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?
 A: 
By Schur's lemma, $f=\lambda \mathrm{Id}_A$ fr some $\lambda \in \mathbb C$.

No, it seems to me that in this step you are begging the question by referencing a piece of Schur's lemma that depends on what you are trying to prove. Schur's lemma definitely says that this is true for some $\lambda \in End_A(A)$, but then you need the result you are proving to conclude that this is equal to $\mathbb C$.
This should not be a big deal to prove anyway, since you presumably have the fundamental theorem of algebra at hand. Let $\alpha\in A\setminus\mathbb C$ and consider the field extension $\mathbb C\subseteq \mathbb C[\alpha]$. Because the extension is finite dimensional, $\alpha$ satisfies a minimal polynomial over $\mathbb C$. But the FTA says this polynomial factors into linear factors, one of which is $x-\alpha$. So the minimal polynomial must be $x-\alpha$ itself, and so $\alpha\in \mathbb C$ and we have reached a contradiction. So the conclusion is that $A=\mathbb C$ already.
The reasoning here applies more generally to finite-dimensional division ring extensions of algebraically closed fields. To properly extend an algebraically closed field, you will have to use something that is transcendental over the algebraically closed field.

Hitting it with Artin-Wedderburn unfortunately does not solve the problem. You've been told it is a division ring, so in particular its Wedderburn decomposition cannot be a product of more than one matrix ring, and it has to be a matrix ring of side-length $1$, because otherwise there would be zero divisors. At this point you've discovered that your ring is a finite dimensional division ring extension of $\mathbb C$: and you're right back at the problem you need to solve: why is a finite dimensional division ring extension of $\mathbb C$ equal to $\mathbb C$?
A: 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.
