A problem from Artin This is a problem from old artin (11.3.10) or new artin (12.3.5):
Consider the map $\varphi : \mathbb{C} [x,y] \rightarrow \mathbb{C} [t]$ defined by $f(x,y)\mapsto f(t^2-t,t^3-t^2)$. Prove that $\ker \varphi$ is a principal ideal and find a generator for this ideal. Prove that the image of $\varphi$ is the set of polynomials $p(t)$ such that $p(0)=p(1)$. Give an intuitive explanation in terms of the geometry of the variety $\{f=0\}$ in $\mathbb{C}^2$.
I dont know how to do the kernel part, for the image part, it is easy to show that image of $f(x,y)$ under $\varphi$ satisfies the condition but i couldn't show the converse.
 A: First, assume that a polynomial $p(t) \in \Bbb C[t]$ satisfies $p(0) = p(1) = 0$. Then there is a polynomial $q(t)\in \Bbb C[t]$ such that $p(t) = t(t-1)q(t)$. Expanding $q(t) = c_1 + c_2t^1 + \cdots c_nt^{n-1}$, we get that
$$
p(t) = \sum_{i = 1}^n c_n t^n(t-1)
$$
In general, the term $t^i(t-1)^j$ is the image of a monomial in $x$ and $y$ iff $j \leq i \leq 2j$ (use enough $y$-terms to get $(i-j)$ right, then use enough $x$ terms to get $i$ right, and you're there). In our case, we will never have any trouble fulfilling $j \leq i$. However, terms such as $t^6(t-1)$ does not fulfill $i \leq 2j$ ("the condition"). I will therefore focus on a general term where the condition is violated and see if it can be rewritten in some way.
We have
$$
t^i(t-1)^j = t^{i-1}(t-1+1)(t-1)^j = t^{i-1}(t-1)^{j+1} + t^{i-1}(t-1)^{j}
$$
where both terms of those we ended up with are closer than the original to fulfilling the condition. Therefore, each term in the new expansion is the image of some monomial in $x$ and $y$, and therefore the whole of $p(t)$ is in the image.
Now, what if $p(0) = p(1) \neq 0$? In that case, apply the above to $p(t) - p(0)$ to get some polynomial $f(x, y)$ such that $\phi(f(x, y)) = p(t) - p(0)$. Then $\phi(f(x, y) + p(0)) = p(t)$.
