# Fiber Products (Friday, October 22) :::{.remark} Last time: we defined and proved the existence of fiber products in $\Sch\slice S$, and for $X,Y,S \in \Aff\Sch$ equal to $\spec A, \spec B, \spec R$ respectively, \[ X\fp{S} Y = \spec (A\tensor_R B) .\] ::: :::{.definition title="Residue field"} For $X \mapsvia{f} Y \in \Sch$ and $p\in Y$, define the **residue field** \[ k(p) \da \OO_{Y, p} / \mfm_{Y, p} .\] ::: :::{.remark} There is a closed immersion $\spec k(p) \injects Y$ if $p$ is a closed point (since it came from a quotient map), and we can take a fiber product \begin{tikzcd} {\spec k(p) \fiberprod{Y} \spec k(p)} && X \\ \\ {\spec k(p)} && Y \arrow[from=1-3, to=3-3] \arrow[from=3-1, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=1-1, to=1-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgayhwKSBcXGZpYmVycHJvZHtZfSBcXHNwZWMgayhwKSJdLFsyLDAsIlgiXSxbMCwyLCJcXHNwZWMgayhwKSJdLFsyLDIsIlkiXSxbMSwzXSxbMiwzXSxbMCwyXSxbMCwxXSxbMCwzLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=) ::: :::{.example title="?"} Consider $\spec k[x,y,t]/ \gens{xy-t} \to \spec k[t]$: \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-22_11-41.pdf_tex} }; \end{tikzpicture} What is the fiber over $p\da\gens{t-1}$ or $q= \gens{t}$? - $k(p) = k[t] / \gens{t-1} \cong k$, - $k(q) = k[t] / \gens{t} \cong k$, so they are abstractly isomorphic. We have the following tensor product in rings: \begin{tikzcd} {k[x,y,t]/\gens{xy=t} \tensor_{k[t]} k} && {k[x,y,t]/\gens{xy-t}} \\ \\ {k\cong k[t]/\gens{t-1}} && {k[t]} \arrow[from=3-3, to=1-3] \arrow[from=3-3, to=3-1] \arrow[from=1-1, to=3-1] \arrow[from=1-1, to=1-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJrW3gseSx0XS9cXGdlbnN7eHktdH0iXSxbMCwwLCJrW3gseSx0XS9cXGdlbnN7eHk9dH0gXFx0ZW5zb3Jfe2tbdF19IGsiXSxbMCwyLCJrXFxjb25nIGtbdF0vXFxnZW5ze3QtMX0iXSxbMiwyLCJrW3RdIl0sWzMsMF0sWzMsMl0sWzEsMl0sWzEsMF0sWzEsMywiIiwxLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d) Generally, pulling back over $k[t] / \gens{t-c}$ has the effect of setting $t=c$ in the tensor product, and thus the fiber products are given by - $k(p) \fiberprod{\spec k[t]} X = \spec k[x, y] / \gens{xy-1}$ - $k(q) \fiberprod{\spec k[t]} X = \spec k[x, y] / \gens{xy}$ ::: :::{.example title="Fiber products aren't quite set products"} Consider \[ X \da \AA^1\slice k\fiberprod{\spec k} \AA^1\slice k \cong \spec (k[s] \tensor_k k[t]) \cong \spec (k[s, t]) \cong \AA^2\slice k .\] However, $X$ is not the set-theoretic product of the two constituent sets, although it does contain the product. Why? Consider $p \da \gens{y^2 - x^3} \in \spec \AA^2\slice k$, which is prime (check irreducibility in each variable!) and thus yields a point which is not the product of any two points in $\AA^1\slice k$. \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-22_12-00.pdf_tex} }; \end{tikzpicture} ::: :::{.example title="Reduction mod $p$"} Let $X\in \Sch \iso \Sch\slice{\spec \ZZ}$ with structure map $X\to \spec \ZZ$, and let $p = \gens{P} \in \spec \ZZ$. Then $k(p) = \ZZ/p = \FF_p$, so consider the fiber over $p$: \begin{tikzcd} {X\times \spec \FF_p} && X \\ \\ {\spec \FF_p} && {\spec \ZZ} \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=3-1, to=3-3] \arrow[from=1-1, to=1-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJYXFx0aW1lcyBcXHNwZWMgXFxGRl9wIl0sWzIsMCwiWCJdLFswLDIsIlxcc3BlYyBcXEZGX3AiXSxbMiwyLCJcXHNwZWMgXFxaWiJdLFsxLDNdLFswLDJdLFsyLDNdLFswLDFdLFswLDMsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) Call this the **reduction mod $p$**, denoted $X_{\FF_p}$. If $X = \spec R$, then $X_{\FF_p} = \spec (R \tensor_\ZZ \ZZ/p) = \spec (R/\gens{p})$. ::: :::{.example title="?"} Take $X \da \spec \ZZ[x,y,z]/\gens{x^5+y^5=z^5}$, so nontrivial $\ZZ\dash$points yield counterexamples to Fermat. Then $X_{\FF_p} = \FF_p[x,y,z] / \gens{x^5 + y^5 + z^5}$, which reduces the coefficients of the equations. ::: > How are these related to models of a scheme? :::{.example title="?"} Take $X = \spec \CC$ to get $\CC \tensor_{\FF_p} \ZZ$ -- what is this ring? One has to regard $\CC$ as a ring over $\ZZ$ first, so write $\CC = \QQbar(T)$ where $T$ is an uncountable basis of transcendental elements. So this yields $\Fpbar(T)$. ::: :::{.remark} Consider $X \in \Var\slice \CC$, e.g. $X = \spec \CC[x,y,z]/\gens{x=\sqrt{2} y, y^2=\pi z^3}$. Then consider the (much smaller) subring generated by the coefficients of the defining equations, so $R\da \ZZ[\sqrt 2, \pi] \cong \ZZ[\sqrt 2][t]$ and consider $\spec R[x,y,z]/ \gens{x=\sqrt y, y^2 = \pi z^3}$. This has the exact same equations but is now defined of $\spec R$. Note that having finitely many equations yields a finitely generated as a $\ZZ\dash$module. Since $R\injects \CC$ we get a morphism $\spec \CC\to \spec R$, and we get a diagram \begin{tikzcd} {Y \da R[x,y,z]/\gens{x=\sqrt 2 y, y^2 = \pi z^3}} && X \\ \\ {\spec R} && \spec\CC \arrow[from=3-1, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=1-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJZIFxcZGEgUlt4LHksel0vXFxnZW5ze3g9XFxzcXJ0IDIgeSwgeV4yID0gXFxwaSB6XjN9Il0sWzIsMCwiWCJdLFswLDIsIlxcc3BlYyBSIl0sWzIsMiwiXFxzcGVjXFxDQyJdLFsyLDNdLFswLDJdLFsxLDNdLFswLDFdLFswLDMsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==) ::: :::{.warnings} So in the literature, *reduction of $X$ mod $p$* generally means $Y_{\FF_p}$ and not $X_{\FF_p}$. ::: :::{.definition title="Base Change"} Given $X, Y\in \Sch\slice S$ with structure maps $f, g$ respectively, the **base change of $f$ along $g$** is defined as the fiber product $X\fiberprod{S} Y \in \Sch\slice Y$. So there is a functor \[ \wait\fiberprod{S} Y: \Sch\slice S &\to \Sch\slice Y .\] ::: > What is the adjoint? > Probably the forgetful functor given by composing along $g$.