Curve has a point which is either singular or has a tangent line parallel to the  y-axis. Suppose $k$ is an algebraically closed field with characteristic 0. Suppose $f(x,y)\in k[x,y]$ is irreducible and viewing $f(x,y)$ as a polynomial over $k[x]$ which is monic in $y$ and of degree>1 in $y$. 

We want to prove that the ideal $(f(x,y),f_y(x,y))\neq k[x,y]$. (if it is true)
Is this statement true for general? For example, let $R$ be a domain of dimension$\geq 1$ with char 0  and suppose $f(y)\in R[y]$ is a monic irreducible polynomial of degree >1. Is it true that the ideal $(f(y),f^{\prime}(y))\neq R[y]$ ?

Thanks.
 A: Under the conditions you state we have indeed  $(f(x,y),f_y(x,y))\neq k[x,y]$  
Else the curve $C=V(f)\subset \mathbb A^2_k$ would be smooth and the projection onto  the $x$-axis would be étale.
But this would yield a connected  étale covering of degree $deg(f)\gt 1$ of $\mathbb A_k^1$, which  is impossible because in characteristic zero $\mathbb A_k^1$ is algebraically simply connected.  
All this  is false in  characteristic $p\gt 0$ : the irreducible polynomial $f(x,y)=y^p-y -x\in k[x,y]$ satisfies $(f(x,y),f_y(x,y))=(y^p-y -x,-1)= k[x,y]$.
And the projection $pr_1:V(f)\to \mathbb A_k^1:(x,y)\mapsto x    $ is a non-trivial  étale covering of degree $p$.
A: If $R$ is an integral domain then $R[x]$ is a Euclidean domain where the metric is just the degree of a polynomial.
Also $f$ being irreducible, the only way $f'$ can have a non-unit common factor with $f$ is that $f|f'$ or $f'=0$ none of which is true here since $f$ has degree more than 1.
$f$ and $f'$ are thus prime to each other and unity in $R[x]$ can be written as an $R[x]$-linear combination of $f$ and $f'$.
So $(f(x),f'(x))\ne R[x]$ is necessarily false!
