Exercise I-$43$(a) in Eisenbud-Harris This is exercise I-$43$(a) in Eisenbud/Harris, Geometry of Schemes. 
Let $k[t] \rightarrow k[x,u]/(xu)$ be a $k$-algebra homomorphism given by $t \mapsto x+u$. For $\alpha \in k$, the fiber ring corresponding to the prime ideal $(t-\alpha)$ of $k[t]$ is given by $k[x]/x(x-\alpha)$. This fiber ring has exactly two prime ideals if $\alpha \neq 0$ and one prime ideal otherwise. The authors say that for $\alpha=0$ the fact that $k[x]/x^2$ is a two-dimensional $k$-vector space reflects the local structure of the map at the point $(x,u) \in \operatorname{Spec}(k[x,u]/(xu))$. However, even if $\alpha \neq 0$, $k[x]/x(x-\alpha)$ is still two-dimensional as a $k$-vector space, so that i don't see how the vector space dimension of the fiber ring captures something special about the point $(x,u)$. On the other hand, i do see that $k[x]/x^2$ can not be isomorphic as a ring to $k \times k$ because the latter has two prime ideals. But now what is the ring isomorphism identifying $k[x]/x(x-\alpha)$ with $k \times k$? The obvious map that identifies these two as $k$-vector spaces seems not to be a ring homomorphism.
 A: As Mohan says in the comments,
\begin{align*}
\frac{k[x]}{(x(x-\alpha))} &\overset{\sim}{\rightarrow} \frac{k[x]}{(x)} \times \frac{k[x]}{(x-\alpha)} \cong k \times k\\
f &\mapsto (f \ \ \text{mod} \ (x), \, f \ \ \text{mod} \ (x-\alpha)) = (f(0), f(\alpha))
\end{align*}
by the Chinese Remainder Theorem, since the ideals $(x)$ and $(x-\alpha)$ are comaximal.
I agree that the dimension does not distinguish the point $(x,u)$, but as you say, the fact that $k[x]/(x^2) \not\cong k \times k$ does.  An easy way to see this is to note that $k[x]/(x^2)$ is not reduced ($x$ is nilpotent), while $k \times k$ is.  I am more comfortable with classical terminology so I may be misusing the term, but I would say that the map is ramified at the point $(x,u)$: the map is generically $2$-to-$1$, but there is only $1$ preimage above the point $(t)$.
The remark about the dimension of $k[x]/(x^2)$ seems a little puzzling until you look at part (b) of the exercise.  There you have a map between two planes in $4$-space that intersect at a point, and a plane.  A very strange thing happens in this case: nearly all the fibers are $2$-dimensional, but the fiber containing the doubled point is $3$-dimensional!  The authors mention how this is due to the fact that the domain of the map is not locally Cohen-Macauley, but I'll let someone who knows more than I do tell you what that means!
