I want to find the number of non-isomorphic semisimple $\mathbb{C}$-algebras of dimension 8. I don't think this is meant to be that hard, but I'm not sure I'm doing it right.
From the Wedderburn theorem we know any semisimple $\mathbb{C}$-algebra is isomorphic to a direct sum of matrix algebras over division $\mathbb{C}$-algebras $D_i$, ie., $A \cong m_{n_1}(D_1) \oplus … \oplus M_{n_r}(D_r)$.
From this question, I can see the finite dimensional division $\mathbb{C}$-algebras are just $\mathbb{C}$, so $D_i = \mathbb{C}$ for all $i$. To find the possible semisimple $\mathbb{C}$-algebras of dimension 8, the only possibilities are:
- $n_i = 1$ for $i \in \{1, 2, …, 8 \}$
- $n_1 = 2, n_2 = n_3 = n_4 = n_5 = 1$
- $n_1 = 2, n_2 = 2$
Does this mean there are 3 isomorphism classses of semisimple $\mathbb{C}$-algebras? Also, obviously 1 and 2/3 are not isomorphic because 1 is commutative and the others aren't, but how do you show 2 and 3 aren't isomorphic?
Thank you
Best Answer
The center is a complex vector space of dimension $8$ in case $1$, dimension $5$ is case $2$, and of dimension $2$ in case $3$, so they are pairwise non isomorphic.