If $X(F) \cap Y$ is dense in $Y$, then $Y$ is defined over $F$. Let $k$ be an algebraically closed field, $F$ a subfield of $k$, $A$ a finitely generated, reduced $k$-algebra, and $A_0$ an $F$-subalgebra of $A$, of finite type over $F$, such that the canonical $k$-algebra homomorphism $$k \otimes_F A_0 \rightarrow A$$ is an isomorphism.  Let $X = \textrm{Spm } A$.  Let $X(F)$ be the set of $F$-algebra homomorphisms of $A_0$ onto $F$.  Every such homomorphism uniquely extends to a $k$-algebra homomorphism of $A$ onto $k$, whose kernel is a maximal ideal of $A$, so we can identify $X(F)$ as a subset of $X$.
Let $Y$ be a closed set in $X$, with defining ideal $I$.  So $$Y = \{ \mathfrak m \in X: I \subseteq \mathfrak m\}$$ and $$I = \bigcap\limits_{\mathfrak m \in Y} \mathfrak m$$ I'm trying to prove the following statement (Springer, Linear Algebraic Groups, Lemma 11.2.2): 

Assume that $X(F) \cap Y$ is dense in $Y$.  Then $I$ is the extension of an ideal of $A_0$.

Attempt: Let $Y \cap D(f_i) : i =1, ... , n$ be an open cover of $Y$ for some $f_i \in A$.  We may assume none of the $f_i$ are in $I$, since then $Y \cap D(f_i)$ is empty.  For each $i$, we know that there is a $k$-algebra homomorphism $\psi_i$ of $A$ into $k$, whose restriction $\phi_i$ to $A_0$ maps $A_0$ onto $F$, with $\psi_i(f_i) \neq 0$.
Working in the ring $A/I$ (whose maximal spectrum we can identify with $Y$), we see that no maximal ideal of $A/I$ contains all the $f_i$, so the cosets $f_i + I$ generate the unit ideal of $A/I$.  So there exist $g_i \in A$ such that $$g_1f_1 + \cdots + g_nf_n - 1 \in I$$ I'm not sure exactly where this is going, but I'm hoping to be able to write $I$ as a finite intersection of extended ideals from $A_0$, and then use the fact that an intersection of extended ideals remains extended for a faithfully flat ring extension as $A_0 \rightarrow A$.
 A: If $I$ is an extension from $A_0$, it better be $JA$ where $J=I\cap A_0$. So, easy to check by replacing $A_0$ by $A_0/J$ and $A$ by $A/JA$, that it suffices to prove that if $I\cap A_0=0$, then $I=0$. 
So, we have $F\subset k$ and let $L$ be the algebraic closure of $F$ in $k$ and then $L$ is algebraically closed, since $k$ is. Let $A_1=L\otimes_F A_0$ and then $I\cap A_1=0$, since $A_0\to A_1$ is algebraic. This says that $X(F)\subset X(L)\subset Y$ and so $X(L)$ is dense in $Y$. Thus, it suffices to show that $X(L)$ is dense in $X(k)$ and then $I=0$.
If not, let $0\neq f\in A$ such that $f$ is zero on $X(L)$ and write $f=\sum f_i\otimes a_i$ where $f_i\in A_1$, $a_i\in k$ and $a_i$ linearly independent over $L$. $f=0$ at a point $p\in X(L)$ means, if $M\subset A_1$ corresponds to the maximal ideal of $p$ (and thus $A_1/MA_1=L$), $f=0$ in $A/MA=A_1/MA_1\otimes_L k$. Since $a_i$s are linearly independent over $L$, this means $f_i\in M$ for all $i$. But, this is true for every maximal ideal $M$ of $A_1$, since $L$ is algebraically closed and thus $f_i=0$ for all $i$, since $A_1$ is reduced.
