# Using pc algebra to bridle if a household of algebras are pair-wise non-isomorphic Answer

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Using pc algebra to bridle if a household of algebras are pair-wise non-isomorphic

Given an innumerable bailiwick $$ok$$, deem a quiver $$Gamma$$ with one vertex and two arrows $$x,y$$ and outline $$R=okGamma/(x,y)^2.$$ This is a three-dimensional $$ok$$-algebra.

Now deem the additive group of two by 2 matrices over $$R$$ denoted by $$M_2(R)$$. To outline the multiplication, deem matrices $$X_t=commence{bmatrix} x & ty y & x aim{bmatrix}$$ with $$tin ok$$. Then we will insert a multiplication $$Acdot B= AX_tB,$$ the place on the privilege we have now the basic matrix product. We will convene $$M_2(R)$$ with this multiplication $$H_t$$. For any $$tin ok$$ the $$H_t$$ is a non-unital $$ok$$-algebra. I’m all in favour of whether or not these algebras are pairwise non-isomorphic.

I’m cognizant of the 1-parameter household of non-isomorphic algebras $$ok{x,y}/(x,y,xy-tyx)$$. For this household, it’s workable to brute-force the proof of the truth that they’re truly non-isomorphic. By the route, if any of you’re cognizant of another methods of proving this reality, delight let me know!

Now the dimension of $$H_t$$ is 12, so taking a look at all linear maps between these algebras and checking whether or not they’re homomorphisms by hand appears not possible. I too could not provide you with any high-brow arguments for why these algebras should breathe non-isomorphic. I did, nonetheless, provide you with the next:

Denote the matrix with a $$1$$ within the $$(i,j)$$ place and 0 in every single place else by $$E_{ij}$$, and say we have now a linear map $$F: H_tto H_m$$.If we would like $$F$$ to breathe a homomorphism, we should 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 solely linear combos of $$x, y$$ as components is annihilated by some other component of $$H_t$$, so with out lack of generality, we could occupy that $$F(E_{ij})$$ is a linear mixture of $$E_{ij}$$. So if it’s a homomorphism, $$F$$ is totally outlined by the pictures of $$E_{ij}$$, that are every outlined by 4 scalars.

When we write $$F(E_{ij})$$ as linear combos of $$E_{ij}$$, the above system of two equations turns into 16 homogeneous equations of organize 2 when it comes to 12 scalar variables with $$t$$ and $$m$$ as parameters. If we might display this method is incongruous (for common $$t$$ and $$m$$), we might have that there are literally no non-zero homomorphisms $$H_t to H_m$$. This is too the root of my speculation that these algebras are non-isomorphic: every homomorphism is outlined by 16 scalars matter to a seemingly very sizable system of homogeneous equations of second organize.

To display that this method is incongruous I made a decision to choose a subsystem of 12 equations in 12 variables and compute the Macaulay resultant (or bridle if it is nonzero, which might denote that it is a polynomial in $$t,m$$ and we’d don’t have any homomorphisms for “generic” $$t,m$$). I attempted doing it in Macaulay2 utilizing 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 ought to give me a matrix of complete row-rank if and provided that the resultant is zero. Here the variables in $$l$$ are simply the coordinates of the $$F(E_{ij})$$ within the foundation of $$E_{ij}$$. Now I do know that the complexity of this code is exponential within the variety of variables and am beginning to doubt that this code will ever terminate executing.

I’m too very uncommon with Macaulay2, that is solely my second time utilizing it. This code has been operating for greater than 24 hours now. So now I lastly come to the query: is the above code inefficient? Or are there other ways to bridle if the resultant is zero? Or hopefully plane a non-computational route of fixing this drawback?

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