Are the $K$-points of an embedded linear algebraic group its intersection with $GL(n,K)$ Let $G \subseteq GL(n,\mathbb{C})$ be a linear algebraic group defined over some subfield $k$ of $\mathbb{C}$, that is, $G$ is the vanishing set of some polynomials in $k[X_{11},\ldots,X_{nn}]$.
For any $k$-algebra $A$, we can define the $A$-points of $G$ as the group
$$
G(A) := \text{hom}_{k-\text{Alg}}(k[X_{11},\ldots,X_{nn}]/I_k(G), A),
$$
where $I_k(G)$ is the ideal in $k[X_{11},\ldots,X_{nn}]$ of all polynomials vanishing on $G$.
If $A=K$ is some field between $k$ and $\mathbb{C}$, is there a natural identification
$$
G(K) \cong G \cap GL(n,K)
$$
and if so, why? I can see how to define such a map, but not if its an isomorphism.
 A: Yes.  Note that if $K$ is a field between $k$ and $\mathbb{C}$, then every variety over $k$ is naturally a variety over $K$.  So there is no loss in assuming $k = K$.
If $X$ is an affine variety over $k$, and $Z$ is a closed subvariety of $X$ which is defined over $k$, then $Z(k) = X(k) \cap Z$.  I'll sketch three ways of seeing this.
In your case, $X = \textrm{GL}_n$, and $Z$ is your closed subgroup.  This is true for nonaffine, not necessarily closed subvarieties as well, but the general proof just reduces to the affine case. 
1 .  Noninstristic way.  View $X$ as the zero set of some polynomials in some affine space $\mathbb{A}^m$.  For example, $\textrm{GL}_n$ is the zero set of a polynomial in $\mathbb{A}^{n^2+1}$.  Then $X(k)$ can be identified with $X \cap k^m$.  To say that $Z$ is a subvariety of $X$ is to say that $Z$ is a subvariety of $\mathbb{A}^m$ contained in $X$, so we get $Z(k) = Z \cap k^m = Z \cap X \cap k^m = Z \cap X(k)$. 
2 .  Intristic way.  To an affine $k$-variety $X$ corresponds its coordinate ring $\mathcal O_X(X)$, which is a $\mathbb{C}$-algebra, and its "relative coordinate ring" $k[X]$, which is a $k$-algebra.  The Nullstellensatz gives you a natural identification of the variety $X$ with $X(\mathbb{C})$, the set of $\mathbb{C}$-algebra homomorphisms of the coordinate ring of $X$ into $\mathbb{C}$.  We can identify $\mathcal O_X(X) = k[X] \otimes_k \mathbb{C}$ as $\mathbb{C}$-algebras.  And of course, $X(k)$ injects into $X$, because you can tensor a $k$-algebra homomorphism of $k[X]$ into $k$ with $1_{\mathbb{C}}$ to get a $\mathbb{C}$-algebra homomorphism of $\mathcal O_X(X)$ to $\mathbb{C}$.
The inclusion morphism $Z \rightarrow X$ corresponds to a surjective $\mathbb{C}$-algebra homomorphism $\mathcal O_X(X) \rightarrow \mathcal O_Z(Z)$.  This homomorphism is obtained by tensoring with $\mathbb{C}$ from a surjective $k$-algebra homomorphism $k[X] \rightarrow k[Z]$.  It's this latter homomorphism which induces a map $Z(k) \rightarrow X(k)$, and the surjectivity of the last homomorphism makes this one injective.  So the claim that $Z(k) = Z \cap X(k)$ comes down to check the commutativity of this diagram of injective maps:
$$\begin{pmatrix} X(k) & \rightarrow & X \\ \uparrow & & \uparrow \\ Z(k) & \rightarrow & Z \end{pmatrix} $$
3 .  Via Galois theory.  A variety over $\mathbb{C}$, defined over $k$, is essentially the same as a variety over $\overline{k}$, the algebraic closure of $k$ in $\mathbb{C}$.  So there is actually no need to work with $\mathbb{C}$.  Being defined over $k$ gives an action of the Galois group $\Gamma = \textrm{Gal}(k_s/k)$ on $X$, where $k_s$ is the separable closure of $k$.  Since we are in characteristic zero, we have $k_s = \overline{k}$.  If $X$ is embedded as a closed $k$-subvariety of affine space, then the Galois action is just the pointwise action.  
Under this action, $X(k)$ is exactly the set of points of $X$ which are fixed by every element in $\Gamma$.  The same for $Z(k)$ and $Z$.  So $Z(k) = Z \cap X(k)$.
