1
$\begingroup$

We fix an algebraically closed field $\Omega$ which has infinite trancendence dimension over the prime subfield. A subfield $k$ of $\Omega$ such that tr.dim $\Omega/k = \infty$ is called an admissible field.

Let $V$ be a Zariski closed subset of $\Omega^n$. Let $I = \{f \in \Omega[X_1,\dots,X_n]|\ f(p) = 0$ for every $p \in V\}$. Let $k$ be an admissible field such that $I$ has a basis in $k[X_1,\dots,X_n]$. Then we say $k$ is a field of definition of $V$.

Let $k$ be an admissible field. Let $V$ be a $k$-closed subset in $\Omega^n$(see this question for the definition of a $k$-closed subset). Let $I_k(V) = \{f \in k[X_1,\dots,X_n]|\ f(p) = 0$ for every $p \in V\}$. Let $A = k[X_1,\dots,X_n]/I_k(V)$. Then what condition should $A$ satisfy for $k$ to be a field of definition of $V$?

  • 0
    ah yes, I $f$orgot that a "$f$ield o$f$ de$f$inition" o$f$ a variety is not just a field over which the variety is definable.2012-11-29

0 Answers 0