Example 89.3.7. Let $p$ be a prime number and let $n \in \mathbf{N}$. Let $\Lambda = \mathbf{F}_ p(t_1, t_2, \ldots , t_ n)$ and let $k = \mathbf{F}_ p(x_1, \ldots , x_ n)$ with map $\Lambda \to k$ given by $t_ i \mapsto x_ i^ p$. Let $A = k[\epsilon ] = k[x]/(x^2)$. Then $A$ is an object of $\mathcal{C}_\Lambda $. Suppose that $D : k \to k$ is a derivation of $k$ over $\Lambda $, for example $D = \partial /\partial x_ i$. Then the map

is a morphism of $\mathcal{C}_\Lambda $. Set $A_1 = A_2 = k$ and set $f_1 = f_{\partial /\partial x_1}$ and $f_2(a) = a$. Then $A_1 \times _ A A_2 = \{ a \in k \mid \partial /\partial x_1(a) = 0\} $ which does not surject onto $k$. Hence the fibre product isn't an object of $\mathcal{C}_\Lambda $.

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