It is clear how to compute the fiber product of two affine schemes. But Rhow can one compute the fiber product $X \times _{R} Spec(k[y])$ where $R$ is an integral domain, $X$ defined by the zero set of $f(x,y,z)$ in the projective plane (i.e. $X=Proj(R[x,y,z]/(f))$ ) and $y$ the generic point of $X$, $k[y]$ the residue field at $y$? Do I have to compute first the fiber product of an affine cover of $X$ and then glue? How can I do this concretely? Thanks for your help!
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3$\begingroup$ By "computing" this fibre product, what exactly do you mean? What kind of data constitutes a computed fibre product, as opposed to writing $X\times_R\mathrm{Spec}(k[y])$? $\endgroup$– Jesko HüttenhainCommented Jul 26, 2012 at 7:19
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$\begingroup$ It means to write X×RSpec(k[y]). My idea was to consider the affine cover U0= {x≠0}∩X and so on and then glue. But I'm not sure how to do this concretely. $\endgroup$– Regula KrapfCommented Jul 26, 2012 at 8:25
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2$\begingroup$ If "it means to write XxRSpec(k[y])", then I'd say you're done. $\endgroup$– Steven LandsburgCommented Jul 26, 2012 at 14:37
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$\begingroup$ I find your notations quite confusing : $y$ is both an indeterminate and a generic point, k[y] is a field ? $\endgroup$– Georges ElencwajgCommented Jul 26, 2012 at 18:00
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$\begingroup$ Yes, that is really confusing. Let $\nu$ be the generic point, then $k[\nu] = \mathcal{O}_{X,\nu} / \mathcal{m}_{\nu}$ where $\mathcal{m}_{\nu}$ is the maximal ideal corresponding to the generic point. $\endgroup$– Regula KrapfCommented Jul 27, 2012 at 8:30
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