Define $φ:\mathbb{C}[x,y]\to\mathbb{C}[x]\times\mathbb{C}[y]\times\mathbb{C}[t]$

Question

I am trying to solve Chapter 11 problem M.5 from Artin's Algebra textbook.

Define $φ:\mathbb{C}[x,y]\to\mathbb{C}[x]\times\mathbb{C}[y]\times\mathbb{C}[t]$ by $φ(f(x,y))=(f(x,0), f(0,y), f(t,t))$.

Determine img($φ$) and find generators for ker($φ$).

I have shown that, if $\sum a_{ij}x^iy^j$ is in ker($φ$), then $a_{00}=a_{10}=a_{20}=...=0$ and $a_{00}=a_{01}=a_{02}=...=0$ and $a_{11}=a_{21}+a_{12}=a_{31}+a_{22}+a_{13}=...=0$.

Now, I see that $(x^2y-xy^2, x^3y-xy^3, x^4y-xy^4,...)\subseteq ker(φ)$. But I am unable to prove $\supseteq$. I suspect that $\supseteq$ does not hold, but am stumped on what to do next. I am guessing that the equation $a_{11}=a_{21}+a_{12}=a_{31}+a_{22}+a_{13}=...=0$ will help.

Also, I do not even know where to start on trying to compute img(φ), other than plugging in stuff. A hint on this would be appreciated.

EDIT 1

It seems to me now that $ker(φ)=(xy^2-x^2y)$. The $\supseteq$ direction is obvious. Now let $k(x,y)=\sum a_{ij}x^iy^j \in ker(φ)$. Then $a_{00}=a_{10}=a_{20}=...=0$ and $a_{00}=a_{01}=a_{02}=...=0$ and $a_{11}=a_{21}+a_{12}=a_{31}+a_{22}+a_{13}=...=0$.

It is easy to show that $k(x,y)=xyq(x,y)$ for some $q\in \mathbb{C}[x,y]$. Say $q(x,y)=\sum b_{ij}x^iy^j$.

Since $k(t,t)=0, q(t,t)=0$. We obtain $b_{00}=b_{10}+b_{01}=b_{20}+b_{11}+b_{02}=...=0$.

I want to show that $q(x,y)$ is a multiple of $y-x$. If I can show that $b_{r0}x^r+b_{r-1,1}x^{r-1}y+...+b_{0r}y^r$ is a multiple of $y-x$ (for arbitrary r), then I should be done. I have done this for r=0, r=1, and r=2. But how do I do it for arbitrary r?

EDIT 2 I have proven that the kernel is $(xy^2-x^2y )$ thanks to Daniel and the division algorithm. I am now proceeding to prove that the image is what he says it is.

EDIT 3 I have finished.

Answer 1

For the question of finding the kernel: the kernel of $\varphi$ is the intersection of the kernels of $\mathbb{C}[x, y] \to \mathbb{C}[x]$, $f \mapsto f(x, 0)$, and so on for the other components. But the kernels are respectively $\langle y \rangle$, $\langle x \rangle$, and $\langle x - y \rangle$. Also, $\mathbb{C}[x, y]$ is a UFD, and the generators $y, x, x-y$ are pairwise relatively prime (being irreducible and not differing by a factor of a unit), which will tell you what the intersection of the three ideals is. (This is essentially just an expansion of the hints in Lubin's answer.)

For finding the image, first note that for $n \ge 2$, we have $$ x y^{n-1} \mapsto (0, 0, t^n); \\ (x-y) x^{n-1} \mapsto (x^n, 0, 0); \\ (y-x) y^{n-1} \mapsto (0, y^n, 0). $$ Therefore, modulo these elements, the problem of finding the image reduces to finding what combinations $(a + bx, c + dy, e + ft)$ are in the image. Also, note that if $i + j \ge 2$, then $\varphi(x^i y^j)$ is a linear combination of the elements above (if $i \ge 1$ and $j \ge 1$, then the image is $(0, 0, t^{i+j})$, whereas $\varphi(x^i) = (x^i, 0, 0) + (0, 0, t^i)$ for $i \ge 2$ and $\varphi(y^j) = (0, y^j, 0) + (0, 0, t^j)$ for $j \ge 2$). Therefore, calculating the images of $1, x, y$ will suffice.

Putting all this together, you should get that $\operatorname{im}(\varphi)$ is:

$$ \{ (a_0 + a_1 x + \cdots + a_n x^n, b_0 + b_1 y + \cdots + b_m y^m, c_0 + c_1 t + \cdots + c_\ell t^\ell) \in \mathbb{C}[x] \times \mathbb{C}[y] \times \mathbb{C}[t] \mid \\ a_0 = b_0 = c_0, ~ a_1 + b_1 = c_1 \}. $$

Answer 2

Hint 1:
The kernel is an ideal of $\Bbb C[x,y]$, which is a noetherian ring, so can be written with finitely many generators. Certainly, something goes to zero in the cross product of the three rings if and only if it goes to zero in each. Aren’t the three separate kernels $(x)$, $(y)$, and $(y-x)$? It looks to me as if the kernel is generated by a single element.

Hint 2:
Describing the image looks tricky. I’d write down what the image of any monomial $ax^my^n$ looks like and try to get an efficient description of the situation, covering all cases.