Klein quartic smooth in Char different from 7 I cannot solve the following exercise:
Let $\mathbb F$ be a field with $char(\mathbb F)\ne 7$. Let $C$ be the projective Klein curve defined by 
$x^3y+y^3z+z^3x =0$.
a) Prove that $C$ s smooth
b) Compute the divisors of the functions $x/y$ and $x/z$.
How can I prove that this curve is smooth only in $char(\mathbb F)\ne 7$?
 A: The polynomial is fixed under the cyclic permutation of variables: $x\mapsto y\mapsto z\mapsto x$, so it suffices to study a single affine chart. I picked the chart with $z=1$.
At a singular point the following three equations must all hold simultaneously
$$
\left\{\begin{array}{lcl}
f=x^3y+y^3+x&=&0,\\
f_x=3x^2y+1&=&0,\\
f_y=x^3+3y^2&=&0.\end{array}\right.
$$
From the second equation we can solve $y=-1/3x^2$. Plugging this into the first gives
$$
\frac{2x}3-\frac1{27x^6}=0,
$$
implying that at a singularity $x^7=1/18$. OTOH, plugging it into the last equation gives
$$
x^3+\frac1{3x^4}=0
$$
Implying that $x^7=-1/3$. So for a singularity to exist we must have $18=-3$.
This holds in characteristics $3$ and $7$ only.


*

*In characteristic $3$ the second equation reads $1=0$ which is absurd.

*In characteristic $7$ we have the singular point $x=2,y=4$. Because there are no seventh roots of unity $x=2$ is the only solution, and this is therefore the only singular point. It may be worth pointing out that $2$ is the primitive third root of unity $\omega$ in characteristic seven, so the singular point can be rewritten in the homogeneous coordinates as
$$[x:y:z]=[1:\omega:\omega^2]$$ showing that this point really is stable under the cyclic permutation of coordinates.

A: a) To prove smoothness it's enough to check it affine locally. So let's take for example the affine chart where $x\neq 0$. Then, in this chart the curve is defined by the equation $f(Y,Z)=Y+Y^3Z+Z^3=0$, where $Y=y/x$ and $Z=z/x$. 
To show smoothness we want to see the Jacobian matrix has rank $=2-1=1$, i.e., for every point $(Y,Z)$ in the curve either $\partial_Yf(Y,Z)=1+3Y^2
Z\neq 0$ or $\partial_Zf(Y,Z)=Y^3+3Z^2\neq 0$.
If $char(K)=7$ then $(2,4)$ is a nontrivial solution to all these equations, so the curve is singular at this point. I don't know how to show that these systems (you need to check the other charts as well!) don't have any solutions over fields of characteristic $\neq 7$.
b) Computing divisors is also something you can do locally as long as you keep track of where are things coming from. For example, let's compute the divisor $\operatorname{div}(x/y)$. I'm going to be a little bit sloppy with the notation below, so be careful while going through it. I'm also working with characteristic 0, I'll let you work the details for the other cases.
Again consider the affine chart where $x\neq 0$, call it $U_0$. In here, $x/y=1/Y$. Clearly the divisor can only have a pole at the point $Y=Z=0$, so we just need to calculate the order. Furthermore, the maximal ideal of this point contains $(Y,Z)\subseteq\mathfrak{m}$, and since $(Y,Z)$ is maximal we see that $\mathfrak{m}=(Y,Z)$.
Now we pass to the localization at $\mathfrak{m}$. Since $Y=-Y^3Z-Z^3$ we see that $Y\in\mathfrak{m}^3\subset\mathfrak{m}^2$. So that $\mathfrak{m}/\mathfrak{m}^2$ is generated by $Z$ and $Z$ is a local parameter by Nakayama's lemma, i.e., $\operatorname{ord}_{(0,0)}(Z)=1$.
By the same equation we get that:
$$
\operatorname{ord}_{(0,0)}(Y)
=\min\{3\operatorname{ord}(Y)+\operatorname{ord}(Z),3\operatorname{ord}(Z)\}
=\min\{3\operatorname{ord}(Y)+1,3\},
$$
from which it follows that $\operatorname{ord}_{(0,0)}(Y)=3$ (since the valuation at $Y$ is positive). Hence, $\operatorname{div}(x/y)|_{U_0}=-3\cdot [1:0:0]$.
Repeat the same procedure over $U_1=\{y\neq 0\}$. In here we have the coordinates $X=x/y$ and $Z=z/y$. In this chart the equation becomes $X^3+Z+Z^3X$ and we see that $Z=-X^3-Z^3X$. The same argument as above shows that $X$ is a local parameter so that $\operatorname{ord}_{(0,0)}(X)=1$ and $\operatorname{div}(x/y)|_{U_1}=1\cdot[0:1:0]$.
Finally we go to $U_2$ with coordinates $X,Y$ and equation $(X^3Y+Y^3+X)$. In this chart the rational function corresponds to $X/Y=-(X^3Y+Y^3)/Y=-(X^3+Y^2)$, which has order $\min\{3\operatorname{ord}(X),2\operatorname{ord}(Y)\}$. Repeating the same argument as above we see that $X$ has order at least $3$, while $Y$ is a local parameter, from which it follows that $\operatorname{ord}(X)=3$ and $\operatorname{ord}(Y)=1$. Hence, $\operatorname{div}(x/y)|_{U_2}=2\cdot[0:0:1]$.
Therefore we can conclude that 
$$
\operatorname{div}(x/y)=-3\cdot[1:0:0]+1\cdot[0:1:0]+2\cdot[0:0:1].
$$
Edit: I followed the accepted answer in here when computing the divisor, Calculating the divisors of the coordinate functions on an elliptic curve
