Finite dimensional division algebra over $\Bbb{C}$

Question

Another abstract algebra question from my university days that has me stumped at where to start!

I know what a division ring is and I think I understand what a division algebra over $\mathbb C$ is. (A division ring $D$ where there is an additional operation of scalar multiplication of elements of $D$ with elements of $\mathbb C$. I have come across this question and don't really know where to start. I'm not asking for a full proof, but a good starting point would be greatly appreciated!

Let $D$ be a finite-dimensional division algebra over $\mathbb C$. Show that $D=\mathbb C$.

Am I right in thinking that a finite-dimensial division algebra is one where $D$ is spanned by a certain (finite) number of (linearly independent?) elements of $D$?

Thanks in advance,

Andy.

Answer 1

Another way to start is to choose a basis $\{d_1,\ldots,d_n\}$ for $D$. Left multiplication by $d\in D$ is an linear transformation $D\to D$. Write $$d.d_i=\sum_j\lambda_{ji}d_j,\;\;\;\lambda_{ji}\in \mathbb{C}$$ This yields a matrix $(\lambda_{ji})$ whose characteristic polynomial satisfies $\chi(d)=0$. The polynomial splits into linear factors over $\mathbb{C}$ so $$0=\chi(d)=\prod_\mu(d-\mu)^{n_\mu}.$$ Since $D$ is a domain, we deduce that $d-\mu=0$ for some eigenvalue $\mu$. Hence $n=1$ and $D\cong \mathbb{C}$.

To connect this with Wedderburn-Artin, observe that $D$ is a simple $\mathbb{C}$-algebra since it has no nontrivial proper ideals. Now, semisimple $\mathbb{C}$ algebras are isomorphic to $$M_{n_1}(\mathbb{C})\oplus\cdots\oplus M_{n_k}(\mathbb{C}).$$ Of these, the simple ones satisfy $k=1$ (i.e. are isomorphic to $M_n(\mathbb{C}))$. Of those, the division algebras satisfy $n=1$.

Answer 2

A finite-dimensional division algebra over $\mathbb{C}$ is a division algebra $D$ over $\mathbb{C}$ which is finite-dimensional as a complex vector space. The way you'll use this hypothesis is the following: every $d \in D$ has the property that the elements $\{ 1, d, d^2, \dots \}$ must be linearly dependent, since there are infinitely many of them, so it follows that there is some nontrivial linear dependence between them. Equivalently, every $d \in D$ satisfies a polynomial $f(d) = 0$ with complex coefficients.

Now, since $\mathbb{C}$ is algebraically closed, this polynomial factors into linear factors...