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, 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 ...

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 ...

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 regular 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, 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 ...

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, 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 ...

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 regular schemes. So here, $$\mathrm{k}((\varepsilon)) = \lim_{i \rightarrow -\infty}\varepsilon^i\mathrm{k}[[\varepsilon]]$$ But why isn't it an algebraic subset of $\prod_{-\infty}^{\infty}\mathrm{k}$?

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 regular 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, 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 ...