Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$ Does anybody know the genus of the following (projective) plane curve?: 
$$\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$$
where the $a_i$'s and the $b_i$'s are complex numbers with $a_j \ne a_i\ne b_i \ne b_j$ for every $i\ne j$.
 A: Even under the conditions on the $a_i$ and $b_i$, the genus cannot be expressed in terms of $n$. For instance for $n=3$, the genus is generically $1$. However, if for instance $a_1b_2=a_2b_1$, then the cubic curve is singular, so has genus $0$.
A: Let me give it a try from a different point of view. I'll work over the complex numbers. I did not really work out the details carefully, but in any case I hope that the idea is useful.
The linear forms $a_jX+b_jY+Z$ are linearly independent when taken by three. Proof: write the relevant 3x3 matrix for three different indices. After some column operations you get the Vandermonde matrix of the real parts of $\zeta^i,\zeta^j,\zeta^k$, which are different.
Suppose that the curve in question has an irreducible component of genus $g$ (I did not check irreducibility, but I don't need it; this is why I use a component). The normalization of it gives a map $F:C\to \mathbb{P}^2$ where $C$ is smooth projective of genus $g$. Taking the coordinates $X,Y,Z$ in $\mathbb{P}^2$ we write $F=[f:g:h]$ with $f,g,h$ in the function field of $C$, and we let $d$ be the height of $[f:g:h]$. Consider the lines $L=\{Z=0\}$ (this is my line at infinity, to fix ideas) and $H_j=\{a_jX+b_jY+Z=0\}$; by the previous paragraph the $H_i$ are in general position. Let $S=F^{-1}(L)$, this is a subset of $C$ and we have $\# S\le d$.
The fact that $F$ has image contained in the zero set of the curve in question gives that $F|_{C\smallsetminus S}$ never hits the lines $H_j$. At this point we invoke some result from Nevanlinna theory. Corollary 4 in J. T.-Y. Wang's paper "The truncated second main theorem of function fields" (JNT 1996) will do it: we get
$$
(n-4)d\le 0 + 3\max\{0,2g-2+\#S\}\le 3\max\{0,2g-2+d\}.
$$
For $n\ge 5$ we are forced to have $2g-2+d>0$ so that
$$
(n-4)d\le 6g-6+3d.
$$
You are interested in avoiding genus $0$ and $1$, so we can assume $g\le 1$ obtaining
$$
(n-4)d\le 3d.
$$
This is not possible for $n>7$, so, such curves don't have components of genus $0$ or $1$. This proves the result for $p> 15$ ($p$ does not need to be prime as far as I can see, taking $\zeta=e^{2\pi i/p}$).
A: Edit. There is a serious gap in the argument below.  The issue, the same one that arose in two similar questions, is that the Euler identity does not hold in positive characteristic $q$.  Therefore, for $q$ a prime dividing $n$, the critical locus of the degree $n$, dehomogenized polynomial $$f(x,y) = \prod_{i=1}^n (1+a_i x + b_iy)$$ on $\mathbb{A}^2$ is not necessarily $\{0\}$ in $\mathbb{A}^1$.  I have computed examples of such polynomials where the critical locus is precisely $\{0\}$ (for instance, $(p,q) = (5,2)$ or $(7,3)$), but I am not certain  for which primes $p>3$ and for which prime divisors $q$ of $n=(p-1)/2$ the critical locus equals $\{0\}$.  I am leaving the following argument for the moment, and I will try to sort out the critical locus of $f$.
In the OP's specific case, $p$ is an odd prime, $p=2n+1$, $\zeta_p$ is a primitive $p^{\text{th}}$ root of $1$, and $a_i=\zeta_p^{2i}-2+\zeta_p^{-2i}$, $b_i=\zeta_p^{i} - 2 + \zeta_p^{-i}$.  Of course the set of primitive roots of $1$ has size $p-1=2n$.  There is an action of $\mathbb{Z}/2\mathbb{Z}$ on this set sending a root $\zeta$ to $\zeta^{-1}$.  This action is fixed point free, since otherwise $2$ divides $p$.  
Although the OP does not quite say this, I assume that the index $i$ ranges over the $n$ orbits for this action.  In particular, since every such primitive root $\zeta$ is the square of another primitive root (squaring is an automorphism of a cyclic group of odd order), I am going to reindex with $i=2j$.  Thus, in my indexing, $$\alpha_j = \zeta_p^{4j}-2 + \zeta_p^{-4j} = (\zeta_p^{2j}-\zeta_p^{-2j})^2, \ \beta_j = \zeta_p^{2j}-2+\zeta_p^{-2j} = (\zeta_p^{j}-\zeta_p^{-j})^2.$$
Now let $q$ be a prime that divides $n$, say $n=qm$, $p=2qm+1$.  In other words, $p\equiv 1 \ (\text{mod}\ 2q)$.  Let $R$ be the Dedekind domain $\mathbb{Z}[1/p][\zeta_p]/\langle \zeta_p^p-1 \rangle$.  It is straightforward to compute that $R$ is finite and étale over $\mathbb{Z}[1/p]$.  The elements $\alpha_p$ and $\beta_p$ are as defined above.  The curve $C_R\subset \mathbb{P}^2_R$ is the $R$-flat Cartier divisor with defining polynomial
$$
f(X,Y,Z) = \left( \prod_j (Z+\alpha_j X + \beta_j Y) \right) - Z^n.
$$
To prove that the generic fiber of $C_R$ over $\text{Frac}(R)$ is smooth, it suffices to prove smoothness after reducing modulo $q$.
Since $q$ is relatively prime to $q$, there exists a least positive integer $r$ such that $q^r\equiv 1\ (\text{mod} \ p)$.  Let $k$ be the finite field with $q^r$ elements.  Then the multiplicative group $k^\times$ is a cyclic group of order $q^r-1$.  Since $p$ divides $q^r-1$, there exists an element $\zeta$ in $k^\times$ that has order $q$.  This gives a ring homomorphism $R\to k$ sending $\zeta_P$ to $\zeta$.    Thus also 
In particular, we can form the images in $k$ of the elements $\alpha_j$ and $\beta_j$. 
As above, since $\zeta$ has order $p$, and since $2$ does not divide $p$, no $\zeta^j-\zeta^{-j}$ equals $0$.  Thus $\beta_j = (\zeta^j-\zeta^{-j})^2$ is not zero.  Similarly, if some $\zeta^j = -\zeta^{-j}$, then $\zeta^j$ has order $4$ in $k^\times$ rather than order $p$.  Thus also $\alpha_j = \beta_j(\zeta^j+\zeta^{-j})^2$ is nonzero.
I claim that for distinct indices $j\neq \ell$, also $\alpha_\ell \beta_j - \alpha_j\beta_\ell$ is nonzero in $k$. Since $j\neq \ell$, both $\zeta^{\ell} \neq \zeta^j$ and $\zeta^{\ell} \neq \zeta^{-j}$.  Using again that squaring is an automorphism of the cyclic group $\langle \zeta \rangle \subset k^\times$ of order $p$, we also have
$$
\zeta^{2\ell} \neq \zeta^{2j}, \ \ \zeta^{2\ell} \neq \zeta^{-2j}.
$$
By direct computation, $$
\alpha_\ell\beta_j - \alpha_j\beta_\ell = \beta_\ell\beta_j((\zeta^{2\ell}-\zeta^{2j}) - (\zeta^{-2j}-\zeta^{-2\ell})).$$
As above, $\beta_\ell$ and $\beta_j$ are nonzero.  Therefore, by way of contradiction, assume that
$$
\zeta^{2\ell} - \zeta^{2j} = \zeta^{-2j}-\zeta^{-2\ell}.
$$
Multiplying both sides by $\zeta^{2\ell+2j}$ gives,
$$
\zeta^{2\ell}\zeta^{2j}(\zeta^{2\ell}-\zeta^{2j}) = (\zeta^{2\ell} - \zeta^{2j}).
$$
Therefore, either we have
$$
\zeta^{2\ell} = \zeta^{2j},
$$
or we have
$$
\zeta^{2\ell} = \zeta^{-2j}.
$$
But both of these contradict the hypothesis that $\ell \neq j$.  Therefore, by way of contradiction, $\alpha_\ell \beta_j - \alpha_j \beta_\ell$ is nonzero.  In particular, the three lines $Z=0$, $Z+\alpha_jX + \beta_jY =0$, and $Z+\alpha_\ell X + \beta_\ell Y$ are not collinear in $\mathbb{P}^2_k$. Since this holds for every $j$ and $\ell$, the curve $C_k$ is smooth.
Therefore, combined with the comments above, the reduction curve $C_k$ in $\mathbb{P}^2_k$ is smooth.  So the curve $C_{\text{Frac}(R)}$ is also smooth over $\text{Frac}(R)$.  Therefore the base change to $\mathbb{C}$ is a smooth curve in $\mathbb{P}^2_{\mathbb{C}}$ of degree $n$ and genus $(n-1)(n-2)/2$, just as abx explained.  
