Triangular Gröbner-basis conjecture for chemical reaction networks

A Counterexample to the Triangular Gröbner-Basis Conjecture for Chemical Reaction Networks

Proposition

The triangular Gröbner-basis conjecture of Ferrari et al. is false as stated.

Consider the chemical reaction network

$$2X \xrightarrow{\;2\;} X, \qquad 2Y \xrightarrow{\;2\;} Y, \qquad X+Y \xrightarrow{\;1\;} 0,$$

where the labels are the positive mass-action rate constants. This network satisfies the three assumptions in the conjecture, but no permutation of its two variables gives a reduced lexicographic Gröbner basis of the conjectured triangular form.

Proof

1. The steady-state ideal

The reaction vectors are

$$\nu_1=(-1,0), \qquad \nu_2=(0,-1), \qquad \nu_3=(-1,-1).$$

Under mass-action kinetics, the reaction-rate vector is

$$v(x,y)=(2x^2,\,2y^2,\,xy).$$

Hence the differential equations are

$$\dot x=-2x^2-xy, \qquad \dot y=-2y^2-xy.$$

Changing the signs of the steady-state equations does not change the ideal they generate. Thus the steady-state ideal is

$$I=(f_1,f_2) =(2x^2+xy,\,2y^2+xy) \subset \mathbb Q[x,y].$$

2. Verification of the conjecture's hypotheses

The network follows mass-action kinetics. Moreover, every reactant complex has total molecularity two, so all interactions are at most pairwise and the steady-state equations are quadratic.

The stoichiometric matrix is

$$N= \begin{pmatrix} -1&0&-1\\ 0&-1&-1 \end{pmatrix}.$$

It has rank two. Therefore its left kernel is zero, so the network has no nonzero linear conservation laws. The requirement that all conservation laws be included is consequently satisfied vacuously.

It remains to verify that the network has no nontrivial stoichiometrically independent decomposition. The three reaction vectors are pairwise nonparallel. In every partition of the three reactions into two nonempty subsets, one subset contains at least two reactions and therefore spans $\mathbb Q^2$. The other subset contains a nonzero reaction vector, so its stoichiometric subspace has nonzero intersection with $\mathbb Q^2$. The two subspaces therefore cannot form a direct sum.

Finally, the two steady-state equations are algebraically independent. Indeed,

$$\det \begin{pmatrix} \partial f_1/\partial x&\partial f_1/\partial y\\ \partial f_2/\partial x&\partial f_2/\partial y \end{pmatrix} = \det \begin{pmatrix} 4x+y&x\\ y&x+4y \end{pmatrix} =4(x^2+4xy+y^2),$$

which is not the zero polynomial.

3. Reduced Gröbner basis for $x>y$

For the lexicographic order $x>y$, set

$$g_1=x^2-y^2, \qquad g_2=xy+2y^2, \qquad g_3=y^3.$$

These polynomials generate $I$. In one direction,

$$g_1=\frac{f_1-f_2}{2}, \qquad g_2=f_2,$$

and

$$3g_3=y g_1+(2y-x)g_2,$$

so $(g_1,g_2,g_3)\subseteq I$. Conversely,

$$f_1=2g_1+g_2, \qquad f_2=g_2,$$

so $I\subseteq(g_1,g_2,g_3)$.

Their leading monomials are

$$\operatorname{LM}(g_1)=x^2, \qquad \operatorname{LM}(g_2)=xy, \qquad \operatorname{LM}(g_3)=y^3.$$

Buchberger's criterion gives

$$S(g_1,g_2) =y g_1-x g_2 =-y^3-2xy^2 \xrightarrow{\;g_2\;}3y^3 \xrightarrow{\;g_3\;}0,$$ $$S(g_1,g_3) =y^3g_1-x^2g_3 =-y^5 \xrightarrow{\;g_3\;}0,$$

and

$$S(g_2,g_3) =y^2g_2-xg_3 =2y^4 \xrightarrow{\;g_3\;}0.$$

Thus $\{g_1,g_2,g_3\}$ is a Gröbner basis. It is reduced: all three polynomials are monic, and the only nonleading monomial appearing in $g_1$ or $g_2$ is $y^2$, which is divisible by none of $x^2,xy,y^3$. Therefore the reduced lexicographic Gröbner basis is

$$G_{x>y} =\{x^2-y^2,\;xy+2y^2,\;y^3\}.$$

4. Reduced Gröbner basis for $y>x$

The ideal is invariant under exchanging $x$ and $y$. Hence, for the lexicographic order $y>x$, the reduced Gröbner basis is

$$G_{y>x} =\{y^2-x^2,\;xy+2x^2,\;x^3\}.$$

These are the only two variable orders.

For $x>y$, the conjectured form would be

$$\{x-h(y),\,q(y)\},$$

whereas the actual reduced basis is $G_{x>y}$. For $y>x$, the conjectured form would be

$$\{y-h(x),\,q(x)\},$$

whereas the actual reduced basis is $G_{y>x}$. Neither reduced basis has the required form.

5. Intrinsic obstruction

There is also a coordinate-independent obstruction. The equations

$$x(2x+y)=0, \qquad y(x+2y)=0$$

have only the solution $(0,0)$ over an algebraic closure: if both coordinates are nonzero, then

$$2x+y=0, \qquad x+2y=0,$$

whose coefficient matrix has determinant $3$, forcing $x=y=0$.

Thus

$$\sqrt I=(x,y).$$

Let

$$A=\mathbb Q[x,y]/I, \qquad \mathfrak m=(x,y)/I.$$

Since $I\subseteq(x,y)^2$,

$$\mathfrak m/\mathfrak m^2 \cong (x,y)/(x,y)^2,$$

and hence

$$\dim_{\mathbb Q}\mathfrak m/\mathfrak m^2=2.$$

Therefore the local scheme defined by $I$ has embedding dimension two at its unique point.

If a Gröbner basis of the conjectured form existed after a variable permutation, then eliminating the linearly occurring variable would give an isomorphism

$$A\cong \mathbb Q[t]/(q(t)).$$

Every localization of a one-variable quotient has embedding dimension at most one, contradicting the embedding dimension computed above. Thus no variable permutation can produce the conjectured triangular shape.

This proves that the network is a counterexample. $\square$