Hartshorne Exercise I 5.14(c) I have some problems on this exercise. Since (c) uses (b), I give the statement of these two as follows:

(b)Generalize the example in the text (5.6.3) to show that if $f=f_{r}+f_{r+1}+\ldots \in$ $k[[x, y]]$, and if the leading form $f_{r}$ of $f$ factors as $f_{r}=g_{s} h_{t}$, where $g_{s}, h_{t}$ are homogeneous of degrees $s$ and $t$ respectively, and have no common linear factor, then there are formal power series
$$
\begin{aligned}
g &=g_{s}+g_{s+1}+\ldots \\
h &=h_{t}+h_{t+1}+\ldots
\end{aligned}
$$
in $k[[x, y]]$ such that $f=g h$.
(c)Let $Y$ be defined by the equation $f(x, y)=0$ in $\mathbf{A}^{2}$, and let $P=(0,0)$ be a point of multiplicity $r$ on $Y$, so that when $f$ is expanded as a polynomial in $x$ and $y$, we have $f=f_{r}+$ higher terms. We say that $P$ is an ordinary $r$-fold point if $f_{r}$ is a product of $r$ distinct linear factors. Show that any two ordinary double points are analytically isomorphic. Ditto for ordinary triple points. But show that there is a one-parameter family of mutually nonisomorphic ordinary 4-fold points.

Now, for the ordinary triple point, I can use $b$ to factor $f$ into $f_1f_2f_3$ where $\{f_i\}$ have pairwise linearly independent linear terms $\{l_i\}$. I want to show $k[[x,y]]/f\simeq k[[x,y]]/(xy(x+y))$(hence its ring structure doesn't depend on the choice of $f$).
I learned the automorphism group of $k[[x,y]]$ from this. Suppose $l_3=al_1+bl_2$, I can easily get an automorphism $\phi$ which send $x$ to $af_1$ and $y$ to $bf_2$, but then $\phi(x+y)=af_1+bf_2$ which may not be equal to $f_3$. Could you provide a way to repair my method or give a correct answer?
Besides I don't know how to give an example of two nonisomorphic 4-fold points. Could you give a hint for me?
Any help is appreciated, thanks!
 A: We'll need one argument from the solution to part (b):
Lemma. Let $P_d$ denote the vector space of homogeneous polynomials in two variables of degree $d$.
The map $P_{t+n}\times P_{s+n} \to P_{s+t+n}$ given by $(a,b)\mapsto ag_s+bh_t$ is surjective assuming $g_s$ and $h_t$ are relatively prime.
Proof. I claim it is enough to prove surjectivity for $n=0$.
Any standard basis monomial $M$ in $P_{s+t+n}$ can be written as $x^iy^j$ times some standard basis monomial $M'$ in $P_{s+t}$.
If we can find $p,q$ so that $pg_s+qh_t=M'$, then $(x^iy^jp)g_s+(x^iy^jq)h_t=M$, and this shows that every standard basis monomial in $P_{s+t+n}$ is in the image of our map, so it is surjective.
To prove surjectivity when $n=0$, consider the matrix of our map in the standard monomial basis. This is exactly the Sylvester matrix associated to the homogeneous resultant of $g_s$ and $h_t$.
But the homogeneous resultant of $g_s$ and $h_t$ is nonzero iff they are coprime, so we are done. $\blacksquare$

I'll show the argument for the ordinary triple point case.
Suppose $f_3=L_1L_2L_3$ for mutually independent linear terms: we get that $f=(L_1+\cdots)(L_2+\cdots)(L_3+\cdots)$ factors by (b), and the automorphism given by $x\mapsto L_1+\cdots,y\mapsto L_2+\cdots$ gives that $k[[x,y]]/(f)\cong k[[x,y]]/(xy(ax+by)+f')$ where $f'$ is a power series with no terms of order less than four and $a,b\neq 0$.
Up to the automorphism sending $x\mapsto bx$ and $y\mapsto ay$, we may assume that $a=b=1$, and now our goal is to eliminate $p$.
Finding an automorphism of $k[[x,y]]$ which does this is equivalent to solving a collection of linear systems in the same fashion as in (b): the degree-$n$ portion after substituting in $x\mapsto x+\sum_{r>1} p_r(x,y)$ and $y\mapsto y+\sum_{r>1} q_r(x,y)$ where $p_r,q_r$ are homogeneous of degree $r$ can be written as a linear combination of products of $p_i$ and $q_j$ for $i,j<n$ plus $p_n(y^2+2xy)+q_n(x^2+2xy)$.
Let $P_n$ be the vector space of homogeneous polynomials of degree $n$ in the variables $x,y$.
The map from $P_n\times P_n\to P_{n+2}$ which gives the contribution of the terms $p_n,q_n$ to the degree-$n+2$ homogeneous part of our power series after substitution is $(p_n,q_n)\mapsto p_n(y^2+2xy)+q_n(x^2+2xy)$ which by the lemma can be seen to be surjective because $y^2+2xy=y(y+2x)$ and $x^2+2xy=x(x+2y)$ are coprime.
So we can always solve for $p_n,q_n$ to eliminate the higher-order terms, and any ordinary triple point is analytically isomorphic to the one given by $xy(x+y)$.
This will generalize to handle the case of ordinary quadruple points as well, in that you can always eliminate the higher-order terms. After that, you'll need to think about how linear automorphisms act on polynomials which are the product of four distinct linear terms in two variables - I'll leave that to you for now, but give you a hint that thinking geometrically here can be helpful.
