Let $Y=\mathrm{Spec}(\mathbb{C}[X,Y]/(X^2-Y^2))$ and $S=\mathrm{Spec}(\mathbb{C}[X])$. The scheme morphism $f$ just "imitate" the "projection onto the $X$ axis", i.e. it corresponds to the ring map $\varphi:\mathbb{C}[X]\hookrightarrow\mathbb{C}[X,Y]\to\mathbb{C}[X,Y]/(X^2-Y^2)$.
Question is what it the generic fiber?
First, I do it like this: The residue field at $(0)$ of $\mathbb{C}[X]$ is simply $\mathbb{C}[X]_{(0)}/(0)\cong\mathbb{C}(X)$. Then the generic fiber is simply $Y_\eta=\mathrm{Spec}(\mathbb{C}(X)\otimes_{\mathbb{C}[X]}\mathbb{C}[X,Y]/(X^2-Y^2))$. But then what? I.e. what is this tensor product?
On the other hand, $Y_\eta=f^{-1}[\{\eta\}]=\{\mathfrak{p}\subseteq \mathbb{C}[X,Y]/(X^2-Y^2)|\mathfrak{p}^c=(0)\subseteq\mathbb{C}[X]\}$.
Best Answer
The geometric intuition here is that you are looking at the union of the two lines $Y = X$ and $Y = -X$ in the Cartesian plane, and you are projecting onto the $X$-axis. Generically, the fiber of this map is the union of two reduced points, so the algebra should confirm that.
Let $K = \mathbb{C}(X)$. Then notice by the Chinese Remainder Theorem that your tensor product is $$K[Y]/(Y^2 - X^2) = K[Y]/(Y - X) \oplus K[Y]/(Y + X).$$ It then follows that $$Y_\eta = \operatorname{Spec}\big( K[Y]/(Y - X) \oplus K[Y]/(Y + X) \big)= \operatorname{Spec} K \sqcup \operatorname{Spec} K,$$ where we used the fact that Specs of products give disjoint unions of Specs.
Thus, the generic fiber looks like the union of two points, as expected.