Let $A$ be a ring (say finitely generated algebra over an algebraically closed field). Then, does $\varinjlim \mathrm{Spec} A[t]/(t^m)$ exist (in the category of schemes)? And if it does, then is it equal to $\mathrm{Spec} A[[t]]$?

Edit based on comment below: The result holds easily when $A=k$ is a field. Then, $\mathrm{Spec} k[t]/(t^n)$ only has one point and hence, for any scheme $X$ such that we have a family of morphisms $\mathrm{Spec} k[t]/(t^n) \to X$, we can assume that $X$ is affine and the result follows from the equivalence of the category $Ring^{op}$ and the category of affine schemes.

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    $\begingroup$ See mathoverflow.net/questions/9961/colimits-of-schemes in particular Emerton's answer. $\endgroup$ Dec 8, 2010 at 1:37
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    $\begingroup$ So, according to the link, it does exist and we do have the equality. How do we actually prove it? $\endgroup$
    – Brian
    Dec 8, 2010 at 1:55
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    $\begingroup$ The proof for the case when $A=k$ (a field) is easy, which is in the 3rd paragraph of Emerton's answer. I'm interested in seeing the proof for the present case or even better, the even more general one mentioned by Emerton in the last paragraph (in the EDIT part). $\endgroup$
    – Brian
    Dec 8, 2010 at 2:13
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    $\begingroup$ @Harry: We are taking colimits in the category of schemes, and in this case, the colimit exists. It is not necessary to pass to the cocomplete settings you suggest, and in fact, we would get different answers if we were to do so. $\endgroup$
    – S. Carnahan
    Dec 8, 2010 at 5:04
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    $\begingroup$ Dear Brian, I am trying to recover the proof of the claim in the EDIT part of that answer; I will report here if successful, and will add a further edit to that answer if not. $\endgroup$
    – Emerton
    Dec 8, 2010 at 5:26


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