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$\DeclareMathOperator\Spec{Spec}$Suppose $X = \Spec A$ is a smooth affine variety over $\mathbb C$ and suppose $L/K$ is a finite extension of its function field. Let $Y = \Spec B$, where $B$ is the set of elements of $L$ which are integral over $A$. Is there a simple geometric/explicit description of $Y\to X$ if we know the field extension explicitly? This is a somewhat vague question so let me given an example of a kind of answer that I would consider geometric/explicit.

Suppose we know that $L$ is obtained by adjoining to $K$ a root of some irreducible polynomial $f\in K[t]$, i.e., we have $L \simeq K[t]/(f)$ as field extensions of $K$. Then, can we geometrically describe $Y\to X$ in terms of this $f$? If this is something well-known, a reference would be very much appreciated.

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    $\begingroup$ TeX note: \operatorname adds operator spacing automatically, so you can use $\operatorname{Spec} A$ \operatorname{Spec} A instead of $\text{Spec }A$ \text{Spec }A. If you will use an operator repeatedly, then you can define it once and for all as \DeclareMathOperator\Spec{Spec}. I have edited accordingly. $\endgroup$
    – LSpice
    Commented Mar 19, 2023 at 0:03
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    $\begingroup$ Short answer: Yes. In general $Y$ will be an $[L:F]$-sheeted branched cover of $X$. The points $p \in X$ where $f(p) \in \mathbb C[t]$ has repeated roots are the branching set, with the branching dictated by how these roots merge. The simplest nontrivial example to keep in mind is an (affine open of a) hyperelliptic curve branched over $\mathbb A^1$, with $f = t^2 - (x-x_1)\cdots (x-x_n)$, where $\{x_i\}$ are the branch points. $\endgroup$ Commented Mar 19, 2023 at 5:34
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    $\begingroup$ @KevinCasto: It can be more subtle that that. If $A=\mathbb{C}[x]$ and $f=t^2-x(x-1)^2$, then a double root occurs at $x=0$ and $x=1$, but the normalization is $\operatorname{Spec}(A[u]/(u^2-x))$ (setting $u=t/(x-1)$) which is ramified only at $0$. $\endgroup$ Commented Mar 19, 2023 at 6:45
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    $\begingroup$ @MohanSwaminathan Oop sorry, this isn't true in general. But it is true for curves, i.e. when $Y$ is 1-dimensional, as in the cases we've been discussing; see mathoverflow.net/questions/9652/… $\endgroup$ Commented Mar 19, 2023 at 21:00
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    $\begingroup$ For just the fact that B is finite over A, see stacks.math.columbia.edu/tag/032L $\endgroup$ Commented Mar 20, 2023 at 6:25

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