4
$\begingroup$

Given an infinite field $k$, consider a quiver $\Gamma$ with one vertex and two arrows $x,y$ and define $R=k\Gamma/(x,y)^2.$ This is a three-dimensional $k$-algebra.

Now consider the additive group of 2 by 2 matrices over $R$ denoted by $M_2(R)$. To define the multiplication, consider matrices $$X_t=\begin{bmatrix} x & ty \\ y & x \end{bmatrix}$$ with $t\in k$. Then we can introduce a multiplication $A\cdot B= AX_tB,$ where on the right we have the classic matrix product. We will call $M_2(R)$ with this multiplication $H_t$. For any $t\in k$ the $H_t$ is a non-unital $k$-algebra. I am interested in whether these algebras are pairwise non-isomorphic.

I am aware of the 1-parameter family of non-isomorphic algebras $k\{x,y\}/(x,y,xy-tyx)$. For this family, it is possible to brute-force the proof of the fact that they are actually non-isomorphic. By the way, if any of you are aware of some other ways of proving this fact, please let me know!

Now the dimension of $H_t$ is 12, so looking at all linear maps between these algebras and checking whether they are homomorphisms by hand seems impossible. I also couldn't come up with any high-brow arguments for why these algebras must be non-isomorphic. I did, however, come up with the following:

Denote the matrix with a $1$ in the $(i,j)$ position and 0 everywhere else by $E_{ij}$, and say we have a linear map $F: H_t\to H_m$.If we want $F$ to be a homomorphism, we must have $$F(E_{11})F(E_{21})=tF(E_{12})F(E_{11}),\ F(E_{11})F(E_{21})=tF(E_{12})F(E_{11}).$$

Notice that any matrix with only linear combinations of $x,\ y$ as elements is annihilated by any other element of $H_t$, so without loss of generality, we may assume that $F(E_{ij})$ is a linear combination of $E_{ij}$. So if it is a homomorphism, $F$ is completely defined by the images of $E_{ij}$, which are each defined by 4 scalars.

When we write $F(E_{ij})$ as linear combinations of $E_{ij}$, the above system of 2 equations turns into 16 homogeneous equations of order 2 in terms of 12 scalar variables with $t$ and $m$ as parameters. If we could show this system is inconsistent (for general $t$ and $m$), we would have that there are actually no non-zero homomorphisms $H_t \to H_m$. This is also the source of my hypothesis that these algebras are non-isomorphic: each homomorphism is defined by 16 scalars subject to a seemingly very large system of homogeneous equations of second order.

To show that this system is inconsistent I decided to pick a subsystem of 12 equations in 12 variables and compute the Macaulay resultant (or check if it's nonzero, which would mean that it's a polynomial in $t,m$ and we would have no homomorphisms for "generic" $t,m$). I tried doing it in Macaulay2 using this code:

loadPackage "EliminationMatrices"
R= QQ[t,m,a,b,c,d,u,v,w,r,p,q,z,s]
l={a,b,c,d,u,v,w,r,p,q,z,s}
f1=u*p+v*z-t*(a*u+b*w)
f2=u*s+v*q-t*(a*v+b*r)
f3=w*p+r*z-t*(c*u+d*w)
f4=w*s+r*q-t*(c*v+d*r)
f5=v*p+m*u*z-t*(b*u+m*a*w)
f6=v*s+m*u*q-t*(b*v-m*a*r)
f7=r*p+m*w*z-t*(d*u+m*c*w)
f8=t*s+m*w*q-t*(d*v+m*c*r)
f9=u*u+v*w-a*p-b*z
f10=u*v+v*r-a*s-b*q
f11=w*u+r*w-c*p-d*z
f12=w*v+r^2-c*s-d*q
M=matrix{{f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12}}
MR=elimanationMatrix(l,M)

This should give me a matrix of full row-rank if and only if the resultant is zero. Here the variables in $l$ are just the coordinates of the $F(E_{ij})$ in the basis of $E_{ij}$. Now I know that the complexity of this code is exponential in the number of variables and am starting to doubt that this code will ever finish executing.

I am also very unfamiliar with Macaulay2, this is only my second time using it. This code has been running for more than 24 hours now. So now I finally come to the question: is the above code inefficient? Or are there alternative ways to check if the resultant is zero? Or hopefully even a non-computational way of solving this problem?

$\endgroup$
6
  • 1
    $\begingroup$ You are looking at an example of a Rees matrix ring sciencedirect.com/science/article/pii/002186938390193X over the contracted monoid algebra of the four element monoid 1,x,y,0 with all products not involving 1 equal to zero. For Rees matrix semigroups there are characterizations of isomorphism but I am not so sure in this context $\endgroup$ Jan 13 at 17:45
  • $\begingroup$ The obvious guess is the algebras are isomorphic iff there are invertible 2x2 matrices P,Q over R and an automorphism $\phi$ of $R$ with $P\phi(X_t)Q=X_m$. At least your is what happens in the semigroup setting $\endgroup$ Jan 13 at 17:48
  • $\begingroup$ In the semigroup case you would look at what the isomorphism does to multiples of matrix units. But in your setting this may not work because there are no units in the matrix $X_t$ which we usually want in the semigroup setting $\endgroup$ Jan 13 at 17:51
  • 1
    $\begingroup$ There's a practical computer-algebra way to determine whether (a) they're all isomorphic up to finitely many exceptions or (b) there "(bounded finite)-to-one" non-isomorphic, i.e., there exists $n$ such that for each $t$ the set of $s$ such that $A_s\simeq A_t$ has cardinal $\le n$. See mathoverflow.net/questions/378149/… $\endgroup$
    – YCor
    Jan 13 at 19:04
  • $\begingroup$ @YCor Thank you! That was exactly what I was looking for. By this criterion, this is a trivial deformation after all. $\endgroup$ Jan 15 at 14:04
1
$\begingroup$

There's a practical computer-algebra way to determine whether

(a) they're all isomorphic up to finitely many exceptions, or

(b) there "(bounded finite)-to-one" non-isomorphic, i.e., there exists $n$ such that for each $t$ the set of $s$ such that $A_s\simeq A_t$ has cardinal $\le n$.

See this question.

By practical, I mean it remains doable in reasonable time for algebras of higher dimension, say $\le 100$.

$\endgroup$
2
$\begingroup$

For this and other Macaulay2-related questions, I highly recommend the Macaulay2 google group.

In general, there are some "exploratory" techniques (i.e. not quite a proof) which are more efficient and may indicate whether your original approach will yield the desired outcome. For your example, I think it makes sense for starters to work with coefficients over a large prime field and plug in random constants for $m$ and $t.$

In the example below, I take $I$ to be the ideal $\langle f_1, \ldots , f_{12}, f_{13} \rangle $ in the polynomial ring $\mathbb{F}_{10007}[a,b,c,d,u,v,w,r,p,q,z,s],$ where $f_{13}$ is a random inhomogeneous polynomial of degree $1$ (this can be interpreted as random chart on the projective space $\mathbb{P}^{11}.$) The computation that $\dim I = 3$ takes about 1s on my laptop. This suggests (but doesn't quite prove) that $V_{\mathbb{C} \, \mathbb{P}^{11}} (f_1, \ldots , f_{12}) \ne \emptyset .$ Your conjecture may still be correct---as you mention, this is only a subsystem of the one you've derived...

pr = 10007
FF = ZZ/pr
t = random FF
m = random FF
R = FF[a,b,c,d,u,v,w,r,p,q,z,s]
f1 = u*p+v*z-t*(a*u+b*w)
f2 = u*s+v*q-t*(a*v+b*r)
f3 = w*p+r*z-t*(c*u+d*w)
f4 = w*s+r*q-t*(c*v+d*r)
f5 = v*p+m*u*z-t*(b*u+m*a*w)
f6 = v*s+m*u*q-t*(b*v-m*a*r)
f7 = r*p+m*w*z-t*(d*u+m*c*w)
f8 = t*s+m*w*q-t*(d*v+m*c*r)
f9 = u*u+v*w-a*p-b*z
f10 = u*v+v*r-a*s-b*q
f11 = w*u+r*w-c*p-d*z
f12 = w*v+r^2-c*s-d*q
f13 = random(1,R)-1
G = groebnerBasis(ideal(f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12,f13), Strategy=>"F4");
dim ideal leadTerm G
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.