/cokernel of $f$ also lies in $T$. Now take their intersection.

If you work with classes instead of universes, this intersection will cause some set-theoretic problems. One can avoid them as follows, thereby giving another construction of $\underline{S}$:

Define full subcategories $S_{\alpha}$ for ordinals $\alpha$ as follows: Let $S_{0} = S \cup \{0\}$. In the limit step, take the union. If $S_{\alpha \omega + 2n}$ is already defined, then adjoin all kernels of all morphisms in this set and optain $S_{\alpha \omega + 2n+1}$. Then, let $S_{\alpha \omega + 2n+2}$ be the closure under cokernels. Finally, define $\overline{S} := \cup_{\alpha} S_{\alpha}$.

As for a specific example, $R$ is "abelian dense" in $\mathrm{Mod}_{fg}(R)$, when $R$ is noetherian.

If $\mathcal{A}$ is some abelian category and $S$ is some set of objects in $\mathcal{A}$, then its "abelian closure" $\overline{S}$ should be the smallest abelian full subcategory of $\mathcal{A}$ containing $S$ such that the inclusion functor is exact (in other words, it's a "sub-abelian category" as Qiaochu calls it). Note that it is automatically closed under isomorphisms (consider $0 \to A \cong A' \to 0$).

It always exists: Consider the set of all full subcategories $T$ of $\mathcal{A}$ containing $S$, which also contain $0$ and are closed with respect to kernels, cokernels and direct sums. By this I mean that if $f$ is some morphism between objects in $T$, then every kernel/cokernel of $f$ also lies in $T$ (similarly with direct sums). Now take their intersection.

If you work with classes instead of universes, this intersection will cause some set-theoretic problems. One can avoid them as follows, thereby giving another construction of $\underline{S}$:

Define full subcategories $S_{\alpha}$ for ordinals $\alpha$ as follows: Let $S_{0} = S \cup \{0\}$. In the limit step, take the union. If $S_{\alpha \omega + 3n}$ is already defined, then adjoin all kernels of all morphisms in this set and optain $S_{\alpha \omega + 3n+1}$. Then, let $S_{\alpha \omega + 3n+2}$ be the closure under cokernels, and let $S_{\alpha \omega + 3n+3}$ be the closure under binary direct sums. Finally, define $\overline{S} := \cup_{\alpha} S_{\alpha}$.

As for a specific example, $R$ is "abelian dense" in $\mathrm{Mod}_{fg}(R)$, when $R$ is noetherian.

If $\mathcal{A}$ is some abelian category and $S$ is some set of objects in $\mathcal{A}$, then its "abelian closure" $\overline{S}$ should be the smallest abelian full subcategory of $\mathcal{A}$ containing $S$ such that the inclusion functor is exact (in other words, it's a "sub-abelian category" as Qiaochu calls it). Note that it is automatically closed under isomorphisms (consider $0 \to A \cong A' \to 0$).

It always exists: Consider the set of all full subcategories $T$ of $\mathcal{A}$ containing $S$, which also contain $0$ and are closed with respect to kernels and cokernels. By this I mean that if $f$ is some morphism between objects in $T$, then every kernel/cokernel of $f$ also lies in $T$. Now take their intersection.

If you work with classes instead of universes, this intersection will cause some set-theoretic problems. One can avoid them as follows, thereby giving another construction of $\underline{S}$:

Define full subcategories $S_{\alpha}$ for ordinals $\alpha$ as follows: Let $S_{0} = S \cup \{0\}$. In the limit step, take the union. If $S_{\alpha \omega + 2n}$ is already defined, then adjoin all kernels of all morphisms in this set and optain $S_{\alpha \omega + 2n+1}$. Then, let $S_{\alpha \omega + 2n+2}$ be the closure under cokernels. Finally, define $\overline{S} := \cup_{\alpha} S_{\alpha}$.

As for a specific example, $R$ is "abelian dense" in $\mathrm{Mod}(R)$.

If $\mathcal{A}$ is some abelian category and $S$ is some set of objects in $\mathcal{A}$, then its "abelian closure" $\overline{S}$ should be the smallest abelian full subcategory of $\mathcal{A}$ containing $S$ such that the inclusion functor is exact (in other words, it's a "sub-abelian category" as Qiaochu calls it). Note that it is automatically closed under isomorphisms (consider $0 \to A \cong A' \to 0$).

It always exists: Consider the set of all full subcategories $T$ of $\mathcal{A}$ containing $S$, which also contain $0$ and are closed with respect to kernels and cokernels. By this I mean that if $f$ is some morphism between objects in $T$, then every kernel/cokernel of $f$ also lies in $T$. Now take their intersection.

If you work with classes instead of universes, this intersection will cause some set-theoretic problems. One can avoid them as follows, thereby giving another construction of $\underline{S}$:

Define full subcategories $S_{\alpha}$ for ordinals $\alpha$ as follows: Let $S_{0} = S \cup \{0\}$. In the limit step, take the union. If $S_{\alpha \omega + 2n}$ is already defined, then adjoin all kernels of all morphisms in this set and optain $S_{\alpha \omega + 2n+1}$. Then, let $S_{\alpha \omega + 2n+2}$ be the closure under cokernels. Finally, define $\overline{S} := \cup_{\alpha} S_{\alpha}$.

As for a specific example, $R$ is "abelian dense" in $\mathrm{Mod}_{fg}(R)$, when $R$ is noetherian.