A standard example of an ind-scheme over a field $\mathrm{k}$ which is not a 
$\mathrm{k}$-scheme is $\mathrm{k}((\varepsilon))$.
My question is how to prove that rigorously? To put it more precisely, 
let $$\mathrm{k}((\varepsilon)) = \{ a \in \prod\_{-\infty}^{\infty}\mathrm{k}: 
a\_i =0, i \ll 0 \}$$
An ind-scheme is an injective limit of schemes. So here, 
$$\mathrm{k}((\varepsilon)) = 
\lim\_{i \rightarrow -\infty}\varepsilon^i\mathrm{k}[[\varepsilon]]$$
But why isn't it an algebraic subset of $\prod\_{-\infty}^{\infty}\mathrm{k}$?


EDIT: I seem to have mixed up some notions, and have asked two different questions at once (or maybe even three) so I'll try to make myself clear. 
My motivation for the question was to be able to justify the following 
"$k((\epsilon))$ is not an algebraic subset of $\prod\_{-\infty}^{\infty}k$ so 
we define it as $k$-points of an ind-scheme". So the original question is: 
why $k((\epsilon)) \subseteq \prod\_{-\infty}^{\infty}k$ isn't algebraic
and it is answered by Jason Starr (though I'm not sure if I understand the answer).
We can also define $k((\epsilon))$ as and ind-scheme by 
$$k((\epsilon)) = colim\_n Spec(k[x\_{-n},x\_{-n+1}, \ldots])$$
Now, one can ask, why isn't the ind-scheme we've constructed a scheme after all
(btw. wouldn't it contradict Jason's argument?), and 
this question is answered by Scott Carnahan below. Finally, there is a question: if the co-limit exists in the category of schemes which Scott Carnahan addresses below as well ...



How to show that an ind-scheme is not a scheme? 
I think you can just consider the decreasing sequence of ideals of the increasing sequence of algebraic subsets $\epsilon^i k[[\epsilon ]]$ of $\prod\_{-\infty}^{\infty}k$. For each $i$, the ideal is $( a\_j | j\leq i-1 )$. The intersection of this sequence of ideals is $\{ 0 \}$. The corresponding algebraic subset of $\prod\_{-\infty}^{\infty} k$ is the whose scheme. Since $k((\epsilon))$ is not the whole scheme, it is not an algebraic subset.


