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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 $V$?

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If $V$ is a $k$-closed subset, shouldn't $k$ automatically be a field of definition of $V$? –  Alex Kruckman Nov 29 '12 at 3:02
    
@AlexKruckman Suppose $char(\Omega) = p > 0$. Let $F$ be the prime subfield of $\Omega$. Let $t$ be an element of $\Omega$ such that $F(t)/F$ is not algebraic. Let $k = F(t)$. Let $\alpha$ be a root of $X^p - t$ in $\Omega$. Since $\alpha$ is a unique root of $X^p - t$, $V = \{\alpha\}$ is $k$-closed. But $k$ is not a field of definition of $V$ because $\alpha$ is not an element of $k$. –  Makoto Kato Nov 29 '12 at 3:29
    
ah yes, I forgot that a "field of definition" of a variety is not just a field over which the variety is definable. –  Alex Kruckman Nov 29 '12 at 5:07
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