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Let $V$ be a projective variety defined over $\mathbb{F}_p$. For any extension of $\mathbb{F}_p$ let $V(k)$ be the $k$-points of $V$, that is, if $V$ is defined by the homogeneous polynomials $f_1,\ldots,f_r\in\mathbb{F}_p[X_0,\ldots,X_n]$, then $V(k)=\{(x_0:\ldots :x_n)\in k^n:f_i(x_0,\ldots,x_n)=0\ \forall i\}$. Then we can talk about dimension of $V(k)$ for any $k$ because dimension is just a concept of topological spaces and all $V(k)$ are topological spaces with the Zariski topology.

1-Are all the numbers $\dim V(k)$ equal? (assume $k$ runs over subfields of $\bar{\mathbb{F}_p}$). If not, when we read the expression "Let $V$ be a projective variety defined over $\mathbb{F}_p$ of dimension $r$", do we have to read $r=\dim V(\bar{\mathbb{F}_p})$?

2-Similarly I have read the expression "Let $V$ be a smooth projective variety defined over $\mathbb{F}_p$". The concept of smoothness I know is in term of dimension, that is, $V$ is smooth if the rank of the matrix $[\partial f_i/\partial x_j]$ in each $(a_0:\ldots:a_n)\in V$ is $n-\dim V$. So my question is which is the definition of that expression?.

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If you take polynomials with no non-trivial roots over $\mathbb{F}_p$, then $V(\mathbb{F}_p)=\emptyset$, but there will always be some field extension $k$ such that $V(k)\neq \emptyset$. –  Matt Feb 11 '13 at 23:43
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up vote 2 down vote accepted

If $k$ is finite, then $V(k)$ is finite, hence of dimension $\le 0$. The dimension of $V$ as an algebraic variety is the dimension of $V(\bar{\mathbb F}_p)$ with its Zariski topology. In the smoothness criterion, the dimension of $V$ is also defined this way.

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