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  pdftitle={A counterexample to the Eisenbud--Mazur conjecture},
  pdfauthor={Chung Pang Mok, Yiqi Xu}}
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\title[A counterexample to the Eisenbud--Mazur conjecture]
{A counterexample to the Eisenbud--Mazur conjecture}



\author{Chung Pang Mok}
\address{Shanghai Institute for Mathematics and Interdisciplinary Sciences (SIMIS),
	Shanghai 200433, China}

\email{cpmok@simis.cn}

\author{Yiqi Xu}
\address{Fakult{\"a}t 8 -- Mathematik und Physik,
	Universit{\"a}t Stuttgart,
	Stuttgart, Germany}
\email{xuyiqi1@gmail.com}




% Authors are listed alphabetically by surname.
\date{October 19, 2026}
\subjclass[2020]{Primary 13A15; Secondary 13H05, 13N05}
\keywords{Symbolic powers, Eisenbud--Mazur conjecture, formal curves,
  Veronese surface, critical values}

\begin{document}
\begin{abstract}
We construct a height-five prime ideal $P$ in the six-variable formal
power series ring $R$ over $\Q$ such that
$P^{(2)}\not\subseteq\m P$, thus providing a prime-ideal counterexample to the Eisenbud--Mazur conjecture (with $\Q$ being the coefficient field). The ideal is the kernel of an explicit quadratic parametrization over
a number field of degree $28$ over $\Q$. A witness has order three, with leading
term the determinant of a symmetric $3\times3$ matrix. We construct
the witness by eliminating critical values to all orders and then
straightening the resulting formal curve. Interpolation at the $28$
nodes of an arrangement of eight lines supplies the required lifting
maps. A rational linear functional on third and fourth jets certifies
nonmembership in $\m P$. The construction also yields an unmixed radical ideal
with $28$ smooth branches over $\C$, and provides a counterexample to the more general form of the Eisenbud--Mazur conjecture stated by More (with $\C$ being the coefficient field). Finally, the prime ideal counterexample
descends to the localized polynomial ring $\Q[x_0,\ldots,x_5]_{(x_0,\ldots,x_5)}$,
where a polynomial witness with the same fourth jet exists.
All finite identities used in the proof admit exact verification.
\end{abstract}
\maketitle

\section{Introduction}
\label{sec:introduction}

The Eisenbud--Mazur conjecture asks whether
\begin{equation}
  P^{(2)}\subseteq \m P
  \label{eq:EM}
\end{equation}
for every prime ideal $P$ in a regular local ring $(R,\m)$ containing
a field of characteristic zero. Here $P^{(2)}=P^2R_P\cap R$ is the symbolic square of $P$.
Equivalently, it predicts that no element of the symbolic square can
be a minimal generator of $P$. Eisenbud and Mazur related
this containment to the existence of nontrivial evolutions of reduced
local algebras \cite{EM97}. The formulation for arbitrary regular local
rings containing a characteristic-zero field, and more generally for
unmixed radical ideals, appears in \cite[Conjecture~1.5]{Mor15}.

We construct a counterexample to the original Eisenbud--Mazur conjecture in the formal power series ring in six variables over $\Q$.
The quotient by $P$ is a one-dimensional complete local domain, and its
normalization has a residue-field extension of $\Q$ of degree $28$.
After extending coefficients to $\C$, the construction gives a counterexample to \cite[Conjecture~1.5]{Mor15} with $28$ smooth branches. We also
give a counterexample to the original Eisenbud--Mazur conjecture in a localized polynomial ring over $\Q$;
this is established in
Theorem~\ref{thm:algebraic}.

There are substantial positive results for special classes of ideals.
Eisenbud and Mazur prove the containment for generically complete
intersection licci ideals, including perfect ideals of height two and
Gorenstein ideals of height three, and for generically complete
intersection almost complete intersections \cite{EM97}. Huneke and Ribbe obtain further positive
results using residual intersections, integral closure, and
differentials \cite{HR98}. More proves the conjecture for prime ideals
obtained by contracting a parametrized smooth curve along the extension
$k[[t^2,x_1,\ldots,x_m]]\subset k[[t,x_1,\ldots,x_m]]$
\cite[Theorem~2.1]{Mor15}. A related uniform bound is
$P^{(h+1)}\subseteq\m P$, where $h$ is the height of $P$:
Hochster and Huneke prove this for regular local rings containing a field
\cite[Theorems~3.5 and~4.2]{HH07}, and Murayama proves a generalization
for arbitrary regular local rings \cite[Theorem~C]{Mur23}.
For the height-five prime ideal constructed below, this bound concerns the
sixth symbolic power and is compatible with the failure of
\eqref{eq:EM}.

Two features of the problem account for both these positive results and
the difficulty of finding a counterexample. First, in characteristic zero
the symbolic square of a radical ideal has a differential description:
$f\in P^{(2)}$ if and only if $f$ and all its partial derivatives lie in
$P$ \cite[Theorem~3]{EM97}. So \eqref{eq:EM} is a statement about
critical loci: a witness is a hypersurface $f=0$ that is singular along
$V(P)$ and is nevertheless a minimal generator of $P$. Second, if $P$ is
stable under a derivation $\sum w_ix_i\partial_i$ with positive integer
weights, the Euler identity $\sum w_ix_i\partial_if=(\deg f)\,f$ places
every such $f$ in $\m P$ \cite[Proposition~10]{EM97}. A counterexample
must therefore be non-quasi-homogeneous in every system of coordinates.
The hypothesis on the characteristic is essential as well: in
characteristic $p$ the differential description breaks down, and
\cite{EM97} exhibits monomial curves in characteristics $2$ and $3$ that
violate \eqref{eq:EM}.

We set
\[
 R=\Q[[x_0,x_1,x_2,x_3,x_4,x_5]],
 \qquad \m=(x_0,x_1,x_2,x_3,x_4,x_5).
\]
The first two nonzero homogeneous terms of our witness are the cubic
\begin{equation}
\begin{split}
 D&=\det\begin{pmatrix}
 x_0&x_1&x_2\\
 x_1&x_3&x_4\\
 x_2&x_4&x_5
 \end{pmatrix}\\
  &=x_0x_3x_5-x_0x_4^2-x_1^2x_5
       +2x_1x_2x_4-x_2^2x_3
\end{split}
\label{eq:D}
\end{equation}
and the quartic
\begin{equation}
\begin{split}
 W={}&x_0^4+x_0^3x_1-x_0^3x_2-2x_0^2x_3^2
       +16x_0x_1x_3x_5-7x_0x_2x_3^2\\
    &+14x_0x_3x_4x_5-20x_0x_3x_5^2
       +8x_1x_5^3-7x_2x_3x_5^2+x_3^4+x_4x_5^3-x_5^4.
\end{split}
\label{eq:W}
\end{equation}
The significance of the quartic $W$ is explained after Theorem~\ref{thm:main}; see also Sections~\ref{sec:curve} and~\ref{sec:critical}.


\begin{theorem}
\label{thm:main}
There is an explicitly defined continuous $\Q$-algebra homomorphism
\[
 \varphi:R\longrightarrow L[[t]],\qquad
 \varphi(x_0)=t,\qquad
 \varphi(x_i)=t\ell_i+t^2w_i\quad(1\leq i\leq5),
\]
where $L$ is a number field of degree $28$ over $\Q$, with $\ell_i,w_i \in L$ for $1 \leq i\leq 5$, such that
$P=\ker\varphi$ is a height-five prime ideal and there exists $f\in R$
satisfying
\begin{equation}
 f\equiv D+W\pmod{\m^5},\qquad
 f\in P^{(2)},\qquad f\notin\m P.
 \label{eq:main-witness}
\end{equation}
The normalization of $R/P$ is $L[[t]]$, whereas the residue field of
$R/P$ is $\Q$.
\end{theorem}

The field $L$ and all coefficients of $\varphi$ are specified by rational
polynomial formulas below. The witness $f$ in Theorem~\ref{thm:main} is a
complete formal power series, given in equation (\ref{eq:formal-witness}). 

The construction combines a geometric lifting argument with a finite
obstruction to membership in $\m P$. The projectivized singular locus
of $D$ is the quadratic Veronese surface: the partial derivatives of $D$
are the $2\times2$ minors, which generate the prime ideal $Q$ of the cone
over this surface, so that $D\in Q^{(2)}$. The Laplace expansion,
however, writes $D$ as a combination of the minors with coefficients in
$\m$, so $D\in\m Q$ and $Q$ itself is no counterexample. We therefore
replace $Q$ by the ideal of a formal arc whose tangent line lies on the
cone but which leaves the cone at second order. Such an ideal no longer
contains all the minors, so the Laplace expansion no longer applies, and
an element of order three with leading form $D$ can be a minimal
generator; what has to be arranged is that its gradient vanishes along
the arc, and the quartic $W$ supplies the critical point that makes this
possible. On its affine chart
\[
 (x_0,x_1,x_2,x_3,x_4,x_5)=(1,U,V,U^2,UV,V^2),
\]
the restriction of $W$ is the product of the eight lines
$U+\alpha V+\alpha^2=0$, where $\alpha$ runs through the roots of
$z^8-z-1$. Their $28$ nodes form a single Galois orbit. Evaluation at
one node identifies the polynomials in $U,V$ of degree at most six with
$L$. Consequently, evaluation at the corresponding point of the
Veronese surface maps homogeneous cubics in the six variables onto
$L$, and the same holds in every higher degree. 

The interpolation has two uses. First, the cubic determinant has a
nondegenerate quadratic form in the directions normal to the Veronese
surface, and the restricted quartic has a nondegenerate critical point
at each node. Two successive applications of the formal implicit
function theorem therefore produce a critical arc. A quintic term
removes the first obstruction to its critical value being zero; the
surjectivity of evaluation then removes all subsequent obstructions,
one homogeneous degree at a time. Second, the same interpolation gives
$\varphi(\m^3)=t^3L[[t]]$. This permits a formal coordinate change with
identity two-jet that replaces the critical arc by the prescribed
quadratic parametrization while preserving $D+W$ modulo $\m^5$. The
two-term shape of the arc and of the witness is what defeats the Euler
argument: no positive grading makes both homogeneous.

The obstruction to $f\in\m P$ can be checked using only the fourth jet. We determine the
image of $P$ modulo $\m^4$ and display a rational linear functional
supported in degrees three and four after quotienting by $x_0$.
It annihilates the image of $\m P$ and takes the value $1$ on $D+W$.
Note that the proof of symbolic-square membership uses the full formal series,
whereas the nonmembership argument depends only on its fourth jet.
The exact computations needed to verify the finite identities are recorded in the appendix.

Finally, Proposition~\ref{prop:complex} describes the extension of $P$
to $\C[[x_0,\ldots,x_5]]$ as the intersection of $28$ distinct smooth
branch primes. This also gives a counterexample to the
extended form of the Eisenbud--Mazur conjecture considered in \cite[Conjecture~1.5]{Mor15}.

\bigskip

\noindent {\bf Declaration of generative AI:} The counterexample was discovered by a ChatGPT model. The authors reviewed and edited the content as needed and take full responsibility for the content of the published article.

\section{The coefficient field \texorpdfstring{$L$}{L}}\label{sec:field}

Throughout the paper, orders and homogeneous components refer to
ordinary total degree in $x_0,\ldots,x_5$. We write
$H_d=\Q[x_0,\ldots,x_5]_d$ for the space of homogeneous forms of
degree $d$. Let
\[
h(z)=z^8-z-1,
\]
choose two distinct roots $\beta,\gamma$ in an algebraic closure of
$\Q$, and put
\begin{equation}\label{eq:field}
u=\beta\gamma,\qquad v=-(\beta+\gamma),\qquad L=\Q(u,v).
\end{equation}

\begin{lemma}\label{lem:field}
The extension $L/\Q$ has degree $28$. Its embeddings into a splitting
field of $h$ are indexed by the unordered pairs of distinct roots of $h$.
\end{lemma}
\begin{proof}
The polynomial $h$ is irreducible modulo $7$. Modulo $19$ it factors as
\begin{equation}\label{eq:mod19}
h(z)=(z-4)
\bigl(z^7+4z^6-3z^5+7z^4+9z^3-2z^2-8z+5\bigr),
\end{equation}
where the degree-seven factor is irreducible. Both reductions are
squarefree; exact certificates are given in Appendix~\ref{app:finite}.
Consequently the Galois group of $h$ is transitive on its eight roots
and contains a $7$-cycle. The stabilizer of the root fixed by this cycle
acts transitively on the other seven roots. Transitivity of the whole
group therefore implies double transitivity, and hence transitivity
on the $28$ unordered pairs of distinct roots.

The pair $(u,v)$ determines $\{\beta,\gamma\}$ as the roots of
$z^2+vz+u$. Its orbit thus has size $28$, which is $[L:\Q]$.
\end{proof}

For each root $\alpha$ of $h$, consider the line
\[
\Lambda_\alpha:\quad U+\alpha V+\alpha^2=0.
\]
The lines corresponding to $\alpha$ and $\alpha'$ meet at
$(\alpha\alpha',-\alpha-\alpha')$. No three are concurrent, since a
monic quadratic cannot have three distinct roots.

\begin{lemma}\label{lem:interpolation}
Evaluation at $(u,v)$ gives an isomorphism of $\Q$-vector spaces
\[
\Q[U,V]_{\leq6}\xrightarrow{\ \sim\ } L
\]
and with:
\begin{equation}\label{eq:ell}
\ell=(1,u,v,u^2,uv,v^2) \in L^6,
\end{equation}
the evaluation map $H_d\longrightarrow L$, $E\longmapsto E(\ell)$,
is surjective for every $d\geq3$.
\end{lemma}
\begin{proof}
Suppose that $g\in\Q[U,V]_{\leq6}$ vanishes at $(u,v)$.
Lemma~\ref{lem:field} implies that it vanishes at all $28$ intersections.
Each line $\Lambda_\alpha$ contains seven distinct intersections, so
the restriction of $g$ to that line is zero. Over the splitting field,
the product of the eight distinct line equations therefore divides
$g$. Since $\deg g\leq6$, we obtain $g=0$. Both vector spaces have
dimension $28$, proving the first assertion.

Every monomial $U^iV^j$ with $i+j\leq6$ is a product of at most three
elements of $\{U,V,U^2,UV,V^2\}$. Padding with powers of $x_0$ gives
a homogeneous cubic restricting to this monomial under
\[
(x_0,\ldots,x_5)=(1,U,V,U^2,UV,V^2).
\]
Multiplication by $x_0^{d-3}$ proves the assertion for all $d\geq3$.
\end{proof}

We use the ordered basis
\begin{equation}\label{eq:basis}
\mathcal B=(U^iV^{a-i}:0\leq a\leq6,\ 0\leq i\leq a),
\end{equation}
where $a$ and then $i$ increase, and where the entries are evaluated
at $(u,v)$ when regarded as elements of $L$. For exact computations,
one can instead use the relations
\begin{equation}\label{eq:r0r1}
\begin{aligned}
r_0&=U^4-6U^3V^2+5U^2V^4-UV^6-1,\\
r_1&=4U^3V-10U^2V^3+6UV^5-V^7-1.
\end{aligned}
\end{equation}
Indeed, division of $h(z)$ by $z^2+Vz+U$ leaves remainder $r_0+zr_1$.
The quotient $\Q[U,V]/(r_0,r_1)$ then identifies with $L$; the verification is given in Appendix~\ref{app:finite}, by giving a Gr\"obner basis and proving that this quotient has dimension $28$,
and thus that its natural surjection to $L$ is an isomorphism.


\section{The explicit formal curve}\label{sec:curve}

Retain the field $L=\Q(u,v)$ from Lemma~\ref{lem:field} and its
ordered basis of monomials of degree at most six. Define the resultant: 
\begin{equation}\label{eq:resultant}
H(U,V)=\Res_z\bigl(z^8-z-1,z^2+Vz+U\bigr).  
\end{equation}
Explicitly:
\begin{eqnarray*}
& & H(U,V) \\ 
&= &    1+U-V -2U^4 +16U^3V^2 
      -7U^4V+14 U^3V^3 \\ & & -20U^2V^4+8UV^6-7U^2V^5+U^8+UV^7-V^8.
\end{eqnarray*}
If $\alpha$ ranges over the eight roots of $z^8-z-1$ in a splitting
field, then
\[
 H(U,V)=\prod_{\alpha}(U+\alpha V+\alpha^2).
\]
We use the affine coordinates
\begin{equation}\label{eq:chart}
 (x_0,x_1,x_2,x_3,x_4,x_5)
 =(1,U,V,U^2+A,UV+B,V^2+C).
\end{equation}
For a homogeneous polynomial $G$ in the variables $x_0,\ldots,x_5$, write $G_{\mathrm{ch}}$ for its
substitution under~\eqref{eq:chart}. Direct expansion gives
\begin{equation}\label{eq:chart-identities}
 D_{\mathrm{ch}}=AC-B^2,
 \qquad W_{\mathrm{ch}}(U,V,0,0,0)=H(U,V).
\end{equation}
The three normal derivatives $a,b,c$ of $W_{\mathrm{ch}}$ on $A=B=C=0$, namely $a=\partial_AW_{\mathrm{ch}}|_{A=B=C=0}$, and similarly for
$b$ and $c$, are
the following polynomials:
\begin{align}
 a(U,V)&=4U^6-14U^2V-4U^2+14UV^3+16UV^2-7V^5-20V^4,
 \notag\\
 b(U,V)&=14U^2V^2+V^6,\label{eq:abc}\\
 c(U,V)&=14U^3V+16U^3-14U^2V^3-40U^2V^2
             +3UV^5+24UV^4-4V^6.\notag
\end{align}
Set
\begin{equation}\label{eq:K-V0}
 \begin{split}
 K&=ac-\frac{b^2}{4},\\
 V_0&=-231U^6+\frac{1169}{2}U^2V+215U^2
      -\frac{833}{2}UV^3-988UV^2+\frac14V^5+411V^4.
 \end{split}
\end{equation}

\begin{lemma}\label{lem:node-data}
At $(u,v)$ one has
\[
 H=H_U=H_V=0,\qquad V_0-K=0,
\]
and $\Hess(H)(u,v)$ is invertible.
\end{lemma}

\begin{proof}
The point $(u,v)=(\beta\gamma,-\beta-\gamma)$ is the intersection
of the two lines indexed by $\beta$ and $\gamma$. No third line passes
through it. Let $c_*$ be the product of the other six line equations,
evaluated at this point. Then $c_*\ne0$, and differentiating the
product of the eight line equations gives
\[
 \det\Hess(H)(u,v)=-c_*^2(\beta-\gamma)^2\ne0.
\]
The same product description gives the three stated vanishings of $H$.
Finally, expansion of $K=ac-b^2/4$ using~\eqref{eq:abc} and then reduction by the two
relations~\eqref{eq:r0r1} gives
\[
 ac-\frac{b^2}{4}\equiv V_0\pmod{(r_0,r_1)}.
\]
The displayed reduction basis in the finite-certificate appendix
makes this a rational polynomial identity; the supplementary verifier
checks it independently. This shows that $V_0-K$ vanishes at $(u,v)$.
\end{proof}

All derivatives in the next definition are taken as polynomial
derivatives in $U,V$ before evaluation at $(u,v)$. Define $u_1,v_1\in L$
by
\begin{equation}\label{eq:u1v1}
 \begin{pmatrix}u_1\\v_1\end{pmatrix}
 =-\Hess(H)(u,v)^{-1}\nabla(V_0-K)(u,v).
\end{equation}
Put
\begin{equation}\label{eq:ell-w}
 \begin{gathered}
 \ell=(1,u,v,u^2,uv,v^2),\qquad w_0=0,\qquad
 w_1=u_1,\quad w_2=v_1,\\
 w_3=2uu_1-c(u,v),\qquad
 w_4=vu_1+uv_1+\frac12b(u,v),\\
 w_5=2vv_1-a(u,v).
 \end{gathered}
\end{equation}
These formulas specify a continuous $\Q$-algebra homomorphism from $R$ to $L[[t]]$, i.e. an arc:
\begin{equation}\label{eq:param}
 \varphi:R=\Q[[x_0,\ldots,x_5]]\longrightarrow L[[t]],
 \qquad x_i\longmapsto t\ell_i+t^2w_i,
 \qquad P=\ker\varphi.
\end{equation}
(Here $\ell_i$ is the $i$-th coordinate of $\ell$.) In particular, $\varphi(x_0)=t$. Geometrically $\varphi$ corresponds to a formal curve in the formal spectrum of $R$.

\begin{lemma}\label{lem:curve}
For every $d\ge3$,
\begin{equation}\label{eq:conductor-surjectivity}
 \varphi(\m^d)=t^dL[[t]].
\end{equation}
The image $A=\varphi(R)$ is a one-dimensional complete local domain,
finite over $\Q[[t]]$, with residue field $\Q$. Its fraction field is
$L((t))$, and its normalization is $L[[t]]$. Consequently, $P$ is a
height-five prime ideal and $P\subseteq\m^2$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:interpolation}, choose homogeneous cubics
$b_1,\ldots,b_{28}$ such that $b_j(\ell)$ form a $\Q$-basis of $L$.
Relative to this basis, the elements $t^{-3}\varphi(b_j)$ form the
columns of a matrix in $\operatorname{Mat}_{28}(\Q[[t]])$ whose
reduction modulo $t$ is invertible. They therefore form a
$\Q[[t]]$-basis of $L[[t]]$. Since $\varphi(x_0)=t$, every element of
$t^3L[[t]]$ lifts to an element of $\m^3$. Multiplication by
$x_0^{d-3}$ proves the reverse inclusion in
\eqref{eq:conductor-surjectivity}; the forward inclusion follows from
$\varphi(x_i)\in tL[[t]]$.

The image $A$ contains $\Q[[t]]$ and is a $\Q[[t]]$-submodule of
the finite free module $L[[t]]$. It is therefore finite over
$\Q[[t]]$. It is a domain of dimension one. As a quotient of the
complete local ring $R$, it is complete and local. Reduction modulo
$t$ takes $A$ onto $\Q$; its kernel is the image of $\m$, so its
residue field is $\Q$. The inclusion $t^3L[[t]]\subseteq A$ implies
that $\operatorname{Frac}(A)=L((t))$. The ring $L[[t]]$ is finite
over $A$ and integrally closed in this field, hence is the
normalization of $A$.

It follows that $\operatorname{ht}P=6-1=5$. The six coordinates of
$\ell$ are linearly independent over $\Q$ by
Lemma~\ref{lem:interpolation}. The coefficient of $t$ in
$\varphi(p)$ therefore excludes a nonzero linear initial form for
any $p\in P$. The constant coefficient also vanishes, so
$P\subseteq\m^2$.
\end{proof}

\section{Eliminating critical values to all orders}\label{sec:critical}

Consider the homogeneous cubic
\begin{equation}\label{eq:Vsharp}
 \begin{split}
 V_0^\sharp={}&-231x_3^3+\frac{1169}{2}x_0x_2x_3+215x_0^2x_3
 -\frac{833}{2}x_0x_4x_5\\
 &-988x_0x_1x_5+\frac14x_2x_5^2+411x_0x_5^2.
 \end{split}
\end{equation}
Its restriction to $A=B=C=0$ under~\eqref{eq:chart} is exactly
$V_0(U,V)$. Set $ E_5=x_0^2V_0^\sharp$.

\bigskip
The following proposition is the technical core of the construction. 


\begin{proposition}\label{prop:critical}
There exist homogeneous forms $E_n\in H_n$ for all $n\ge6$ and a
continuous $\Q$-algebra homomorphism $\psi:R\to L[[t]]$, i.e. an arc, such that with:
\begin{equation}\label{eq:critical-F}
 F:=D+W+E_5+\sum_{n\ge6}E_n, 
\end{equation}
one has:
\begin{equation}\label{eq:psi-jet}
 \psi(x_0)=t,\qquad
 \psi(x_i)=t\ell_i+t^2w_i+O(t^3)\quad(1\le i\le5),
\end{equation}
the critical condition:
\begin{equation}\label{eq:critical-vanishing}
  \psi(\partial_iF)=0\quad(0\le i\le5),
\end{equation}
and
\[
\psi(F)=0.
\]
\end{proposition}

\begin{proof}
We first associate to every (tentative)
$F=D+W+E_5+O(\m^6)$ an arc $\psi_F:R \rightarrow L[[t]]$ that satisfies the condition:
\[
\psi_F(\partial_i F) =0\quad(1\le i\le5).
\]

First, make the following substitution (the affine cone version of (\ref{eq:chart})):
\[
 x_0=t,\qquad (x_1,\ldots,x_5)
 =t(U,V,U^2+A,UV+B,V^2+C)
\]
into $F$, and then divide by $t^3$; the resulting formal series $G_F$ is then:
\begin{equation}\label{eq:GF}
 G_F=AC-B^2+tW_{\mathrm{ch}}+t^2(E_5)_{\mathrm{ch}}+O(t^3).
\end{equation}
We work over $L[[t,U-u,V-v,A,B,C]]$ (and thus $G_F \in L[[t,U-u,V-v,A,B,C]]$). The Hessian of $AC-B^2$
in the three variables $A,B,C$ is invertible. The formal
implicit function theorem then gives unique critical solutions in $tL[[t,U-u,V-v]]$ satisfying:
\begin{equation}\label{eq:normal-solutions}
 A_F=-tc+O(t^2),\qquad
 B_F=\frac12tb+O(t^2),\qquad
 C_F=-ta+O(t^2).
\end{equation}
Here $a,b,c$ are the polynomials in $U,V$ as before; the corresponding critical equations are:
\begin{eqnarray*}
 \partial_A G_F(t,U,V,A_F,B_F,C_F) &=& 0,\\ 
 \partial_B G_F(t,U,V,A_F,B_F,C_F)&=&0,  \\ \partial_C G_F(t,U,V,A_F,B_F,C_F)&=&0.
\end{eqnarray*}

Let $\widetilde G_F \in tL[[t,U-u,V-v]]$ be the result of substituting these three
solutions into $G_F$. A calculation through degree two in $t$ gives
\begin{equation}\label{eq:reduced-critical-function}
 \widetilde G_F(t,U,V)
 =tH(U,V)+t^2(V_0-K)(U,V)+O(t^3).
\end{equation}

Put $\Psi_F:=t^{-1}\widetilde G_F \in L[[t,U-u,V-v]]$.
Lemma~\ref{lem:node-data} and the formal implicit function theorem
give unique series $U_F,V_F\in L[[t]]$ with constant terms $u,v$
satisfying the critical equations:
\[
 \partial_U\Psi_F(t,U_F,V_F)
 =\partial_V\Psi_F(t,U_F,V_F)=0.
\]
Their first terms are
\begin{equation}\label{eq:tangential-solutions}
 U_F=u+tu_1+O(t^2),\qquad V_F=v+tv_1+O(t^2),
\end{equation}
by~\eqref{eq:u1v1}.


For abbreviations we write:
\begin{eqnarray*}
\widetilde{A}_F &=& A_F(t,U_F,V_F), \\
\widetilde{B}_F &=& B_F(t,U_F,V_F), \\
\widetilde{C}_F &=& C_F(t,U_F,V_F),
\end{eqnarray*}
for the elements of $L[[t]]$ obtained by substituting $U_F,V_F$ into the variables $U,V$.

Substituting $U_F,V_F,\widetilde{A}_F,\widetilde{B}_F,\widetilde{C}_F \in L[[t]]$ into the variables $U,V,A,B,C$, we then have the formulas 
\[
 x_0=t,\qquad (x_1,\ldots,x_5)
 =t(U_F,V_F,U_F^2+\widetilde{A}_F,U_FV_F+\widetilde{B}_F,V_F^2+\widetilde{C}_F)
\]
which then defines an arc
$\psi_F:R \rightarrow L[[t]]$. 


Now (in the affine chart) the change of coordinate map
\[
 (U,V,A,B,C)\longmapsto(U,V,U^2+A,UV+B,V^2+C)
\]
has determinant-one Jacobian. The five critical equations that the five critical solutions $A_F,B_F,C_F,U_F,V_F$ satisfy, thus give:
\begin{equation}\label{eq:spatial-gradient}
 \psi_F(\partial_iF)=0\qquad(1\le i\le5).
\end{equation}
 


Note also that equations~\eqref{eq:normal-solutions} and
\eqref{eq:tangential-solutions} give precisely the coefficients
$w_i$ in~\eqref{eq:ell-w}. Thus every such $\psi_F$ satisfies
\eqref{eq:psi-jet}.

Now define the critical value $ \mathcal V(F)$ of $\psi_F$ by:
\[
 \mathcal V(F)=\psi_F(F) \in L[[t]].
\]
We thus have by definition:
\[
 \mathcal V(F)=t^3 \widetilde{G}_F (t,U_F,V_F).
\]
The vanishings in Lemma~\ref{lem:node-data} give:
\[
H(U_F,V_F)=O(t^2),\qquad (V_0-K)(U_F,V_F)=O(t)
\]
and hence
\eqref{eq:reduced-critical-function} then implies
\begin{equation}\label{eq:critical-value-start}
 \mathcal V(F)\in t^6L[[t]].
\end{equation}
Next we claim, for a homogeneous $E\in H_n$,
$n\ge6$, that
\begin{equation}\label{eq:critical-value-response}
 \mathcal V(F+E)-\mathcal V(F)
 \equiv t^nE(\ell)\pmod{t^{n+1}}.
\end{equation}
For this, introduce a parameter $s$ and carry out the two recursive
implicit constructions for $F+sE$. Each coefficient of the resulting
arc lies in $L[s]$: the constant matrices to be inverted are
unchanged, and all subsequent recursion uses polynomial operations
and these fixed inverse matrices. Thus the chain rule applies
coefficientwise in $L[s][[t]]$. Since $x_0=t$ is fixed and the five
spatial derivatives vanish, it gives
\[
 \frac{d}{ds}\mathcal V(F+sE)
 =\psi_{F+sE}(E)
 \equiv t^nE(\ell)\pmod{t^{n+1}}.
\]
Integrating from $s=0$ to $s=1$ proves
\eqref{eq:critical-value-response}. Note that this is also a crucial point that makes use of the fact that the coefficient field has characteristic zero. 

Now start with $F^{[5]}=D+W+E_5$. Inductively suppose that $F^{[n-1]}$ has been
chosen and that its critical value has no term of degree less than
$n$. Let $c_n\in L$ be the coefficient of $t^n$. By
Lemma~\ref{lem:interpolation}, choose $E_n\in H_n$ with
$E_n(\ell)=-c_n$. Equation~\eqref{eq:critical-value-response}
shows that $F^{[n]}=F^{[n-1]}+E_n$ kills the coefficient of $t^n$
while preserving all previous vanishings.

Now adding a form of degree $n$ changes $G_F$ and its normal
equations by $O(t^{n-3})$. The fixed invertible normal Hessian then
shows that the normal critical solutions, as functions of $t,U,V$,
change by $O(t^{n-3})$. After reduction and division by $t$, the
tangential equations change by $O(t^{n-4})$. Their fixed invertible
constant Hessian shows that $U_F,V_F$ change by $O(t^{n-4})$.
Since the normal solutions and their $U,V$ derivatives are divisible
by $t$, their evaluations on the final tangential arc change by
$O(t^{n-3})$. The original coordinates $x_1,\ldots,x_5$ therefore
change by $O(t^{n-3})$. These orders tend to infinity with $n$.

It follows that $F^{[n]}$ converges $\m$-adically to the series
$F$ with the shape in~\eqref{eq:critical-F}, and the arcs converge coefficientwise
to an arc $\psi=\psi_F$ satisfying~\eqref{eq:psi-jet}. Substitution and
formal differentiation are continuous for these filtrations, so
the induction and~\eqref{eq:spatial-gradient} give
\[
 \psi(F)=0,\qquad \psi(\partial_iF)=0\quad(1\le i\le5).
\]
Finally to complete the proof of~\eqref{eq:critical-vanishing}, it remains to show that $\psi(\partial_0F) =0$. Differentiate $\psi(F)=0$ with respect to $t$. The chain
rule and $\psi(x_0)=t$ give
\[
 0=\psi(\partial_0F)
   +\sum_{i=1}^5\psi(\partial_iF)\frac{d\psi(x_i)}{dt}
  =\psi(\partial_0F),
\]
which thus completes the proof of~\eqref{eq:critical-vanishing}.
\end{proof}

\section{Straightening and symbolic-square membership}\label{sec:symbolic}

\begin{lemma}\label{lem:straightening}
Let $\psi:R\to L[[t]]$ be any arc that satisfies~\eqref{eq:psi-jet}.
There is a continuous $\Q$-algebra automorphism $\theta$ of $R$
such that
\[
 \theta(x_0)=x_0,\qquad
 \theta(x_i)-x_i\in\m^3\quad(1\le i\le5),\qquad
 \varphi\circ\theta=\psi.
\]
\end{lemma}

\begin{proof}
Each difference $\psi(x_i)-\varphi(x_i)$ belongs to $t^3L[[t]]$.
By~\eqref{eq:conductor-surjectivity}, choose $g_i\in\m^3$ whose
image under $\varphi$ is this difference. Define
$\theta(x_i)=x_i+g_i$ for $1\le i\le5$ and
$\theta(x_0)=x_0$. This substitution has identity linear part, so
recursive solution of its homogeneous terms gives a continuous
inverse. Its composite with $\varphi$ agrees with $\psi$ on the
six variables and hence on $R$ by continuity.
\end{proof}

Apply this lemma to the arc of Proposition~\ref{prop:critical}, and
put
\begin{equation}\label{eq:formal-witness}
 f=\theta(F).
\end{equation}
Changing variables by terms of order at least three changes a cubic
only in degrees at least five. Therefore
\begin{equation}\label{eq:fourjet}
 f\equiv D+W\pmod{\m^5}.
\end{equation}
The chain rule and $\varphi\circ\theta=\psi$ imply
\begin{equation}\label{eq:witness-gradient}
 \varphi(f)=0,\qquad \varphi(\partial_i f)=0
 \quad(0\le i\le5).
\end{equation}

\begin{proposition}\label{prop:symbolic}
The series $f$ in~\eqref{eq:formal-witness} belongs to $P^{(2)}$.
More generally, if $g\in R$ satisfies
$\varphi(g)=\varphi(\partial_i g)=0$ for $1\le i\le5$, then
$g\in P^{(2)}$.
\end{proposition}

\begin{proof}
Let $S=L[[x_0,\ldots,x_5]]$ and set
\[
 Q_{\mathrm{arc}}
 =(x_i-x_0\ell_i-x_0^2w_i:1\le i\le5)S.
\]
The same formulas that define $\varphi:R \rightarrow L[[t]]$ as before then give a surjective homomorphism
$S\to L[[t]]$ with kernel $Q_{\mathrm{arc}}$, and
$Q_{\mathrm{arc}}\cap R=P$. Use $x_0$ and
$y_i :=x_i-x_0\ell_i-x_0^2w_i$ ($1 \leq i \leq 5 $) as formal coordinates on $S$.
In terms of these new coordinates, we have $S=L[[x_0]][[y_1,\ldots,y_5]]$, and then $Q_{\mathrm{arc}}$ is the ideal of $S$ generated by $y_i$ for $1 \leq i \leq 5$, and hence $Q_{\mathrm{arc}}^2$ is the ideal of $S$ generated by $y_i y_j$ for $1 \leq i,j \leq 5$. One then sees immediately, by using Taylor expansion for example (with respect to the original coordinates $x_1,\ldots,x_5$, with $x_0$ being regarded as fixed), that if $g\in R$ satisfies $\varphi(g)=\varphi(\partial_i g)=0$ for $1\le i\le5$, then $g\in Q_{\mathrm{arc}}^2$.

Since $L/\Q$ is finite and separable,
$S=R\otimes_{\Q}L$ is finite \emph{\'etale} over $R$.
The local map $R_P\to S_{Q_{\mathrm{arc}}}$ is faithfully flat.
Its closed fiber is a localization of a finite \emph{\'etale}
algebra over $\kappa(P)$, hence a field. Consequently,
\[
 PS_{Q_{\mathrm{arc}}}
 =Q_{\mathrm{arc}}S_{Q_{\mathrm{arc}}}.
\]
Faithful flatness therefore gives
\[
 g\in P^2S_{Q_{\mathrm{arc}}}\cap R_P=P^2R_P.
\]
Intersecting with $R$ proves $g\in P^{(2)}$. Finally the particular series
$f$ satisfies the hypotheses by~\eqref{eq:witness-gradient}.
\end{proof}

\section{A finite certificate for nonmembership}\label{sec:certificate}

Put
\[
T=\Q[[x_1,\ldots,x_5]],\qquad
\mathfrak q=(x_1,\ldots,x_5)T,\qquad
J=(P+(x_0))/(x_0)\subset T.
\]
A bar denotes the image under the substitution $x_0=0$ (here we identify $R/(x_0)$ with $T$). We shall construct a
$\Q$-linear functional $\lambda:T\longrightarrow\Q$ such that
\begin{equation}\label{eq:separator}
\lambda(\mathfrak qJ)=0,\qquad
\lambda(\mathfrak q^5)=0,\qquad
\lambda(\overline D+\overline W)=1.
\end{equation}
Together with \eqref{eq:fourjet}, these identities prove the required
nonmembership in $\m P$.

Consider the six quadrics:
\begin{equation}\label{eq:quadrics}
\begin{aligned}
q_1&=x_0x_3-x_1^2,&
q_2&=x_0x_4-x_1x_2,&
q_3&=x_0x_5-x_2^2,\\
q_4&=x_1x_4-x_2x_3,&
q_5&=x_1x_5-x_2x_4,&
q_6&=x_3x_5-x_4^2
\end{aligned}
\end{equation}
with their accompanying homogeneous cubic terms given by:
\begin{equation}\label{eq:corrections}
\begin{aligned}
C_1={}&16x_0x_1x_3+14x_0x_3x_4-40x_0x_3x_5
       +24x_1x_5^2\\
     &{}-14x_2x_3x_5+3x_4x_5^2-4x_5^3,\\
C_2={}&-7x_0x_3x_5-\tfrac12x_5^3,\\
C_3={}&-4x_0^2x_3+16x_0x_1x_5-14x_0x_2x_3+14x_0x_4x_5\\
     &{}-20x_0x_5^2-7x_2x_5^2+4x_3^3,\\
C_4={}&-\tfrac12x_0^3-\tfrac72x_0x_3^2-\tfrac72x_3x_5^2,\\
C_5={}&-\tfrac12x_0^3-8x_0x_3x_5-4x_5^3,\\
C_6={}&4x_0^3+3x_0^2x_1-3x_0^2x_2-4x_0x_3^2+16x_1x_3x_5\\
     &{}-7x_2x_3^2+14x_3x_4x_5-20x_3x_5^2.
\end{aligned}
\end{equation}

\begin{lemma}\label{lem:jets}
The image of $P$ in $R/\m^4$ is the rational span of the classes of
\begin{equation}\label{eq:thirdjets}
q_i+C_i\quad(1\leq i\leq6),\qquad
x_jq_i\quad(0\leq j\leq5,\ 1\leq i\leq6).
\end{equation}
\end{lemma}

\begin{proof}
Let $\rho$ denote substitution of
$(1,U,V,U^2,UV,V^2)$ into homogeneous forms. The degree-two kernel of $\rho$
is spanned by the six $q_i$, and its degree-three kernel is spanned by their
linear multiples. These statements can be seen by homogenizing the substitution
to the map
\[
(x_0,x_1,x_2,x_3,x_4,x_5)
\longmapsto(s^2,sr,ss',r^2,rs',s'^2).
\]
The six quadrics in \eqref{eq:quadrics} generate all relations among these
degree-two monomials: any two products representing the same monomial in
$s,r,s'$ are related by changing the pairing of four factors. The quotient
dimensions in degrees two and three are therefore $15$ and $28$,
respectively. Since $\rho(H_d)$ consists of polynomials of degree at most
$2d$, Lemma~\ref{lem:interpolation} shows that evaluation at $\ell$ has
these same kernels for $d=2,3$.

By the definition of $\varphi$ as given in \eqref{eq:param}, the coefficient of $t^3$ in $\varphi(q_i)$ is
$\sum_{j=0}^5\partial_jq_i(\ell)w_j$. The negatives of these six
coefficients, in order, are
\begin{equation}\label{eq:correctionvalues}
c,\quad -\tfrac12 b,\quad a,\quad
-\bigl(\tfrac12ub+vc\bigr),\quad
ua+\tfrac12vb,\quad
u^2a+uvb+v^2c,
\end{equation}
where $a,b,c$ are evaluated at $(u,v)$. The terms containing $u_1,v_1$
cancel. Substitution into the displayed $C_i$, followed by reduction using
\eqref{eq:r0r1}, gives exactly the values in \eqref{eq:correctionvalues}.
Consequently,
\begin{equation}\label{eq:correctedquadrics}
\varphi(q_i+C_i)\in t^4L[[t]]\qquad(1\leq i\leq6).
\end{equation}

For $p\in P$, Lemma~\ref{lem:curve} gives
$p=p_2+p_3+O(\m^4)$ with $p_d\in H_d$. The coefficient of $t^2$ in
$\varphi(p)$ gives $p_2=\sum_i\alpha_iq_i$ for some $\alpha_i\in\Q$.
By \eqref{eq:correctedquadrics}, the cubic
$p_3-\sum_i\alpha_iC_i$ evaluates to zero at $\ell$. It is therefore a
linear combination of the $x_jq_i$, proving one inclusion.

Conversely, every expression in \eqref{eq:thirdjets} maps into
$t^4L[[t]]$. Lemma~\ref{lem:curve} gives
$\varphi(\m^4)=t^4L[[t]]$, so its image can be canceled by an element
of $\m^4$. Its class modulo $\m^4$ thus belongs to the image of $P$.
\end{proof}

Now define the functional $\lambda$ on monomials by Table~\ref{tab:lambda}, and set it to zero
on every monomial not listed. It extends to all of $T$ by taking this finite
linear combination of coefficients; in particular,
$\lambda(\mathfrak q^5)=0$.

\begin{table}[htbp]
\caption{All nonzero values of $\lambda$. An exponent tuple
$e=(e_1,\ldots,e_5)$ denotes $x_1^{e_1}\cdots x_5^{e_5}$.}
\label{tab:lambda}
\centering
\renewcommand{\arraystretch}{1.12}
\begin{tabular}{@{}cr@{\qquad}cr@{}}
\hline
Degree-three tuple & Value & Degree-four tuple & Value\\
\hline
$(0,0,0,2,1)$ & $-5632/49$  & $(0,0,0,0,4)$ & $3$\\
$(0,0,0,3,0)$ & $-10240/343$ & $(0,0,0,1,3)$ & $8$\\
$(0,1,0,1,1)$ & $-12$       & $(0,0,0,2,2)$ & $384/49$\\
$(0,1,0,2,0)$ & $-32$       & $(0,0,0,3,1)$ & $512/343$\\
$(0,1,1,0,1)$ & $-192/7$    & $(0,0,1,0,3)$ & $384/49$\\
$(0,1,1,1,0)$ & $-1536/49$  & $(0,0,1,1,2)$ & $512/343$\\
$(0,2,0,1,0)$ & $-3/2$      & $(0,0,3,1,0)$ & $-3/8$\\
$(0,2,1,0,0)$ & $-8$        & $(0,1,0,2,1)$ & $8/7$\\
$(1,0,0,2,0)$ & $-1280/49$  & $(0,1,0,3,0)$ & $64/49$\\
$(1,1,0,0,1)$ & $-3/2$      & $(0,1,1,0,2)$ & $8/7$\\
$(1,1,0,1,0)$ & $-4$        & $(0,1,1,1,1)$ & $64/49$\\
$(1,1,1,0,0)$ & $-192/49$   & $(1,0,0,1,2)$ & $8/7$\\
$(2,0,0,0,1)$ & $-4$        & $(1,0,0,2,1)$ & $64/49$\\
$(2,0,0,1,0)$ & $32/49$     & $(1,0,1,0,2)$ & $64/49$\\
$(2,0,1,0,0)$ & $1536/343$  & &\\
$(3,0,0,0,0)$ & $24/7$     & &\\
\hline
\end{tabular}
\end{table}

\begin{proposition}\label{prop:nonmembership}
The functional $\lambda$ satisfies \eqref{eq:separator}. Every series
$f\equiv D+W\pmod{\m^5}$ consequently lies outside $\m P$.
\end{proposition}

\begin{proof}
The displayed coefficients give the rational identities
\begin{align}
\lambda\bigl(\overline{x_j(q_i+C_i)}\bigr)&=0
&& (1\leq j\leq5,\ 1\leq i\leq6),\label{eq:lambdaquadratic}\\
\lambda\bigl(\overline{x_jx_kq_i}\bigr)&=0
&& (1\leq j\leq5,\ 0\leq k\leq5,\ 1\leq i\leq6).
\label{eq:lambdacubic}
\end{align}
These are $30+180$ finite identities; all are checked by the exact-arithmetic
verifier included with the supplementary material. Note that of course for (\ref{eq:lambdacubic}) it is trivial when $k=0$, so that the number of non-trivial identities in (\ref{eq:lambdacubic}) is 150, and the total number of non-trivial identities that have to be checked in (\ref{eq:lambdaquadratic})--(\ref{eq:lambdacubic}) is 180. 

Note also how the dimension count works here: the dimension of the space of cubics in five variables is 56, and the dimension of the space of quartics in five variables is 126; as $(56 + 126)-180=2 $, so this leaves two degrees of freedom to define the functional $\lambda$ satisfying (\ref{eq:lambdaquadratic})--(\ref{eq:lambdacubic}) (these two remaining degrees of freedom are used to make $\lambda(\overline{D})=4$ and $\lambda(\overline{W})=-3$ as shown below; incidentally this also explains why $\lambda$ is defined this way). 

For $g\in\mathfrak qJ$, choose $p_j\in P$ such that
$g=\sum_{j=1}^5x_j\overline{p_j}$. The terms of $g$ through degree four
depend only on the classes of the $p_j$ modulo $\m^4$.
Lemma~\ref{lem:jets} and
\eqref{eq:lambdaquadratic}--\eqref{eq:lambdacubic} therefore give
$\lambda(g)=0$. This proves vanishing of $\lambda$ on the whole ideal $\mathfrak qJ$.

Finally, \eqref{eq:D} and \eqref{eq:W} give
\[
\overline D=2x_1x_2x_4-x_1^2x_5-x_2^2x_3
\]
and
\[
\overline W=8x_1x_5^3-7x_2x_3x_5^2+x_3^4+x_4x_5^3-x_5^4.
\]
Table~\ref{tab:lambda} gives
$\lambda(\overline D)=4$ and $\lambda(\overline W)=-3$. Thus
$\lambda(\overline f)=1$ whenever
$f\equiv D+W\pmod{\m^5}$. If $f\in\m P$, then
$\overline f\in\mathfrak qJ$, a contradiction.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
The field $L$ and $\varphi:R \rightarrow L[[t]]$ are specified in
\eqref{eq:field} and \eqref{eq:u1v1}--\eqref{eq:param}.
Lemma~\ref{lem:curve} proves the assertions about the height,
residue field, and normalization. Proposition~\ref{prop:critical}
and Lemma~\ref{lem:straightening} construct a series $f$ given in equation (\ref{eq:formal-witness}), with
$f\equiv D+W\pmod{\m^5}$. Proposition~\ref{prop:symbolic} gives
$f\in P^{(2)}$, and Proposition~\ref{prop:nonmembership} gives
$f\notin\m P$.
\end{proof}
\section{Extension of the coefficient field}\label{sec:complex}

For an ideal $I$ in a Noetherian ring, recall the symbolic square:
\[
I^{(2)}=\bigcap_{\mathfrak q\in\operatorname{Min}(I)}
\bigl(I^2R_{\mathfrak q}\cap R\bigr).
\]


The prime ideal $P$ of $R$ in the previous constructions acquires a particularly simple branch description
after extension of the coefficient field.

\begin{proposition}\label{prop:complex}
Let $R_{\C}=\C[[x_0,\ldots,x_5]]$, let
$\m_{\C}=(x_0,\ldots,x_5)R_{\C}$, and put $I=PR_{\C}$. Then $I$ is an unmixed radical ideal, and we have the primary decomposition:
\begin{equation}\label{eq:branches}
I=\bigcap_{\sigma:L\hookrightarrow\C}
\bigl(x_i-x_0\sigma(\ell_i)-x_0^2\sigma(w_i):1\leq i\leq5\bigr)R_{\C}.
\end{equation}
These $28$ ideals are the distinct smooth branch primes of $I$, and
the series $f$ of equation (\ref{eq:formal-witness}) satisfies
\[
f\in I^{(2)}\setminus\m_{\C}I.
\]
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:curve}, the inclusion
$A\subset L[[t]]$ is an inclusion of finite $\Q[[t]]$-modules.
The maximal-ideal topology on $A$ agrees with its $t$-adic topology:
the quotient $A/tA$ is a finite-dimensional local $\Q$-algebra.
Completed extension of the coefficient field consequently gives
\begin{equation}\label{eq:completed-basechange}
R_{\C}/PR_{\C}
\cong A\otimes_{\Q[[t]]}\C[[t]].
\end{equation}
For completeness, this identification can be seen by reducing the
finite $\Q[[t]]$-module $A$ modulo $t^n$, extending coefficients to
$\C$, and taking the inverse limit. Finiteness identifies that limit
with the tensor product in \eqref{eq:completed-basechange}; the
equivalence of topologies identifies it with the completed
presentation $\C[[x]]/P\C[[x]]$.

The extension $\Q[[t]]\longrightarrow\C[[t]]$ is flat, since its
target is torsion-free over the discrete valuation ring $\Q[[t]]$.
It follows that
\[
R_{\C}/I\ \lhook\joinrel\longrightarrow\
\C[[t]]\otimes_{\Q[[t]]}L[[t]]
\cong\prod_{\sigma:L\hookrightarrow\C}\C[[t]].
\]
The image contains $t^3$ times the product on the right. The coordinate
maps have the kernels displayed in \eqref{eq:branches}; their
intersection is therefore $I$. Each quotient is $\C[[t]]$, and
distinct embeddings give distinct pairs $(\sigma(u),\sigma(v))$,
hence distinct tangent directions. Thus these are precisely the
minimal primes of $I$.

The identities $\varphi(f)=\varphi(\partial_i f)=0$ persist on every
branch. Taylor expansion in the five transverse branch coordinates
therefore places $f$ in the square of each branch ideal, and hence
in $I^{(2)}$. Extend the functional of Table~\ref{tab:lambda}
$\C$-linearly on fourth jets. The identities in the proof of
Proposition~\ref{prop:nonmembership} remain valid under this extension.
The extended functional annihilates the image of $\m_{\C}I$ modulo
$x_0$ and takes value $1$ on $f$, proving the nonmembership.
\end{proof}

\begin{corollary}\label{cor:jacobian}
There is a formal power series $f\in\C[[x_0,\ldots,x_5]]$ without
constant term such that
\[
f\notin\m_{\C}\sqrt{(f,\partial_0f,\ldots,\partial_5f)}.
\]
\end{corollary}
\begin{proof}
Take the series given by equation (\ref{eq:formal-witness}). The radical ideal $I$ of
Proposition~\ref{prop:complex} contains $f$ and all its partial
derivatives, so the displayed radical is contained in $I$. The
assertion follows from $f\notin\m_{\C}I$.
\end{proof}

This is the formal-power-series problem in \cite[Introduction and
Corollary~2]{EM97}. Proposition~\ref{prop:complex} then also gives a counterexample to the general 
form of the Eisenbud--Mazur conjecture stated in \cite[Conjecture~1.5]{Mor15} (with $\C$ being the coefficient field).


\section{A counterexample in a localized polynomial ring}\label{sec:algebraization}

The formulas that define $\varphi:R \rightarrow L[[t]]$ also define an algebraic curve.
The construction of $f$ uses a complete formal power
series, but by using flatness and a differential characterization of the symbolic
square, we are able to construct a counterexample to the original Eisenbud--Mazur conjecture in a localized polynomial ring. 


Let
\[
 S_0=\Q[x_0,\ldots,x_5]_{\mathfrak n},
 \qquad \mathfrak n=(x_0,\ldots,x_5),
\]
where the same notation $\mathfrak n$ denotes the maximal ideal
after localization. Let $\mathfrak p_0$ be the kernel of the
polynomial homomorphism
\[
 \Q[x_0,\ldots,x_5]\longrightarrow L[t],
 \qquad x_i\longmapsto t\ell_i+t^2w_i,
\]
and put $\mathfrak p=\mathfrak p_0S_0$.

\begin{theorem}\label{thm:algebraic}
The ideal $\mathfrak p$ is a height-five prime of $S_0$. Its extension
to the completion $R=\widehat{S_0}$ is $P$, and
\begin{equation}\label{eq:symbolic-completion}
 \mathfrak p^{(2)}R=P^{(2)}.
\end{equation}
There exists a polynomial $g\in\Q[x_0,\ldots,x_5]$ such that
\begin{equation}\label{eq:algebraic-witness}
 g\equiv D+W\pmod{(x_0,\ldots,x_5)^5},\qquad
 g\in\mathfrak p^{(2)}\setminus\mathfrak n\mathfrak p.
\end{equation}
In particular, the counterexample exists in a regular local ring
essentially of finite type over $\Q$.
\end{theorem}

\begin{proof}
We first identify the completed prime. Write
\[
 B=\Q[t,t\ell_1+t^2w_1,\ldots,t\ell_5+t^2w_5]
 \subseteq L[t].
\]
The ring $B$ contains $\Q[t]$ and is a $\Q[t]$-submodule of the
finite free module $L[t]$. It is therefore finite over $\Q[t]$.
The fiber $B/tB$ has only one prime ideal, and its residue field is
$\Q$. Indeed, each generator $b_i=t\ell_i+t^2w_i=t a_i(t)$, with
$a_i(t)\in L[t]$, satisfies a monic equation
\[
 b_i^{28}+t c_1(t)b_i^{27}+\cdots+t^{28}c_{28}(t)=0,
 \qquad c_j(t)\in\Q[t],
\]
obtained from the characteristic polynomial of multiplication by
$a_i(t)$ on $L[t]$. Thus each $b_i$ is nilpotent modulo $tB$.
The quotient by these generators is $\Q$, proving the assertion
about the fiber.

Let $\mathfrak b=(t,b_1,\ldots,b_5)B$. It follows that
$B\otimes_{\Q[t]}\Q[t]_{(t)}$ is local and equals $B_{\mathfrak b}$.
The map from $S_0$ has image $B_{\mathfrak b}$ and kernel
$\mathfrak p$. In particular, $S_0/\mathfrak p$ is a
one-dimensional domain, and $\operatorname{ht}\mathfrak p=5$.
The $\mathfrak b$-adic and $t$-adic topologies on this finite local
algebra are equivalent. Hence
\[
 \widehat{S_0/\mathfrak p}
 =B_{\mathfrak b}\otimes_{\Q[t]_{(t)}}\Q[[t]].
\]
Flatness of $\Q[[t]]$ over $\Q[t]_{(t)}$ embeds this ring in
$L[[t]]$. On the other hand, completion of the quotient presentation
identifies it with $R/\mathfrak pR$. The resulting map
$R\to L[[t]]$ is precisely $\varphi$, since it has the same values
on all six variables. Thus
\begin{equation}\label{eq:completed-prime}
 \mathfrak pR=P.
\end{equation}

We next justify~\eqref{eq:symbolic-completion} by an explicit
finite-module kernel. The map
\[
 \delta:\mathfrak p\longrightarrow(S_0/\mathfrak p)^6,
 \qquad h\longmapsto(\partial_0h,\ldots,\partial_5h)
                    \bmod\mathfrak p
\]
is $S_0$-linear by the product rule, and
\begin{equation}\label{eq:algebraic-differential-kernel}
 \ker\delta=\mathfrak p^{(2)}.
\end{equation}
This is the differential description of the symbolic square in
characteristic zero; in the present setting it also has the
following direct proof. If $h\in\mathfrak p^{(2)}$, choose
$s\in S_0\setminus\mathfrak p$ with $sh\in\mathfrak p^2$.
Differentiation and reduction modulo $\mathfrak p$ show
$s\partial_i h\in\mathfrak p$, hence
$\partial_i h\in\mathfrak p$ for every $i$.

Conversely, suppose that $h\in\mathfrak p$ and all its partial
derivatives belong to $\mathfrak p$. Write $h=a/b$ with
$a,b\in\Q[x_0,\ldots,x_5]$ and $b\notin\mathfrak n$.
Then $a$ and all its partial derivatives vanish under the polynomial
parametrization into $L[t]$. The polynomial change of variables
\[
 y_i=x_i-x_0\ell_i-x_0^2w_i\quad(1\le i\le5)
\]
therefore shows that $a$ belongs to $(y_1,\ldots,y_5)^2$ in
$L[x_0,\ldots,x_5]$: both its constant term and all its linear
coefficients in the $y_i$ vanish.
The extension $S_0\to S_0\otimes_{\Q}L$ is finite \emph{\'etale}.
The ideal generated by the $y_i$ is a prime contracting to
$\mathfrak p$. Localizing at that prime gives a faithfully flat
local extension of $(S_0)_{\mathfrak p}$ with closed fiber a
field. As in Proposition~\ref{prop:symbolic}, contraction of its
squared maximal ideal gives $h\in\mathfrak p^2(S_0)_{\mathfrak p}$.
This proves~\eqref{eq:algebraic-differential-kernel}.

Since completion is flat, tensoring the kernel description with $R$
and using~\eqref{eq:completed-prime} gives
\begin{equation}\label{eq:completed-differential-kernel}
 \mathfrak p^{(2)}R
 =\ker\bigl(P\longrightarrow(R/P)^6,\quad
             h\longmapsto(\partial_i h\bmod P)_{i=0}^5\bigr).
\end{equation}
Here the tensor-extended map is indeed formal differentiation
modulo $P$: if $h=\sum_j a_jp_j$ with $p_j\in\mathfrak p$ and
$a_j\in R$, the terms $p_j\partial_i a_j$ vanish modulo $P$.
Proposition~\ref{prop:symbolic} identifies the kernel on the right
with a submodule of $P^{(2)}$. The reverse inclusion follows from
the same product-rule argument applied to $sh\in P^2$ with
$s\notin P$. This proves~\eqref{eq:symbolic-completion}.

Choose polynomials $g_1,\ldots,g_r$ in
$\Q[x_0,\ldots,x_5]$ that generate
$\mathfrak p^{(2)}$ as an $S_0$-ideal. Such generators are obtained
from any finite generating set by clearing denominators that are
units in $S_0$. By~\eqref{eq:symbolic-completion}, write the formal
witness as
\[
 f=\sum_{j=1}^r a_jg_j,\qquad a_j\in R.
\]
Choose polynomials $a_j^{\flat}$ with
$a_j-a_j^{\flat}\in\m^5$, and set
$g=\sum_j a_j^{\flat}g_j$. Then $g\in\mathfrak p^{(2)}$ and
$g-f\in\m^5$. In particular,
$g\equiv D+W\pmod{\m^5}$. Proposition~\ref{prop:nonmembership}
gives $g\notin\m P$. If $g$ belonged to
$\mathfrak n\mathfrak p$, its extension to $R$ would belong to
$\m P$, a contradiction. This proves~\eqref{eq:algebraic-witness}.
\end{proof}

\begin{remark}\label{rem:polynomial-witness}
Theorem~\ref{thm:algebraic} proves the existence of a polynomial
witness $g$ with the displayed fourth jet. Its proof approximates the
coefficients of a representation of $f$ in generators of the
algebraic symbolic square. It does not identify a degree bound,
display the higher coefficients of $g$, or assert that a finite
truncation of the particular formal series $f$ is a witness.
\end{remark}


\appendix
\section{Elementary finite certificates}\label{app:finite}

\subsection{The two finite-field irreducibility checks}
Let $\overline h=z^8-z-1\in\mathbf F_7[z]$. Repeated modular powering gives
\[
z^{7^8}\equiv z\pmod{\overline h},\qquad
z^{7^4}-z\equiv a_7\pmod{\overline h},
\]
where
\[
a_7=z^7-3z^6+2z^5+z^4-3z^3+2z^2.
\]
A B\'ezout certificate is
\[
A_7a_7+B_7\overline h=1\quad\text{in }\mathbf F_7[z],
\]
with
\begin{align*}
A_7&=3z^7+3z^5+3z^4-z^3+2z^2+z,\\
B_7&=-3z^6+2z^5-2z^4+3z^3-z^2+z-1.
\end{align*}
The finite-field irreducibility criterion proves that $\overline h$ is
irreducible: the only prime divisor of $8$ is $2$.

In $\mathbf F_{19}[z]$, put
\[
q_{19}=z^7+4z^6-3z^5+7z^4+9z^3-2z^2-8z+5.
\]
Then
\[
\overline h=(z-4)q_{19},\qquad
z^{19^7}\equiv z\pmod{q_{19}},
\]
and
\[
z^{19}-z\equiv a_{19}=z^5+2z^4+z^3-z\pmod{q_{19}}.
\]
Here $A_{19}a_{19}+B_{19}q_{19}=1$, where
\begin{align*}
A_{19}&=4z^6+2z^5-7z^4+4z^3+5z^2-4z+9,\\
B_{19}&=-4z^4+6z^3+z^2-7z+4.
\end{align*}
Since $7$ is prime, these identities prove that $q_{19}$ is irreducible.
Moreover, $q_{19}(4)=9$ in $\mathbf F_{19}$, so the displayed factorization
of $h$ is squarefree. All powers in these checks are computed by repeated
powering with reduction after each step.

\subsection{Rational reduction in the coefficient field}
For graded reverse lexicographic order with $U>V$, a Gr\"obner basis of
the ideal $(r_0,r_1)$ from \eqref{eq:r0r1} is
\begin{equation}\label{eq:groebner}
\begin{aligned}
g_0={}&U^7-3U^3V-U^3+4U^2V^3+6U^2V^2-UV^5-5UV^4+V^6,\\
g_1={}&U^6V-U^3+3U^2V^2+3U^2V-UV^4-4UV^3+V^5,\\
g_2={}&U^5V^2-U^6+2U^2V+U^2-UV^3-3UV^2+V^4,\\
g_3={}&U^4V^3-2U^5V+U^2-UV^2-2UV+V^3,\\
g_4={}&U^3V^4+U^5-3U^4V^2-UV-U+V^2,\\
g_5={}&U^2V^5+3U^4V-4U^3V^3-U+V,\\
g_6={}&UV^6-U^4+6U^3V^2-5U^2V^4+1,\\
g_7={}&V^7-4U^3V+10U^2V^3-6UV^5+1.
\end{aligned}
\end{equation}
Their leading monomials are precisely the monomials of total degree seven.
Thus $\Q[U,V]/(r_0,r_1)$ has dimension $28$, and its natural surjection
onto the degree-$28$ field $L$ is an isomorphism. This also gives the normal
form used for the identities involving $V_0-K$ and the $C_i$.

\subsection{Reproduction of the finite checks}
The supplementary Python program uses exact rational
arithmetic and finite-field arithmetic to check the displayed certificates.
It checks all $28$ S-polynomials of \eqref{eq:groebner}, verifies equality
of the ideal they generate with $(r_0,r_1)$, checks
\eqref{eq:correctionvalues} and all $210$ identities
\eqref{eq:lambdaquadratic}--\eqref{eq:lambdacubic}, and verifies the
quadratic and cubic Veronese relation-space dimensions. It also checks the
normal derivative formulas, the congruence $V_0\equiv K$, and the stated
homogeneous lift of $V_0$.

A separate Singular input in the supplementary material reconstructs
the resultant and checks its equality with the restriction of $W$, together
with the normal derivative identities and the nonvanishing of the tangential
Hessian determinant in $L$. These finite calculations use no floating-point
arithmetic and do not replace the all-orders argument in the body of the paper.


\section{Gradient interpolation in higher degrees}\label{app:gradient}

The proof of Theorem~\ref{thm:main} does not require the calculations
in this appendix. The purpose here is to describe a direct continuation procedure on
the fixed quadratic parametrization and provide additional finite
checks of the interpolation mechanism.

For $d\geq1$, define the $\Q$-linear map
\[
J_d:H_d\longrightarrow L^6,\qquad
E\longmapsto\bigl(\partial_0E(\ell),\ldots,\partial_5E(\ell)\bigr).
\]
Using the ordered basis \eqref{eq:basis} in each of the six coordinate
blocks, the target has dimension $168$ over $\Q$. Exact rational
row reduction gives
\begin{center}
\begin{tabular}{c r r r}
\toprule
$d$ & $5$ & $6$ & $7$\\
\midrule
$\dim_{\Q}H_d$ & $252$ & $462$ & $792$\\
$\rank_{\Q}J_d$ & $144$ & $160$ & $168$\\
\bottomrule
\end{tabular}
\end{center}
Here monomial columns are ordered lexicographically by their
six-tuples of exponents, starting with the smallest tuple. Each matrix entry is obtained by differentiating a
monomial, substituting $\ell$, and reducing by the Gr\"obner basis
in Appendix~\ref{app:finite}.

\begin{lemma}\label{lem:gradient-surjectivity}
If $J_7$ is surjective, then $J_d$ is surjective for every $d\geq7$.
\end{lemma}
\begin{proof}
For $G\in H_7$, Euler's identity and $\ell_0=1$ give
\[
J_d(x_0^{d-7}G)=T_dJ_7(G),\qquad
T_d(a)=a+\frac{d-7}{7}
\left(\sum_{j=0}^5\ell_j a_j\right)e_0.
\]
The $L$-linear map $T_d$ has determinant $d/7\ne0$, so it is
bijective. Surjectivity of $J_7$ over $\Q$ therefore implies that of
$J_d$.
\end{proof}

\begin{proposition}\label{prop:gradient-continuation}
Suppose $J_7$ is surjective and that forms $G_5\in H_5$ and $G_6\in H_6$
satisfy
\begin{equation}\label{eq:gradient-start}
\varphi\bigl(\partial_i(D+W+G_5+G_6)\bigr)
\equiv0\pmod{t^6}\qquad(0\leq i\leq5).
\end{equation}
Then there are forms $G_n\in H_n$ for all $n\geq7$ such that
\[
f=D+W+\sum_{n\geq5}G_n
\]
satisfies $\varphi(f)=\varphi(\partial_i f)=0$ for every $i$.
Consequently $f\in P^{(2)}\setminus\m P$.
\end{proposition}
\begin{proof}
Suppose that the gradient of the sum through degree $n-1$ vanishes
modulo $t^{n-1}$, where $n\geq7$. By
Lemma~\ref{lem:gradient-surjectivity}, choose $G_n\in H_n$ whose image
under $J_n$ is the negative of the remaining coefficient of $t^{n-1}$.
Since
\[
\nabla G_n(t\ell+t^2w)=t^{n-1}J_n(G_n)+O(t^n),
\]
this kills the next coefficient without changing the preceding ones.
The resulting sum converges in the $\m$-adic topology, and all six
derivatives belong to $P$. The chain rule gives
$\frac{d}{dt}\varphi(f)=0$; the constant term is zero, so
$\varphi(f)=0$ as we are in characteristic zero. Propositions~\ref{prop:symbolic}
and~\ref{prop:nonmembership}, applied through their proofs, give the
last assertion.
\end{proof}

The initial forms in \eqref{eq:gradient-start} exist by taking the
degree-five and degree-six parts of the series already constructed
in Sections~\ref{sec:critical} and~\ref{sec:symbolic}.
Thus the continuation statement is consistent with the main proof.
The rank calculations alone do not prove the initial lifting
condition, and no separate explicit coefficient lists for $G_5,G_6$
are needed or asserted here. This appendix is not used as an
independent proof of the initial lifting step.

\subsection*{Supplementary material}
The accompanying supplementary material consists of a Python program
and a Singular input, which carry out the finite checks described in
Appendix~\ref{app:finite}. The Python program requires only Python~3
with its standard library; the Singular input performs the resultant
check in Section~\ref{sec:curve}. The recorded runs used Python~3.14.7
and Singular~4.4.1. The checks concern the finite identities; the
existence of the full witness is proved by the formal arguments in the
text.
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\end{thebibliography}

\end{document}