Finite dimensional division algebra over $\mathbb{C}$ must be equal to $\mathbb{C}$ Let $A$ be a finite dimensional $\mathbb{C}$-algebra such that for any $0 \neq a \in A,$ there exists $b\in A$ such that $ab = ba =1.$ I want to show that $A = \mathbb{C}.$
First, let $U$ be a nonzero $A$-submodule of $A$. (So considering $A$ as a module over itself.) Take $0 \neq u \in U,$ then there exists $v \in A$ such that $uv = vu = 1,$ so $1 \in A.u \subseteq U,$ implying $A = U.$ In other words, $A$ is a simple $A$-module.
Now consider the $A$-endomorphism of $A$, i.e. a map $f \colon A \to A.$ By Schur's Lemma, $f \equiv \lambda \mathrm{Id}_A$ for some $\lambda \in \mathbb{C}.$ Hence $\mathrm{End}_A(A) \cong \mathbb{C}.$ But $A \cong (\mathrm{End}_A(A))^\mathrm{op} \cong \mathbb{C}^\mathrm{op} \cong \mathbb{C}$ as $\mathbb{C}$ is commutative (where $A^\mathrm{op}$ is the opposite algebra). Is this reasoning correct?

I know this has been answered but I can't help giving another proof as it is purely topological. 
The map $A^* \to A^*, z \mapsto z^2$ is a covering of degree $2$, where $A^* := A \backslash \{0\}$. Now the total space of this covering is connected so this covering can't be trivial. But the base space is simply connected so this covering should be trivial, contradiction.
