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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.

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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 $p$.

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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!

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    Dear Aneesh, if $R$ has dimension $\geq 1$, then $R[x]$ is guaranteed *not* to be principal ideal domain and is thus even less a euclidean domain.2012-07-10