$R$-points on a variety: explain to me like I'm 5 If I have an affine variety $X$ in $K^n$, and a subring $R$ in $K$, $X(R)$, the set of $R$-points, is just $X \cap R^n$, by definition (right?).  What is the highbrow way to view $X(R)$ so it doesn't depend on the imbedding of $X$ in $K^n$?  And in what sense is the resulting definition independent of the imbedding? Do you have to use the word "scheme" to explain it to me? Its ok if you do, but I barely know anything about schemes except the definition -- though i'm generally ok (not great) with classical affine algebraic geometry
 A: Let us assume for now that $K$ is alg. closed and $R$ is a subfield of $K$.
We know that $X$ is affine, and $X$ is defined by $(f_1,...,f_p)$ in $\mathbb{A}^n$. Take $a=(a_1,...,a_n)$ a point in $X$.
You get a $K$-linear map $K[T_1,...,T_n]\to K$ mapping $T_i$ to $a_i$. Saying that $a$ is on $X$ is saying that this map factors through $K[T_1,...,T_n]\to K[T_1,...,T_n]/(f_1,...,f_p)\to K$. In other words giving a point of $X$ is giving a $K$-algebra morphism from $A_K=K[T_1,...,T_n]/(f_1,...,f_p)$ to $K$. 
Now assume that such a morphism is given. You can look at the image $T_i$ in $K$ and this gives you an $a_i$ and $(a_1,...,a_n)$ is an element of $X$ (as $f_k(a_i)$ has to be zero). Thus you can view points of $X$ as $K$-morphisms from $A_K$ to $K$.
To prove that the correspondence defined is a bijection you may use the Nullstellensatz.
Let us assume now that the polynomials are defined over $k$ which is a subfield of $R$ (it may be $R$). Giving a $K$-morphism from $A$ to $K$ is the same thing as giving a $k$-morphism from $A_k=k[T_1,...,T_n]/(f_i)$ to $K$. This morphism may, or may not land in $R$. If it does, in particular the point you get in $X$ will have "coordinates" in $R$ (recall the construction). In other words if the polynomials are defined over $k$, a point with coefficients in $R$ is a $k$-morphism from $A_k$ to $R$. Or an $R$-morphism from $A_R$ to $R$.
To sum up, if your polynomials are defined over $k$, the points of the variety in $K^n$ which have coefficients in $R$ are identified with $k$-morphisms $A_k\to R$. But every $k$-automorphism of $A_k$ will map bijectively such automorphisms to themselves. In other words it only depends on the structure of $A_k$ as a $k$-algebra.
