TITLE: Show that $DF(x,y)$ is invertible in a dense and open subset of $\mathbb {R^2}$
QUESTION [5 upvotes]: My problem is the following: 
Let $p$ be a non constant polynomial over $\mathbb {R}$ and define $F(x,y)=(p(x+y),p(x-y))$.Show that $DF(x,y)$  is invertible in a dense and open subset of $\mathbb {R^2}$.
I've been thinking a lot on this one, but couldn't get far... I'm really stuck... any help is much appreciated!

REPLY [5 votes]: The Jacobi matrix of $F$ at $(x,y)$ is
$$DF(x,y)\left(\begin{matrix}
p'(x+y) & p'(x-y)\\
p'(x+y) & -p'(x-y)
\end{matrix}\right)$$
and its determinant is given by $-2p'(x+y) p'(x-y)$. So $DF$ is not invertible at $(x,y)$ if and only if $x+y \in Z$ or $x-y \in Z$ where $Z$ is the set of roots of $p'$. Since $p$ is non-constant, $p'$ is not the zero polynomial, so $Z$ is finite. The set of points where $DF$ is not invertible is therefore given by
$$\bigcup_{a \in Z} \left(\operatorname{Graph}(y = a+x) \cup \operatorname{Graph}(y = a-x)\right),$$
a finite union of lines in $\mathbb R^2$. Clearly, the complement of this is open and dense.