I am confused about something.
When introducing the Lying Over Theorem -- namely, that if $f \colon R \subset S$ is an integral extension then $f^* \colon \mathrm{Spec}(S) \to \mathrm{Spec}(R)$ is a surjection -- my professor said that the intuition ''$X \to Y$ is a 'cover' implies it is a surjection'' is not always true.
His example was $\mathbf{C}^* \hookrightarrow \mathbf{C}$ and the induced map $\mathbf{C}[x] \hookrightarrow \mathbf{C}[x, y]/(xy - 1)$, where you visualize $\mathbf{C}^*$ as the hyperbola $xy = 1$ and see that its projection onto the $x$-axis is not surjective. But is it not true that $\mathbf{C}[x] \hookrightarrow \mathbf{C}[x, y]/(xy-1)$ is integral? Since $y$ satisfies the polynomial $xt - 1 \in \mathbf{C}[x][t]$ and the ring is finitely generated by $y$ over $\mathbf{C}[x]$ [Incorrect, see comments]? So why doesn't this violate what the Lying Over Theorem is trying to say?
Does anyone see what my professor meant and/or where I'm confused? I apologize for the hazy question.
As you state in the comments, $\mathbb{C}[x,y]/(xy-1)$ is finitely generated as a $\mathbb{C}[x]$-algebra, but not as a $\mathbb{C}[x]$-module. (Recall that, given an extension of rings $A \subseteq B$, one can show that $\alpha \in B$ is integral over $A$ iff $A[\alpha]$ is finitely generated as an $A$-module.) Note that $\mathbb{C}[x,y]/(xy-1) \cong \mathbb{C}[x,1/x]$ so every element is of the form $f(x)/x^e$ for some $f \in \mathbb{C}[x]$.
For contradiction, assume $\mathbb{C}[x,1/x]$ is generated by $\frac{f_1(x)}{x^{e_1}}, \ldots, \frac{f_r(x)}{x^{e_r}}$ as a $\mathbb{C}[x]$-module. Choose $m > \max\{e_1, \ldots, e_r\}$. Then $$ \frac{1}{x^m} = a_1(x) \frac{f_1(x)}{x^{e_1}} + \cdots + a_r(x) \frac{f_r(x)}{x^{e_r}} $$ for some $a_i(x) \in \mathbb{C}[x]$. Then $$ 1 = x^{m-e_1}a_1(x) f_1(x) + \cdots + x^{m-e_r} a_r(x) f_r(x) $$ and since $m > e_i$ for all $i$, substituting $x = 0$ yields $1 = 0$, contradiction.