Are finitely presentable modules closed under extensions? 
If $0 \to A \to B \to C \to 0$ is an exact sequence of modules, and $A$ and $C$ are finitely presentable, then is $B$ finitely presentable? 

The answer is "yes" if we replace modules with groups, as shown here.
The answer is also "yes" if we replace "finitely presentable" by "finitely generated": take a set-theoretic splitting of $B \to C$ to view $B$ as $A \times C$ with the addition given by twisting the addition on $A \oplus C$ by a cocyle: then generators for $B$ are given by pairs of generators for $A$ and generators for $C$. When we're working over a Noetherian ring, this means the same goes for finite presentability. How about in non-Noetherian situations? I'm particularly interested in non-commutative rings, especially group rings.
The obvious way to get finitely many relations would be to take the relations for $A$, the relations for $C$, and add each relation of the form $(a_i,b_j) + (a_{i'},b_{j'}) = (a_i + a_{i'}, b_j + b_{j'} + \omega(a_i,a_{i'}))$ (where the $a_i$'s and $b_j$'s are our generators and $\omega$ is our cocycle). But this doesn't obviously work because $\omega$ is typically not bilinear.
My intuition is that because the obvious approach doesn't work, the answer should be "no". But knowing that the answer is "yes" for groups makes me less confident.

You can proceed as usual by starting with $F\stackrel{f}\to A\to 0$ and $H\stackrel{h}\to C\to 0$, where $F$ and $H$ are free of finite rank. Then show that there is an exact sequence $G=F\oplus H\stackrel{g}\to B\to 0$. Now consider $F'=\ker f$ and so on. You have a short exact sequence $0\to F'\to G'\to H'\to 0$. Now use the result for finitely generated modules.
