Elementary method for finding $I(Y)$ for the curve $Y$ defined parametrically by $x=t^{3}$, $y=t^{4}$, $z=t^{5}$ In order to motivate some of the theory we will be learning in a computational commutative algebra course, my professor assigned a number of computational problems that are [seemingly] quite difficult without more advanced methods. The problem I chose to address was the following:

Let $k$ be a field, let $t,X,Y,Z$ be variables over $k$, and let $\phi:k[X,Y,Z]\to k[t]$ be  the ring homomorphism uniquely determined by $\phi|_k = id$, $\phi(X) = t^3$, $\phi(Y) =  t^4$, and $\phi(Z) = t^5$. Find the generators of the kernel of $\phi$.

I made some progress (the assignment has already been collected), but most of my conclusions were not rigorously justified. I've been wondering what an elementary method for this type of problem looks like. By elementary, I mean no Grobner basis, no Buchberger's algorithm, etc. Thanks so much for your help!
Edit:  I ended up with $\ker\phi = (XZ-Y^2, X^3-YZ, X^2Y-Z^2)$. Might be totally wrong -- mostly based on heuristic observations. 
 A: Here I will try to give an elementary proof of why you have found all the generators.
Let $P_n$ be the set of polynomials whose monomials map to a multiple (in $k$) of $t^n$, i.e. each monomial is of the form $cX^aY^bZ^c$ with $c\in k, 3a+4b+5c=n$ and let $K_n$ be the intersection with the kernel. We wish to show that $$K_n = XK_{n-3} + YK_{n-4} + ZK_{n-5}.$$ What's clear already is that $P_n = XP_{n-3} + YP_{n-4}+ZP_{n-5}$ since every term of a polynomial in $P_n$ either has an $X$, $Y$ or $Z$ as a factor.
Example for kernel: For $X^4 - Y^3$ in $K_{12}$, $X^4 - Y^3 = X(X^3-YZ) + Y(XZ-Y^2)$ where $X^3 - YZ \in K_{9}$ and $XZ-Y^2 \in K_8$.
Let us first show this for every "basic (two-term)" polynomial in $K_n$. Suppose we have $Xp + Yq \in K_n$, so that $p \in P_{n-3}$ and $q \in P_{n-4}$. The goal is to add $Xp' + Yq' = 0$ so that $p+p' \in K_{n-3}$ and $q+q' \in K_{n-4}$. Analogously to the $X^4 - Y^3$ example above, first find $a$ where $\phi(Xp) = at^n$ and $b$ where $\phi(Yq) = bt^n$ so $a+b=0$. 
Given $s \in P_{n-7}$ such that $\phi(s) = t^{n-7}$, form $p' = bYs$ and $q' = aXs$ so that $Xp' + Yq' = 0$, but now $p+p' \in  K_{n-3}$ and $q+q' \in K_{n-4}$ by construction. This implies $Xp+Yq \in XK_{n-3}+YK_{n-4}$.
Now in general, suppose $Xp+Yq+Zr \in K_n$. Let $\phi(Xp) = at^n, \phi(Yq) = bt^n$ and $\phi(Zr) = ct^n$. Want $Xp'+Yq' + Zr' = 0$ with $p+p',q+q',r+r'$ in the kernel.
Take $v \in P_{n-12}$ with $\phi(v) = t^{n-12}$, and let
$p' = (b+c)YZv$, $q' = (a+c)XZv$ and $r' = (a+b)XYv$.
Now $Xp'+Yq'+Zr' = 0$, and $p+p',q+q',r+r'$ are in the kernel, and
$Xp+Yq+Zr = X(p+p')+Y(q+q')+Z(r+r')$, so that $K_n  \in XK_{n-3} + YK_{n-4}+ZK_{n-5}$. This argument works for $n\geq 12$ (to obtain $v$).
Since we started with generators in $K_{8},K_{9},K_{10}$, all that's left is to show that $K_{11}$ is generated, and induction handles all cases $K_n$ for $n\geq 12$.
One thing left out: every element of the kernel can be expressed as arbitrary linear combinations of elements from $K_n$ by linear algebra considerations ($\{t^n, n\geq 0\}$  are linearly independent in $k[t]$).
