Consider the map ψ:C[x, y]→C[t] defined by f(x, y)⇝f(t^2, t^3). Prove that its image is the set of polynomials p(t) such that dp/dt(0)=0

Question

So it is as the title says written here again for ease of reading:

Consider the map $ψ:C\left[x, y\right]\to C\left[t\right]$ defined by $f(x, y) \mapsto f(t^2, t^3)$. Prove that its image is the set of polynomials $p(t)$ such that $\frac {dp}{dt}(0)=0$.

What I have done so far:

$$\frac {dp}{dt}(0)=0\; \therefore\; p(t)=a_0+a_1t+a_2t^2+...+a_nt^n,\; \text {where }a_1=0.$$

Where I get stuck: if $p(t)=\frac yx\rightarrow \frac {t^3}{t^2}=t\;\therefore\; dp/dt=1$, so the image doesn't have the derivative $0$ at $t=0$ in this case. Therefore the image can't be the set of polynomials with derivative $0$ at $t=0$.

I think my problem is that I don't know what "$\to$" implies outside of converting each $x$ to $t^2$ and each y to $t^3$.

Answer 1

The problem is equivalent to the following one:

The image of the map of polynomial rings $\Bbb C[x,y]\to\Bbb C[t]$ determined by $x\to t^2$, $y\to t^3$ is the set / vectorial $\Bbb C$-subspace $V$ of polynomials of the shape $a_0+a_2t^2+\dots+a_nt^n$ which have a zero coefficient for $t^1$.

Proof.

(Double inclusion.)

First, the image is in $V$ because every monomial $x^ky^n$ is mapped to a monomial $t^{2k+3n}$ in $V$, since $2k+3n=1$ has no solution in natural numbers. (For a potential solution of $2k+3n\le 1$ we must have $k,n\le 1$, thus $k=n=0$, but then we get $t^0$.)

Then we note that any $t^m$ with $m\ge 0$, $m\ne 1$, can be written as an image of a monomial $x^ky^n$, because (for any fixed $m$ as above) the equation $m=2k+3n$ has solutions in natural numbers $k,n$. Indeed, for the first values we have: $$ \begin{aligned} 0 &= 2\cdot 0+3\cdot 0\ ,\\ 2 &= 2\cdot 1+3\cdot 0\ ,\\ 3 &= 2\cdot 0+3\cdot 1\ ,\\ 4 &= 2\cdot 2+3\cdot 0\ ,\\ 5 &= 2\cdot 1+3\cdot 1\ ,\\ 6 &= \dots \qquad\text{(solution obtained from the $0$ solution)}\\ 7 &= 2\cdot 2+3\cdot 1\ , \end{aligned} $$ and any further $m$ can be easily obtained by induction (with step $6$) from the $(m-6)$-solution. This shows the converse inclusion.