Gerko's conjecture on semidualizing dimensions

A cyclic counterexample to Gerko’s conjecture on semidualizing dimensions

The conjecture is false.

We construct a commutative Artinian local ring (S), a semidualizing (S)-module (C), and a cyclic (S)-module (N) such that

$$\mathrm{G}\text{-}\dim_S N=0$$

but

$$\mathrm{G}_C\text{-}\dim_S N=\infty.$$

The mechanism is simple: a rank defect between an (A)-linear matrix and its matrix transpose is converted into persistent Ext by dual numbers and idealization.

1. The asymmetric seed

Let

$$k=\mathbf F_2.$$

Define

$$A=k\oplus V\oplus W,$$

where

$$V=\langle a,b,c,d,e\rangle_k$$

and

$$W=\langle p,q,r,s\rangle_k.$$

Set

$$VW=W^2=0,$$

and declare the following to be the only nonzero products between the displayed basis elements of (V):

$$ab=ce=p,$$ $$ad=bc=q,$$ $$b^2=r+s,$$ $$cd=r,$$

and

$$ae=s.$$

All unspecified products in (V) are zero.

Since

$$V^2\subseteq W$$

and

$$(V\oplus W)W=0,$$

the multiplication is associative. Moreover,

$$\mathfrak m_A=V\oplus W$$

is nilpotent and

$$A/\mathfrak m_A\cong k.$$

Thus (A) is an Artinian local (k)-algebra with

$$\dim_kA=10.$$

Define an (A)-linear map

$$D:A^2\longrightarrow A^2$$

by

$$D(u,v)=\bigl(au+bv,;eu+(a+d)v\bigr).$$

Its matrix transpose over (A) is the map

$$D^{\mathsf T}(u,v) \bigl(au+ev,;bu+(a+d)v\bigr).$$

Here (D^{\mathsf T}) means the transpose of the (2\times2) matrix over (A); it is not the ordinary transpose of the resulting (20\times20) matrix over (k).

A direct calculation from the multiplication table gives

$$D^2=0$$

and

$$(D^{\mathsf T})^2=0.$$

The decisive ranks are

$$\operatorname{rank}_kD=10$$

and

$$\operatorname{rank}_kD^{\mathsf T}=8.$$

To verify them, give (A) the grading

$$A_0=k,\qquad A_1=V,\qquad A_2=W.$$

The map

$$D:A_0^2\longrightarrow A_1^2$$

has rank (2). On (A_1^2=V^2), the following images give the standard basis of (W^2):

$$D(a,0)=(0,s),$$ $$D(b,0)=(p,0),$$ $$D(c,0)=(0,p),$$ $$D(d,0)=(q,0),$$ $$D(e,0)=(s,0),$$ $$D(0,a)+D(b,0)=(0,q),$$ $$D(0,b)+D(e,0)+D(c,0)=(r,0),$$

and

$$D(0,c)+D(d,0)=(0,r).$$

Thus (D:V^2\to W^2) has rank (8). Since (D(W^2)=0),

$$\operatorname{rank}_kD=2+8=10.$$

For (D^{\mathsf T}), the degree-zero contribution again has rank (2). Every vector

$$D^{\mathsf T}(u,v) \bigl(au+ev,;bu+(a+d)v\bigr), \qquad u,v\in V,$$

satisfies two independent linear conditions in (W^2):

the (r)-coefficient of the first component is zero, and the (p)-coefficient of the first component equals the (r)-coefficient of the second component.

Hence the image of (V^2) has dimension at most (6). It has dimension exactly (6), since it contains the six independent vectors

$$D^{\mathsf T}(a,0)=(0,p),$$ $$D^{\mathsf T}(c,0)=(0,q),$$ $$D^{\mathsf T}(d,0)=(q,0),$$ $$D^{\mathsf T}(e,0)=(s,0),$$ $$D^{\mathsf T}(0,c)=(p,r),$$

and

$$D^{\mathsf T}(b,0)+D^{\mathsf T}(0,c)=(0,s).$$

Therefore

$$\operatorname{rank}_kD^{\mathsf T}=2+6=8.$$

2. Turning the matrix into one scalar

Let

$$B=A[\tau]/(\tau^2).$$

Make

$$M=A^2$$

into a (B)-module by letting (A) act componentwise and letting (\tau) act as (D):

$$\tau m=Dm.$$

This is well defined because (D^2=0).

Form the idealization

$$S=B\ltimes M,$$

with multiplication

$$(b_1,m_1)(b_2,m_2) \bigl(b_1b_2,;b_1m_2+b_2m_1\bigr).$$

Then (S) is a commutative Artinian local ring. As an (A)-module,

$$S\cong B\oplus M\cong A^4,$$

so

$$\dim_kS=40.$$

Set

$$x=(\tau,0)\in S.$$

On the decomposition (S=B\oplus M), multiplication by (x) is

$$x|S=L\tau^B\oplus D.$$

Multiplication by (\tau) on (B=A\oplus A\tau) has rank (10). Hence

$$\operatorname{rank}_k(x|_S) 10+\operatorname{rank}_kD$$

Also,

$$x^2=0.$$

Now let

$$\omega=D_k(A)=\operatorname{Hom}_k(A,k)$$

and set

$$C=S\otimes_A\omega.$$

For every (A)-module (L), there is a natural isomorphism

$$\operatorname{Hom}_A(L,\omega) \cong \operatorname{Hom}_k(L,k).$$

Thus (\omega) is injective over (A). Moreover,

$$\operatorname{End}_A(\omega) \cong A^{\mathrm{op}} A,$$

and this is the homothety isomorphism. Therefore (\omega) is semidualizing over (A).

Since (S) is free over (A), the module

$$C=S\otimes_A\omega$$

is semidualizing over (S). Indeed, if (F_\bullet\to\omega) is a degreewise finite free (A)-resolution, then (S\otimes_AF_\bullet\to C) is an (S)-free resolution and

$$\operatorname{Hom}S \bigl(S\otimes_AF\bullet,C\bigr) \cong S\otimes_A \operatorname{Hom}A(F\bullet,\omega).$$

It follows that

$$\operatorname{Ext}_S^i(C,C)=0 \qquad \text{for all }i>0,$$

and that the homothety map

$$S\longrightarrow\operatorname{End}_S(C)$$

is an isomorphism.

As an (A)-module,

$$C \cong (\omega\otimes_AB) \oplus (\omega\otimes_AM).$$

On the first summand, multiplication by (x) has rank (10). On the second summand, it is

$$1_\omega\otimes_AD.$$

If (\rho(u)) denotes the regular multiplication matrix of (u\in A), then (u) acts on (\omega=A^*) by

$$\rho(u)^{\mathsf T}.$$

Consequently, the full (k)-matrix of (1_\omega\otimes_AD) is the ordinary transpose of the full regular (k)-matrix associated to the (A)-matrix transpose (D^{\mathsf T}). Therefore

$$\operatorname{rank}k(1\omega\otimes_AD) \operatorname{rank}_kD^{\mathsf T}$$

Hence

$$\operatorname{rank}_k(x|_C) 10+\operatorname{rank}_kD^{\mathsf T}$$

3. The rank defect becomes homology

For any finite-dimensional (k)-space (L) on which (x^2=0), define

$$H_x(L) \frac{\ker(x:L\to L)} {\operatorname{im}(x:L\to L)}.$$

Since (\operatorname{im}x\subseteq\ker x),

$$\dim_kH_x(L) \dim_kL 2\operatorname{rank}_k(x|_L).$$

For (S),

$$\dim_kH_x(S) 40-2\cdot20$$

Thus

$$\ker(x:S\to S) \operatorname{im}(x:S\to S),$$

or equivalently,

$$(0:_Sx)=xS.$$

For (C),

$$\dim_kH_x(C) 40-2\cdot18$$

Equivalently,

$$\dim_k\frac{(0:_Cx)}{xC}=4.$$

The entire numerical mechanism can therefore be written as

$$\dim_kH_x(C) 2\left( \operatorname{rank}_kD \operatorname{rank}_kD^{\mathsf T} \right) 2(10-8)$$

Thus the same square-zero scalar (x) is exact on (S), but has four-dimensional homology on the semidualizing module (C).

4. The counterexample

Set

$$N=S/xS.$$

Since

$$(0:_Sx)=xS,$$

the period-one complex

$$\cdots \xrightarrow{x}S \xrightarrow{x}S \xrightarrow{x}S \xrightarrow{x}S \longrightarrow N \longrightarrow0$$

is exact.

Its (S)-dual is the same exact complex. Hence the associated two-sided periodic free complex is totally acyclic. Therefore (N) is totally reflexive and

$$\mathrm{G}\text{-}\dim_S N=0.$$

Applying (\operatorname{Hom}_S(-,C)) gives the period-one cochain complex

$$0 \longrightarrow C \xrightarrow{x}C \xrightarrow{x}C \xrightarrow{x}C \xrightarrow{x}\cdots.$$

For every (n\geq1),

$$\operatorname{Ext}_S^n(N,C) \cong H_x(C).$$

Hence

$$\dim_k\operatorname{Ext}_S^n(N,C)=4 \qquad \text{for every }n\geq1.$$

Suppose that

$$\mathrm{G}_C\text{-}\dim_S N=d<\infty.$$

Then (N) admits a length-(d) resolution by totally (C)-reflexive modules. Every totally (C)-reflexive module (G) satisfies

$$\operatorname{Ext}_S^i(G,C)=0 \qquad \text{for every }i>0.$$

Dimension shifting would therefore give

$$\operatorname{Ext}_S^n(N,C)=0 \qquad \text{for every }n>d.$$

This contradicts

$$\operatorname{Ext}_S^n(N,C)\cong H_x(C)\neq0 \qquad \text{for every }n\geq1.$$

Consequently,

$$\mathrm{G}_C\text{-}\dim_S N=\infty.$$

We have therefore constructed

$$\mathrm{G}\text{-}\dim_S N=0$$

and

$$\mathrm{G}_C\text{-}\dim_S N=\infty.$$

Thus Gerko’s conjectured inequality

$$\mathrm{G}_C\text{-}\dim_S N \leq \mathrm{G}\text{-}\dim_S N$$

would read

$$\infty\leq0.$$

Therefore Gerko’s conjecture on semidualizing dimensions is false.

Equivalently,

$$N\in\mathcal G_S$$

but

$$N\notin\mathcal G_C.$$

Hence

$$\mathcal G_S\nsubseteq\mathcal G_C,$$

so (\mathcal G_S) cannot be a thick subcategory of (\mathcal G_C).

The finite certificate consists only of

$$D^2=0, \qquad \operatorname{rank}_kD=10, \qquad \operatorname{rank}_kD^{\mathsf T}=8.$$

Dual numbers and idealization convert the rank defect

$$10-8=2$$

into four-dimensional persistent Ext:

$$\operatorname{Ext}_S^n(N,C)\cong k^4 \qquad \text{for every }n\geq1.$$