This is exercise II.6.4 in Shafarevich's BAG. I am pretty sure I solved it, but my answer doesn't align with the hint, and I wonder whether that's an error in the book or on my part.
The task is to find the branch locus of the map $p: X\to \mathbb{P}^n$, where $X$ is the normalization of $\mathbb{P}^n$ in the quadratic extension of $k(\mathbb{P}^n)=k(x_1,\ldots,x_n)$ defined by the equation $y^2=f(x_1,\ldots,x_n)$, where $f$ is of degree $m$. A hint is given: the answer depends on the parity of $m$. However, I was unable to find this dependency.
Denote $F=k(\mathbb{P}^n)[y]/(y^2-f)$. $X$ and $p$ are then defined up to isomorphism by these three properties:
- $X$ is normal,
- $f$ is finite,
- $k(X)=F$, $f^*:k(\mathbb{P}^n)\to F$ is the inclusion into extension.
In one of the previous exercises, I showed that such normalizations of affine varieties exist, are unique, and I also defined them explicitly. So let's consider an affine chart of $\mathbb{P}^n$ and its normalization $Y$.
To get this normalization, we consider $k[\mathbb{A}^n]$ as a subring of $F$ and find its integral closure, which then uniquely defines an algebraic variety with the above properties. Hence we need to find all the elements of $k(x_1,\ldots,x_n)[y]/(y^2-f)$ integral over $k[x_1,\ldots,x_n].$ We can write an arbitrary element $t$ as $\alpha y+\beta$, $\alpha,\beta\in k(x_1,\ldots,x_n).$ Note that
$$t^2=2\beta t+\alpha^2 f - \beta^2,$$
so by Gauss's lemma, $t$ is integral if and only if the coefficients of this polynomial are in $k[x_1,\ldots,x_n].$ Hence we have $\beta\in k[x_1,\ldots,x_n],\ \alpha^2 f\in k[x_1,\ldots,x_n]$. But $f$ is irreducible, so $\alpha\in k[x_1,\ldots,x_n]$.
Hence the integral closure is $k[x_1,\ldots,x_n,y]/(y^2-f)$, and the normalization itself is the zero locus of $y^2-f$ in $\mathbb{A}^{n+1}$ with coordinates $x_1,\ldots,x_n,y$. The map $p$ is then just the projection to $\mathbb{A}^n$ with the $x$ coordinates. The preimage of any point $(x_1,\ldots,x_n)$ consists of the points $(x_1,\ldots,x_n,y)$ with $y^2=f(x_1,\ldots,x_n)$, so the branch locus is defined by $f(x_1,\ldots,x_n)=0$.
Assuming the above reasoning is correct, the zero locus of the original map $p:X\to\mathbb{P}^n$ can be then found by using projectivization. But then the answer is completely independent of $m$. And I can't find any errors in the above reasoning.
EDIT: I just realized that for the quadratic extension to make sense $f$ does not have to be irreducible, it only has to not be a square. But then it will be perfectly OK for $\alpha$ to be a fraction which complicates things, and I also can't see how the reasoning in this case can be made dependent only on $m$ and not on the factors of $f$ itself.