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?
