From "Abelian CAtegories and its application to Rings and Modules " by Popescu N, par. 3.5 pag 88:

Popescu using the names *"small"* and *"finitely presented"* for the yours *"sumcompact"* and *"compact"* respectively.

Popescu call a object $X$ (of a Grothendick abelian category $\mathcal{C}$)

of *"finite type"* if for any direct union of subobjets $Y=\cup_{i\in I}Y_i$ the natural morphism $Colim_{i\in I} \mathcal{C}(X, Y_i)\to \mathcal{C}(X, Y))$ is a isomorphism, this is equivalent to:

for any directed union of subobjets $X=\cup_{i\in I}X_i$ there is a $i_o\in I$ such that $X=X_{i_0}$.

In a category of modules *finitely presented* is equivalent to the usal definition (there is a exact $0\to A\to X\to C\to 0$ with $A,\ B$ finitely generated), and *finite type* is equivalent to *finitely generated*.

From 5.4 of Popescu book a finitely generated module is small (sumcompact).
And of course exist finitely generated modules that aren't finitely presented.
then we have the implications:

*finitely presented* $\Rightarrow$ *finitely generated* $\Rightarrow$ *small (suncompact)* and *finitely .generated*$\not\Rightarrow$ *finitely presented*

Then cannot have that *small (sumcompact)*$\Rightarrow$ *finitely presented*.

$EDIT$ I get a mistake simply gived a answer to another question (sorry, mistake), I find the answer as exercises in *"Rings os Quotients"* B. Strenstrom, Springer Verlag 1975, pag 134 n.13.

**I try to do a proof (I hope):**

THEOREM) for a abject $C$ in a Grothendieck category (we think simply to a module category) the following are equivalent:

1) For any sequence of subobject like $C_1\subset C_2\subset\ldots C$ we have $C=C_m$ for some $m$.

2) For any sequence of subobject like $M_1\subset M_2\subset\ldots M$ with union $M$ we have that $(C, M)=\cup_n(C, M_n)$ (naturally).

3) The functor $(C, -)$ commute by denumerable coproducts.

4) The functor $(C, -)$ commute by coproducts.

5) The functor $(C, -)$ commute by directed unions (i.e. $C$ is $f.g.$).

PROOF. $(1\Rightarrow 2):$ we have to proof that any $f: C\to M $ has image in some $M_m$, if we put $C_n:=f^{-1}(M_n)$ we done.

$(2\Rightarrow 3):$ of course $(C,-)$ commute by finite coproduts (they are biproducts), we have to prove that a $f: C\to M$, with $M=\coprod_n X_n$ as a factorization on a finite summands, let $M_n:=\coprod_{i\leq n} X_i$ we done.

$(3\Rightarrow 4):$ we have to prove that a $f: C\to M$, with $M=\coprod_{i\in I} X_i$ as a factorization on a finite summands, suppose the opposite: then we have an infinite denumerable set of indices $i_0, i_1\ldots \in I$ such that for any integer $n$ exist a $x_n\in C$ with $f(x_n)_{i_n}\neq 0$

then we consider $J:= I\setminus ${$i_0, i_1\ldots$} and the quotient map $\pi: \coprod_{i\in I}X_i\to (\coprod_{i\in I}X_i)/(\coprod_{j\in J}X_j)\cong\coprod_n X_{i_n} $ and the composition $\pi\circ f: C\to \coprod_n X_{i_n}$, this map isnt factorizable to a finite summands (absurd).

$(4\Rightarrow 5):$ Let $M=\cup_{i\in I} M_i$ where $I$ is a directed order. We can suppose $I$ cofinite i.e. for any $i\in I$ exist only finite $j$ such that $j\leq i$ (e.g. *"Shape Theory"* Sibe MArdiesic NH 1982 T.2 pag. 10). Then the natural map $\pi: M\to \coprod_{i\in I} M/M_i$ with $(\pi(x))_i=\pi_i(x)$, $\pi_i: M\to M/M_i$ natural, is well defined. We have to proof that any $f: C\to M$ has a image on some $M_j$, considering $\pi\circ f: C\to \coprod_i M/M_i$ then this map has a factorization on finite summands $M/M_{i_1},\ldots M/M_{i_N}$, if some $M_{i_n}$ is $M$ the assert is trivial

if no we can have a $j\in I$ strictly greater of any $i_1,\ldots i_N$, then $f\circ \pi_j: C\to M\to M/M_j $ is the $0$ map, then the image of $f$ is in $M_j$.

$(5\Rightarrow 2):$ Trivial

$(2\Rightarrow 1):$ Let $M:=M,\ M_n:=C_n $ and considerind $1_C$

presented. – Fernando Muro Mar 23 '11 at 11:34definea compact object to be an object $X$ such that $Hom(X,-)$ preserves coproducts. – Martin Brandenburg Mar 23 '11 at 17:22littleonly requires that $\mathrm{Hom}(X,{-})$ preserves finite coproducts. I think this is a typo, since in an additive category, finite coproducts are "the same" as finite products, so $\mathrm{Hom}(X,{-})$ commutes with finite coproducts, for any $X$. However, this also shows that if $\mathrm{Hom}(X,{-})$ commutes with filtered colimits, it commutes with arbitrary coproducts (by the argument I gave above). – user2035 Nov 20 '11 at 19:09connectednessof X. (Think about the category of topological spaces.) There's some flexibility about whether you ask for preservation of all small coproducts or just the finite ones. See for instance ncatlab.org/nlab/show/connected+object – Tom Leinster Nov 22 '11 at 18:41