Help understanding the map $\varphi : \mathbb{C}[x, y] \rightarrow \mathbb{C}[x] \times \mathbb{C}[y] \times \mathbb{C}[t]$

Question

This map is defined by $f(x, y)\rightsquigarrow (f(x, 0), f(0, y), f(t, t)$

I understand this map to mean that, for example, if I have the polynomial $f(x, y) = x^2 + xy + y^2$, then $\varphi(f) = (x^2, y^2, 3t^2)$

So, when talking about the image of this map, I'm struggling to understand how to describe it.

Also, as far as the kernel of $\varphi$ goes, I know this means we need to describe all the elements where $f(x, 0) = 0, f(0, y) = 0, f(t, t) = 0$. I originally thought "well, polynomials that look like $xy$ certainly evaluate to $0$ when $x = 0$ or $y = 0$, so that's a good candidate, however we still need to deal with $f(t, t)$.

So, I'm looking for some direction on these two questions - the image, and the kernel generators.

Answer 1

An element $(f_1(x),f_2(y),f_3(t))$ of the image must satisfy $f_1(0)=f_2(0)=f_3(0)$. Also it must satisfy $f_3'(0)=f_1'(0)+f_2'(0)$. As far as I can see these are the only constraints on the image.

Answer 2

As user Lord Shark the Unknown pointed out, the elements $(f_1(x),f_2(y),f_3(t))$ in the image must satisfy $f_1(0)=f_2(0)=f_3(0)$ and $f'_3(0) = f'_1(0) + f'_2(0)$. I'll show that these two conditions are sufficient for an arbitrary element to lie in the range.

So let $f_1,f_2,f_3$ satisfy the above; we aim to find $f$ for which $f(x,0) = f_1(x)$, $f(0,y) = f_2(y)$, and $f(t,t) = f_3(t)$. WLOG assume $f_1(0) = f_2(0) = f_3(0) = 0$ (if not, we can subtract to get them to all equal $0$, in which case we just adjust the final $f$ accordingly). Let's try to find an answer of the form of $f(x,y) = f_1(x) + f_2(y) + g(x,y)$, where $g(x,0) = g(0,y) = 0$. Notice for such $g$ that $f(x,0) = f_1(x)$ and $f(0,y) = f_2(y)$, so the only other constraint on $g$ we want is for $$ f_3(t) = f(t,t) = f_1(t) + f_2(t) + g(t,t),\quad\text{i.e.}\quad g(t,t) = f_3(t)-f_1(t)-f_2(t). $$ By assumption, we have $f_3(0)-f_1(0)-f_2(0) = 0$ and $f'_3(0)-f'_1(0)-f'_2(0) = 0$, so we can write $f_3(t) - f_1(t) - f_2(t) = t^2h(t)$ for some polynomial $h$. Letting $g(x,y) = xyh(x)$ then does the job.