I'm working on understanding the following proof: https://dl.dropboxusercontent.com/u/17606191/proof.gif but I'm having some trouble understanding some of the author's terminology. We're asked to consider a commutative subspace F of D, where D is a division algebra. It's then shown that if x is in F, then so is x inverse, and so F is a field. Does subspace mean subalgebra here? Would any subalgebra also have to be a division algebra? Or could you have a subalgebra that includes an element but not it's inverse?
Thanks for any replies.
To answer your first question, yes, $F$ is a subalgebra of $D$. We see this because the author has asked for $F$ to be a maximal commutative subspace of $D$ that contains $\mathbb{R}\langle d \rangle$; since the author has insisted upon commutivity of $F$, we assume that (s)he is referring to the noncommutative operation in $D$; that is, we take $F$ to have the property that $f,g \in F$ if and only if $fg=gf$, $\mathbb{R}\langle d \rangle \subseteq F$, $1_F=1_D$, and $F\subseteq D$.
In order to answer your second and third questions, we must ask what axioms you require for a subset $B$ of an algebra $A$, when placed under the operations of $A$, to be a subalgebra. If you do not insist on a subalgebra $B$ containing the identity of $A$ (when $A$ is a unitary algebra), then the short answer is no, not every subalgebra of a division algebra is itself a division algebra. This is seen as follows:
Let us regard $\mathbb{Q}$, the field of rational numbers, as a $\mathbb{Z}$-algebra. Then it is easy to check that $\mathbb{Q}$ is a division ring (since fields are commutative division rings) and that $\mathbb{Q}$ is a $\mathbb{Z}$-algebra. Then let us note that $2\mathbb{Z}$ (the set of even integers) is a subring of $\mathbb{Q}$ without identity (to see this, look at the canonical ring monomorphism $\iota:2\mathbb{Z} \to \mathbb{Q}$). Then, since $2\mathbb{Z}$ is a $\mathbb{Z}$ ideal, it is clear that $2\mathbb{Z}$ is also a $\mathbb{Z}$-algebra. Thus $2\mathbb{Z}$ is a subalgebra of $\mathbb{Q}$ that is not a division ring.
However, if we insist that for a structure $A$ to be an algebra, it must be a unitary algebra, then any subalgebra $B \subseteq A$ has the property $1_B = 1_A$, where $1_A$ is the multiplicative unit of $A$. However, we may still find some subalgebras of division algebras that are not division algebras; for instance, let us consider $\mathbb{Q}$ again as a $\mathbb{Z}$-algebra, this time under the assumption that all (sub)algebras are unitary. Then we observe that $\mathbb{Z}$ is a subalgebra of $\mathbb{Q}$, since $1_{\mathbb{Z}}=1_{\mathbb{Q}}$. However, $\mathbb{Z}$ is not a division ring, and so we see that we may still construct a subalgebra of a division algebra that is not itself a division algebra.
So the answer to your questions is no, not every subalgebra of a division algebra is itself a division algebra. What really matters here is the commutative ring $K$ over which you define your algebra $A$, for that has a very, $very$ strong influence over the structure of $A$ itself. I'm sorry that this is rather long answer, but hopefully it helps you out and has been informative! Best of luck understanding the rest of the proof!