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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.

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

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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.

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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 $C$.

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