Geometric interpretation of forking and non-forking extensions in ACF; Concrete examples involve non geometrically irreducible varieties? Suppose $A\subseteq B\subseteq L$ for an algebraically closed field $L$. Make the identification of the model-theoretic type space $S_n(A)=SpecK[A][X_1,...,X_n]=\mathbb A^n_{K[A]}$ as the scheme theoretic affine space where $K[A]$ is the field generated by $A$ and similarly so for $B$. I wish to understand the meaning of nonforking and forking extensions in this language. 
I think I reduced the illustration to an analysis about varieties which are irreducible but not geometrically irreducible, and finding components in a base change which have strictly smaller degree, but am not sure. 
Algebraic geometry question (also see a more precise version below): Can someone give me an example of such a phenomena?
Model theory question: Is this a correct and complete description?
Given a type $p\in S_n(A)$, we can consider $p$ as an irreducible affine subvariety $X$. An extension of types $p\subseteq q$ where $q\in S_n(B)$ then corresponds to a pre-image $Y$ of $X$ under the morphism $\mathbb A_{K[B]}^n\rightarrow \mathbb A_{K[A]}^n$ induced by the inclusion $K[A]\hookrightarrow K[B]$.This means to take $X$ in $\mathbb A^n_{K[A]}$ and consider the larger set of points in $\mathbb A^n_{K[B]}$ and $Y$ is an irreducible component of that larger set of points.
It is a theorem that Morley rank equals the geometric (and Krull) dimension in $ACF$. So I guess for $q$ to be a nonforking extension of $p$, this means $Y$ has the same dimension as $X$. So in order for there to be interesting example of both forking and non-forking extensions of $p$, the base change of $X$ to an algebraic closure must contain many irreducible component, of many different dimensions. Can someone give me an example of such a variety?
Furthermore, it is a theorem that types have unique non-forking extensions. That means there is a unique $Y$ of same dimension as $X$. How to identify this $Y$?
Precise geometry question: I am looking for an irreducible variety $X$ such that the base change to some field extension $Y$ is no longer irreducible, and contains components of many different dimensions. Furthermore (at least under certain conditions, see paragraph below), it seems like there must be a unique component of $Y$ of maximal dimension, equal to that of $X$. How to find the $Y$?
But I seem to find some contradiction when considering a basic example. The Morley rank of $q$ is supposed to be bounded by the Morley rank of $p$. Specifically, take $A=\mathbb R$ and $B=\mathbb C$ and $X$ to be any variety without real points but having complex points, for example the curve $x^2+y^2=-1$. Then it is not even true that $\dim(Y)\leq \dim(X)$. But perhaps this is just due to some technical problem about the real points being empty?
Reference on Forking: I used the definitions and facts from David Marker's book 'An introduction to Model theory'. But here is an article on the subject (albeit it does not list all the required theorems): http://modeltheory.wikia.com/wiki/Forking
 A: There are a number of misconceptions in your question. Most importantly, the idea that examples of forking in ACF always come from varieties which are not geometrically irreducible is wrong. Every type over Morley rank $>0$ (i.e. every non-algebraic type) has a forking extension over some larger set of parameters. See the parabola example in point 1.


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*Your description of extensions of types is wrong. Here is a correct description: We're given $A\subseteq B$ and the map $f\colon \mathbb{A}^n_{K[B]}\to \mathbb{A}^n_{K[A]}$. If you identify a type $p\in S_n(A)$ with an irreducible variety $X\subseteq \mathbb{A}^n_{K[A]}$, then an extension $p\subseteq q\in S_n(B)$ is an irreducible subvariety $Y\subseteq f^{-1}(X)\subseteq \mathbb{A}^n_{K[B]}$ which projects generically onto $X$ under $f$. In particular, $Y$ does not have to be an irreducible component of the preimage of $X$. For a very simple example, consider the variety defined by $y = x^2$ in $\mathbb{A}^n_{K[A]}$, suppose $B$ is not an algebraic extension of $A$, and consider any $b\in B$ which is transcendental over $A$. Then the closed point $(b,b^2)$ is an irreducible subvariety of $f^{-1}(X)$ and its image under $f$ is the generic point of $X$. The type $q$ isolated by $(x=b)\land (y = b^2)$ is a forking extension of the generic type of the parabola, $p$, which contains $y=x^2$ and $r(x,y)\neq 0$ for all polynomials $r$ which are not in the ideal of $K[A][x,y]$ generated by $y-x^2$. It might be easier to visualize this if you identify types with scheme-theoretic points instead of their closures (irreducible varieties). Given a type/point $p$ in $\mathbb{A}^n_{K[A]}$, the extensions of $p$ to $B$ are exactly the types/points in $f^{-1}(\{p\})$. In my example, $f^{-1}(\{p\})$ contains the generic point of the parabola over $B$ (the unique non-forking extension of $p$) together with all of the closed points on the parabola whose coordinates are not in $A$ (the forking extensions of $p$).

*"It is a theorem that Morley rank equals the geometric (and Krull) dimension in ACF." This is correct, but the key thing is in ACF. You're not going to be able to see the Morley rank of a type geometrically by looking in the affine space $\mathbb{A}^n_{K[A]}$ unless $K[A]$ is algebraically closed. You have to base change to an algebraically closed field. This answers your question about $x^2 + y^2 = -1$. This formula has Morley rank $1$, and while it has no $\mathbb{R}$-points, it picks out a $1$-dimensional curve in $\mathbb{A}^2_\mathbb{C}$. Incidentally, this fact is related to my claim above that every type of Morley rank $>0$ has a forking extension. If you take a type $p$ of positive Morley rank over $A$ and base change to a sufficiently large algebraically closed field, then (a) you'll see that the variety $X$ corresponding to $p$ has positive dimension, so there are points on it, and (b) there will be some points on $X$ whose coordinates are transcendental over $A$. Picking one of these points gives a forking extension of $p$.

*"It is a theorem that types have unique non-forking extensions." This is false. What is true is that if $A$ is algebraically closed, then every type over $A$ has a unique non-forking extension to a type over $B$. (Technical note: in a general theory, one has to assume that $A$ is algebraically closed in $\mathbb{M}^{\text{eq}}$, but ACF eliminates imaginaries, so there's no difference between $\text{acl}$ and $\text{acl}^{\text{eq}}$). See Marker Theorem 6.3.2 and Corollary 6.3.13. For a concrete example, consider the type over $\mathbb{R}$ isolated by $x^2 = -1$. This type has two extensions to types over $\mathbb{C}$ (isolated by $x = i$ and $x = -i$), both of which are non-forking extensions. 
