Define $k:R[x,y]\to R[x,y]$ by $f(x,y)\mapsto\frac{\partial^2f}{\partial x\partial y}$. What is the kernel of $k$? Let $R[x,y]$ be a set of all real polynomial with two variables, $x$ and $y$. Define
a homomorphism $k:R[x,y] \to R[x,y]$ by $f(x,y) \mapsto \frac{\partial^2 f}{\partial x \partial y}$. Find $Ker \ k$.
I know that
$Ker (k) = \lbrace f \in R[x,y] \mid k(f) = e \rbrace$.
But, what's the identity element of the codomain? Any idea or hints? Thanks in advanced.
 A: Hint: elements of $R[x,y]$ are polynomials, thus composed of monomials, so look at what $\frac{\partial^2}{\partial x \partial y}$ does to monomials.
A: The operator $k$ is not a ring homomorphism (because $k(fg) \neq k(f)k(g)$) so it is a group homomorphism,where the group property is addition (because the double derivative is additive and scaling, certainly).
Therefore, we must find all elements of $R[x,y]$ such that $\frac{\partial^2f}{\partial x \partial y} = 0$ as a function. But $f$ is a polynomial : so let us collect all powers of $y$ together, and write $f(x,y) = a_0(x)+a_1(x)y+a_2(x)y^2 + ... + a_n(x)y^n$ where $a_i$ are polynomials only in $x$.
The derivative of this, with respect to $y$ is $a_1(x) + 2a_2(x)y + ... + na_n(x)y^{n-1}$. The derivative of that with respect to $x$ is $\frac{da_1}{dx} + 2y\frac{da_2}{dx} + ... + \frac{da_n}{dx}ny^{n-1}$. We have to find when this is a zero polynomial.
However, for it to be zero, there's only one way : every coefficient of $y^i$, and the constant term has to be $0$. So, each of $\frac{da_i}{dx} = 0$ for $i>0$, and therefore $a_i$ are constants not depending on $x$!
EXCEPT for $a_0$. So we get $f(x,y) = a_0(x) + a_1y+a_2y^2+...+a_ny^n = a_0(x) + y(a_1+2a_2+...+a_ny^{n-1})$ is of the form $h(x) + yg(y)$ for one-variable polynomials $h$ and $g$.

NOTE : Just caught me off guard this one, but if people are wondering why the answer looks "asymmetric" in $x$ and $y$ (every polynomial of the form $h(x)+y(g(y))$, why is there a $y$ multiplied with $g(y)$ while nothing is multiplied with $h(x)$?) while we know that the mixed partial is symmetric in $x$ and $y$, then the answer is that the aforementioned polynomial class is symmetric in $x$ and $y$.
That is because a function of the form $h(x)+yg(y)$ is actually of the form $C + xh'(x) + yg(y)$ where $C$ is a constant. By moving the $C$ to the $yg(y)$ term, we just get $xh'(x) + g'(y)$ where $h',g'$ are some polynomials. So, if we were to make this set look symmetric in both $x$ and $y$, we could write it as $$
f(x,y) = C + xh(x) + yg(y) 
$$
where $C\in R$ and $h,g$ are any one-variable polynomials.
A: What if like this?
$\frac{\partial^2(f)}{\partial x \partial y} = 0$
$\frac{\partial(f)}{\partial x} = \int 0 dy$
Since $f \in R[x,y]$, then
$\frac{\partial(f)}{\partial x} = a_0 + a_1x + a_2x^2 + \dots + a_{n-1}x^{n-1} + a_nx^n + C$
$f(x,y) = \int a_1x + a_2x^2 + \dots a_{n-1}x^{n-1} + a_nx^n dx$
$f(x,y) = \frac{a_1}{2}x^2 + \frac{a_2}{3}x^3 + \dots + \frac{a_{n-1}}{n}x^n + \frac{a_n}{n+1}x^{n+1} + b_1y + b_2y^2 + \dots + b_{n-1}y^{n-2} + b_ny^n + C$
For $a_i,b_j \in \mathbb{R}, i=0,1,2,\dots,n$ and a constant $C$.
