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Let $S=Spec(A)$ and $S'=Spec(B)$ be two affine schemes for some rings $A$ and $B$ such that there is a morphism of schemes $f:S'\rightarrow S$. For any $S$-scheme $X$, one can consider the fiber product $X\times_S S'$ of $X$ and $S'$ over $S$.

If we assume that $X$ is given by a set of equations $(E)$ in $A$, what are the equations which define the $S'$-scheme $X\times_S S'$? is it the equations in $B$ which are obtained by applying to $(E)$ the morphism of rings induced by $f$ ? I can this be written properly?

Another construction which is even more simple : assuming that $Y$ is an $S'$-scheme, $Y$ can be considered as an $S$-scheme via $Y\longrightarrow S'\longrightarrow S$ (composing by $f$). I have two questions about this construction : first in the same way i did for fiber products, is it possible to find the equations which define $Y$ as an $S$ variety from whose which define it as an $S'$ variety ?

Finally something that seems reasonnable to me : $Z$ is an $S$-scheme, and you consider the fiber product $T=Z\times_S S'$ as an $S'$-scheme. Is the scheme $T$ consider as an $S$-scheme with the previous construction isomorphic to $Z$ as an $S$-scheme? I think its the case just because of the definition of the fiber product, but i would like to be sure.

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I think there`s a typo in your last paragraph. As for your 2nd paragraph -- yes. For example, for everything affine, the fiber product is Spec of the tensor product $\mathcal O_X \otimes_A B$. – user54535 Dec 30 '12 at 19:00
up vote 5 down vote accepted

First paragraph: yes. Second paragraph: getting equations is going to be gross in general, although it depends on the nature of $S'$ and $S$. Third paragraph: no.

Maybe this will be a motivating example. Let $S$ be Spec of an algebraically closed field, and let $S'$ be Spec of a much larger algebraically closed field (e.g. adjoin a transcendental to your field, then algebraically close that). If you take an $S'$ scheme and "think of it as an $S$-scheme" you will get a monstrosity that's not of finite type (e.g. in the last paragraph you are asserting that $\mathbb{C}[t]$ is isomorphic to $\overline{\mathbb{Q}}[t]$ as a $\overline{\mathbb{Q}}$-algebra).

The lesson to take away is that base change is the reasonable, good functor, and the "forgetful functor" is the ugly one, the opposite of the way you'd think it would be on first glance.

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Just to add to the excellent answer of countinghaus, in regarding your first question, your morphism $f$ is going the wrong way. You want a morphism $f:S^\prime\rightarrow S$ to talk about $X\times_SS^\prime$, not $S\rightarrow S^\prime$. Anyway, when $X$ is the spectrum of an $A$-algebra $R$, on choosing generators, and assuming this algebra is finitely presented, it is isomorphic to $A[x_1,\ldots,x_n]/(f_1,\ldots,f_m)$ for some $n\geq 0$ and some $f_j\in A[x_1,\ldots,x_n]$. The base change is the spectrum of $R\otimes_AB$, tensor product with respect to the ring map $\psi:A\rightarrow B$ corresponding to $S^\prime\rightarrow S$. The canonical $B$-algebra map

$R\otimes_AB=A[x_1,\ldots,x_n]/(f_1,\ldots,f_m)\otimes_AB \rightarrow B[x_1,\ldots,x_n]/(\psi(f_1),\ldots,\psi(f_m))$

is an isomorphism, so indeed, the base change is "cut out" by the polynomials $f_j$, viewed via $\psi$ as having coefficients in $B$.

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