Concept of an exact ideal of a module category Let $R$ be a ring and $\text{Mod}\,R$ the category of (left) $R$-modules. Consider an ideal $\mathcal{I}$ of $\text{Mod}\,R$. For $R$-modules $X$ and $Y$ let $\mathcal{I}(X,Y)$ be the collection of all morphisms $X\rightarrow Y$ in $\mathcal{I}$. Then $\mathcal{I}(X,-)$ is a subfunctor of $\text{Hom}_R(X,-)$ and $\mathcal{I}(-,Y)$ a subfunctor of $\text{Hom}_R(-,Y)$. In general $\mathcal{I}(X,-)$ and $\mathcal{I}(-,Y)$ need not to be left-exact. If they are for all $X$ and $Y$, we call $\mathcal{I}$ an exact ideal. Are exact ideals studied in the literature?
If we consider an Artin algebra $A$ and the category of finitely generated $A$-modules $\text{mod}\,A$, one can show the following: There is a bijection between exact ideals of $\text{mod}\,A$ and ideals of $A$.
Edit: Lets call $\mathcal{I}$ covariantly exact if $\mathcal{I}(X,-)$ is left-exact for all $X$ and contravariantly exact if $\mathcal{I}(-,Y)$ is left-exact for all $Y$. In general $\mathcal{I}$ is neither covariantly exact nor contravariantly exact.
However, given $\mathcal{I}$ we can construct the smallest covariantly exact ideal $\mathcal{J}$ that contains $\mathcal{I}$ as follows. Let $X\rightarrow Y$ be in $\mathcal{J}$ if for all injective $R$-modules $I$ and morphisms $Y \rightarrow I$ the composition $X \rightarrow Y \rightarrow I$ is in $\mathcal{I}$. Similarly, we can construct the smallest contravariantly exact ideal $\mathcal{K}$ that contains $\mathcal{I}$ as follows. Let $X\rightarrow Y$ be in $\mathcal{K}$ if for all projective $R$-modules $P$ and morphisms $P \rightarrow X$ the composition $P \rightarrow X \rightarrow Y$ is in $\mathcal{I}$.
 A: Exact ideals $\mathcal{I}$ in an abelian category are in 1:1 correspondence with full subcategories $\mathcal{C}$ closed under isomorphisms, direct sums, subobjects and quotients.
Namely, if $\mathcal{I}$ is an exact ideal, let $\mathcal{C}$ be the subcategory of objects $X$ whose identity endomorphism is in $\mathcal{I}$.
If $X,Y$ are in $\mathcal{C}$, then the endomorphisms of $X\oplus Y$ given by projection onto $X$ or $Y$ factor through the identity endomorphisms of $X$ and $Y$, so are in $\mathcal{I}$. Since $\mathcal{I}$ is closed under addition, the sum of these endomorphisms is in $\mathcal{I}$, so $X\oplus Y$ is in $\mathcal{C}$.
If $X$ is in $\mathcal{C}$ and $U$ is a subobject of $X$, then since $\mathcal{I}$ is an ideal, the inclusion of $U$ in $X$ is in $\mathcal{I}$, and then the exactness of $0\to \mathcal{I}(U,U)\to \mathcal{I}(U,X)\to \mathcal{I}(U,X/U)$ implies that $U$ is in $\mathcal{C}$. Similarly any quotient of $X$ is in $\mathcal{C}$.
Conversely if $\mathcal{C}$ is a full subcategory closed under isomorphisms, direct sums, subobjects and quotients, let $\mathcal{I}$ be the ideal generated by (the identity endomorphisms of objects in) $\mathcal{C}$. Since $\mathcal{C}$ is closed under direct sums it consists of all morphisms factoring through an object in $\mathcal{C}$.
In fact $\mathcal{I}$ consists of all morphisms with image in $\mathcal{C}$, for if $f:X\to Y$ and $g:Y\to Z$ with $Y$ in $\mathcal{C}$, then $\mathrm{Im}(gf) \cong \mathrm{Im}(f) /(\mathrm{Im}(f)\cap \mathrm{Ker}(g))$, which is a subquotient of $Y$, so in $\mathcal{C}$.
It follows immediately that $\mathcal{I}$ is exact. For example suppose $0\to U\to V\to W$ is exact and $f:X\to V$ is a morphism whose composition with the morphism $V\to W$ is zero. Then $f$ factors through the morphism $U\to V$, and $f$ and the morphism $X\to U$ have isomorphic images.
Now these constructions are inverse. To see this we need to show that if $f:X\to Y$ is any morphism in an exact ideal $\mathcal{I}$, then the identity endomorphism of its image $\mathrm{Im}(f)$ is in $\mathcal{I}$. But using that $\mathcal{I}(X,-)$ is left exact with the exact sequence $0\to \mathrm{Im}(f)\to Y\to Y/\mathrm{Im}(f)\to 0$ we see that the morphism $X\to \mathrm{Im}(f)$ is in $\mathcal{I}$, and then using that $\mathcal{I}(-,\mathrm{Im}(f))$ is left exact with the sequence $0\to \mathrm{Ker}(f) \to X \to \mathrm{Im}(f) \to 0$ we deduce that $1_{\mathrm{Im}(f)}$ is in $\mathcal{I}$.
