Past exam of the algebra course of the University of Oxford / 2008 / A / 1 / ii / Solution

A noncommutative algebra may have a commutative quotient: take $A=K\times M_2(K)$, $B=K$, and $\phi (\lambda ,M)=\lambda$. This is a surjective unital homomorphism. Neither requested change of commutativity can occur for an isomorphism, since an isomorphism transfers the identity $ab=ba$ in both directions.

For the matrix classifications, write $E_{ij}$ for a matrix unit and $N=E_{12}+E_{23}$. Then $N^2=E_{13}$ and $N^3=0$. The complete square-zero sets and commutativity results areIndeed $A_1$ is diagonal. In $A_2$ and $A_5$ the two off-diagonal generators have all products zero, so multiplication has the form $(x,u)(x^{\prime },u^{\prime })=(x{x}^{\prime },x{u}^{\prime }+x^{\prime }u)$. The diagonal entries of a square force $x=0$, after which the off-diagonal part squares to zero. In $A_3=K[N]$, a square-zero element first has scalar coefficient zero and then $(yN+z{N}^2{)}^2=y^2{N}^2$, forcing $y=0$. This argument works also in characteristic two. In $A_4$, a square first forces both diagonal parameters to vanish, leaving precisely $K{E}_{12}$. Noncommutativity follows from $E_{22}{E}_{12}=0$ but $E_{12}{E}_{22}=E_{12}$.

The coordinate map $xI+y{E}_{12}+z{E}_{13}\mapsto xI+y{E}_{13}+z{E}_{23}$ is an isomorphism $A_2\to A_5$, by the common multiplication formula. Every other pair is nonisomorphic: $A_4$ is the only noncommutative member, and among the commutative members the dimensions of the square-zero sets are respectively zero, two, one, and two. These sets are vector spaces in these examples, and an algebra isomorphism must carry one onto the other. Thus the only distinct isomorphic pair is $A_2\cong A_5$.