# 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.

• Kernel is ∫f(y) dy + g(x), f is some function in y and g is some function in x, – Subhajit Jun 30 at 3:55
• @Subhajit How? Is the homomorphism under multiplication? – user795084 Jun 30 at 3:58
• No, I think this is under addition, so, 0 is the identity! – Subhajit Jun 30 at 4:01
• @Subhajit Hmmm... me too. So, $Ker (k) = \lbrace f \in R[x,y] \mid k(f) = 0 \rbrace$. Then, what's next? – user795084 Jun 30 at 4:03
• Because,by multiplication operation,,,,the map can't be a homomorphism – Subhajit Jun 30 at 4:03

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 $$g$$.

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.

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 $$C$$.