# Show that $DF(x,y)$ is invertible in a dense and open subset of $\mathbb {R^2}$

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!

## 2 Answers

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.

• Thank you very Much!very helpful! – HipsterMathematician Nov 19 '12 at 19:08

Very carefully write out the matrix $DF(x,y)$ using the chain rule. You know that the matrix is invertible if and only if its determinant is nonzero, so write out an expression for the determinant and set it equal to zero. You should end up with something like $$p'(x+y)p'(x-y) = 0 .$$ When is this possible? Remember that $p'$ is itself just a polynomial over $\mathbb{R}$, so think about the set of zeroes of that polynomial and how they related to the zeroes of the expression $p'(x+y)p'(x-y).$

• "know that the matrix is invertible if and only if its determinant is zero" isn't it non-zero? – HipsterMathematician Nov 19 '12 at 19:31
• Yeah, made a typo. Should be fixed now. :) – Zach L. Nov 20 '12 at 0:04