Fibers of extension of scalars to algebraic closure of an affine variety is of dimension 0. 
Let $X$ be an affine variety over $k$ and $\overline{k}$ be the algebraic closure of $k$. Then the projection morphism $X_\overline{k} = X \times _k \overline{k} \to X$ has dimension 0 fibers. 

Say $X = \text{Spec } A$. Then $X_\overline{k} = \text{Spec} (A \otimes _k \overline{k})$ 
The fiber of the homomorphism at point $p \in X$ is
$$X_p = X_\overline{k} \times _X \text{Spec} k(p) = \text{Spec} (A \otimes_k \overline{k} \otimes_A k(p)) = \text{Spec}(A_p/m_p A_p \otimes _k \overline{k}) $$
Thus, it is equivalent to say that the ring $A_p / m_p A_p \otimes_k \overline{k}$ has (Krull) dimension $0$. As far as I know, the tensor of two fields is not necessarily a field, thus ruling out any trivial dimension considerations. I am not really sure how to think of prime ideals in such a tensor product. Any ideas to tackle this would be appreciated. 
An idea I'm formulating right now is that $k(p) = A_p / m_p A_p$ is a field extension of $k$. Thus the tensor is simply an extension of scalars, from one field to its algebraic closure. Is this perhaps equivalent to considering the same field $k(p)$ but now explicitly embedded in this larger field, namely the closure?
 A: It suffices to show that if $K$ and $L$ are two fields extending a field $k$ and $L$ is algebraic over $k$, then $\dim K\otimes_k L=0$ (let $K=A_p/m_pA_p$ and $L=\overline{k}$).  Let $R=K\otimes_k L$; to show that $R$ has dimension $0$, it suffices to show that for any homomorphism $R\to F$ where $F$ is a field, the image of $R$ in $F$ is a field (if $p\subset q$ are primes in $R$, then you could take $F$ to be the field of fractions of $R_p/q$ and $R\to F$ to be the canonical map).
So let $\varphi:R\to F$ be a homomorphism to a field.  Then $\varphi$ restricts to a homomorphism $K\to F$, and the image $\varphi(K)$ is a subfield of $F$.  Since $L$ is algebraic over $k$, $R$ is generated as an $K$-algebra by elements that are algebraic over $K$, and hence every element of $R$ is algebraic over $K$.  It follows that every element of $\varphi(R)$ is algebraic over $\varphi(K)$.  It follows that for any $\alpha\in\varphi(R)$, $\varphi(K)[\alpha]$ is actually a field, and so $\varphi(R)$ is a field.
A: Since $k\to \bar k$ is integral so is $k(p)\to k(p)\otimes_k\bar k$ (Atiyah-Macdonald:  Chap.5, Ex.3).
But then $\dim  k(p)=\dim k(p)\otimes_k \bar k$ (Matsumura: Exer. 9.2), so that indeed $\dim k(p)\otimes_k \bar k=0$. 
Edit: the fibers are actually finite
Since the ring  $k(p)\otimes_k \bar k$ is noetherian of dimension zero, it is artinian and has thus finite discrete spectrum.
So all the fibers of $X_\overline{k} \to X$ are actually finite discrete schemes.
This was not obvious, because the scheme morphism itself  $X_\overline{k} \to X$ is far from finite in general: think of $X=\operatorname {Spec}(k)=$  a point!  
