# Dimension of a kernel

It was mentioned in my class that ${\partial\over\partial x}+{\partial\over\partial y}$ acting on the space of polynomials $p(x,y)$ where each individual variable --$x,y$ can have degree up to $n$ -- such that $x^ny^n$ is permissible but not $x^{n+1}$ -- has kernel of dimension $n+1$, why is that?

Thank you.

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$(\partial/\partial x + \partial/\partial y)f=0$ iff $f$ is a function of $x-y$. To see it, you can e.g. make the change of variables $u=x+y$, $v=x-y$, and compute that $\partial/\partial x + \partial/\partial y=2\partial/\partial u$. If $f(u)$ is a polynomial of degree $k$ then $f(x-y)$ has your restriction on degrees iff $k\leq n$, so you get dimension $n+1$ (of the space of polynomials in one variable, of degree $\leq n$). Notice, however, that we need characteristic $0$ (I suppose your polynomials have coefficients in some field). Otherwise we might have $(\partial/\partial u) u^n=0$, and the dimension would increase.