Let $M$ be a module over the polynomial ring $\mathbb C[x]$ such that $x+n$ is invertible in $M$ for every integer $n$. Let 
$N=\oplus \_{n>0} \mathbb C\_n$ where $\mathbb{C}\_n := \mathbb C[x]/x+n$.


$\mathbf{Question:}$ Is it true that $Ext^1(M,N)=0$? 


It is clear that $Ext^1(M,\mathbb C\_n)=0$ but this is not enough since $M$ is not finitely generated and therefore $Ext(M,?)$ does not commute with direct sums.



$Ext^1$ for some modules over the polynomial ring in one variable
$\newcommand{\bC}{\mathbb{C}}\newcommand{\bZ}{\mathbb{Z}}$The resolution constructed by Dylan Wilson seems to show that $Ext^1(M,N)$ is non-zero already for $M=\bC[x][\frac{1}{x+n},n\in\bZ]$. 



> 
> There is an identification $$Hom(\bC[x][\frac{1}{x+n},n\in\bZ],\bigoplus\limits\_{n\in\bZ}\bC[x,\frac{1}{x+n}]/\bC[x])=\prod\limits\_{n\in\bZ}'\bC((x+n))$$ where the restricted tensor product is taken with respect to the submodules $\bC[[x+n]]\subset\bC((x+n))$
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Proof. Given a homomorphism $f:\bC[x][\frac{1}{x+n},n\in\bZ]\to\bigoplus\limits\_{n\in\bZ}\bC[x,\frac{1}{x+n}]/\bC[x]$ associate to it the element of the restricted product $(a\_n)\_{n\in\bZ}$ given by $a\_n=\lim\limits\_{k\to\infty}(x+n)^k\widetilde{f(\frac{1}{(x+n)^k})\_n}$ where $\widetilde{f(\frac{1}{(x+n)^k})\_n}$ denotes an aribtrary lift of the $n$-th component of $f(\frac{1}{(x+n)^k})$ to an element of $\bC[x,\frac{1}{x+n}]$. For almost all $n$ the element $a\_n$ lies in $\bC[[x+n]]$ because $f(1)\_n$ is zero for almost all $n$. The map $f\mapsto(a\_n)$ from the LHS to the RHS is injective because a homomorphism $f:\bC[x][\frac{1}{x+i},i\in\bZ]\to\bC[x,\frac{1}{x+n}]/\bC[x] $ is completely determined by its restriction to $\bC[x][\frac{1}{x+n}]$. The map is surjective because a Laurent series $a\_n\in\bC((x+n))$ gives a homomorphism $\bC[x,\frac{1}{x+n}]\to\bC[x,\frac{1}{x+n}]/\bC[x]$ which then extends uniquely to a homomorphism from the whole of $M$. $\square$


This yields a formula for the desired RHom $$RHom(M,\bigoplus \bC[x]/(x+n))=\\ RHom(M,\mathrm{fib}(\bigoplus\limits\_{n\in\bZ}\bC[x,\frac{1}{x+n}]/\bC[x]\to \bigoplus\limits\_{n\in\bZ}\bC[x,\frac{1}{x+n}]/\bC[x]))=\\ \mathrm{fib}\left(\prod\limits\_{n\in\bZ}'\bC((x+n))\xrightarrow{(x+n)\_{n\in\bZ}} \prod\limits\_{n\in\bZ}'\bC((x+n))\right)$$ So, for instance, the "adele" $(1,1,\dots)$ gives a representative of a non-zero class in the Ext^1.


