Neat way to find the kernel of a ring homomorphism

This question is more or less a duplicate of Show the set of points $(t^3, t^4, t^5)$ is closed in $\mathbb A^{3}$.

If we have a surjective ring homomorphism $A\stackrel{f}\to B$, where $A,B$ are noetherian rings, and $I\subset A$ an ideal, then this gives rise to an exact sequence of $A$-modules $$0\to\ker f\to A\to B\to0$$ and by tensorizing with $\hat A$ (the completion of $A$ in the $I$-adic topology) we get $$0\to\ker f\otimes_A\hat A\to A\otimes_A\hat A\to B\otimes_A\hat A\to0.$$ This leads to the exact sequence $$0\to (\ker f)\hat A\to\hat A\stackrel{\hat f}\to\hat B\to0,$$ so $\ker\hat f=(\ker f)\hat A$. (The completion of $B$ is taken in the $f(I)$-adic topology.)

Now use the result for polynomial rings which I've mentioned above and extend it to the corresponding power series rings (which are the completions of the polynomials rings involved in our question).

$\require{AMScd}$

With $A=k[[x,y,z]]$, $B=k[[t]]$ and $I = (x^3-yz, y^2-xz, z^2 -x^2 y)$ and $\bar{A} = A/I$ consider the diagram

$$ \begin{CD} \bar{A}_y @>{f}>> B_{t^4}\\ @AAiA @AAA \\ \bar{A} @>{}>> B \end{CD} $$

The map $f$ has a left-inverse $g$ given by $g(t) = z/y$. One can calculate

$x \to t^3 \to z^3/y^3 = x$ because $z^3 - x y^3 \in I$

$y \to t^4 \to z^4/y^4 = y$ because $z^4 - y^5 \in I$

$z \to t^5 \to z^5/y^5 = z$ because $z^5 - y^5 z \in I$

(I checked the $\in I$ with the ideal $J \subseteq R$ having the same generators as $I$, but in $R=k[x,y,z]$)

Now it remains to prove that $i$ is injective which follows from $(I:y^s) = I$. So we need only prove $(I:y) = I$. Now give $x,y,z$ the weights $3,4,5$, so $I$ is homogenous. Then $y h \in I$ implies $y h_d \in I$ for all weighted homogeneous parts $h_d$ of $h$. But this, I think, is equivalent to $y h_d \in J$ with the ideal $J$ introduced above. As $J$ is prime in $R$, we have $h_d \in J$, therefore $h_d \in I$. Going from the $h_d$ with lowest $d$ upward proves $h \in I$.