Show that the center of a division ring is a field.

Question

I tried the proof, but wasn't able to proceed further: let a and b be any elements belonging to Z(center), Now a.b=b.a hence Z is commutative.

Now proving that Z is a division ring:(then we can show that since every commutative division ring is field, Z is a field) NO CLUE...

Answer 1

Hopefully you can already prove:

The center of a ring is a ring, in fact, a commutative ring.

Furthermore, if $D$ is a division ring, then for all $x\in Z(D)$, if $x\neq 0$, then $x^{-1}$ exists somewhere in $D$.

Now to show the commutative ring $Z(D)$ is a field, you'd have to show that $x^{-1}\in Z(D)$, because inverses are unique, and uniqueness of inverses in a division ring means you only have that candidate for the inverse in $Z(D)$.

So, the task is clear: if $0\neq x\in Z(D)$, prove $x^{-1}\in Z(D)$.

To prove a nonzero element $a$ commutes with all nonzero elements $b$, you can just check that $a^{-1}ba=b$ for all $b$.

Well, what do you think about $(x^{-1})^{-1}bx^{-1}=xbx^{-1}$?

Answer 2

Let me just extend and complete the above answer.

It is easy to show that $Z(D)$ is a commutative subring of $D$. It suffices to show that if $x \neq 0, x \in Z(D),$ then $x^{-1} \in Z(D)$, that is, every non-zero element in $Z(D)$ has a multiplicative inverse.

Clearly, as $D$ is a division ring, for any $0 \neq x \in D$, $\exists$ an element y such that $xy = yx = 1$. (In other words, every non-zero element in $D$ is a unit). Let such an $x \in Z(D)$. We have to show that $y$ commutes with every element in $D$, so that $y \in Z(D)$. Indeed, we have for all $r \in D$:

$$1\cdot r = r \cdot 1$$ $$\\\implies (xy)r = r(xy) \;\;\;\;\; (as \space xy = 1)$$ $$\implies (xy)r = (xr)y \;\;\;\;\; (as \space x \in Z(D), so \space xr = rx \space \forall \space r \in D)$$ $$\implies yr = ry, \space \forall \space r \in D \;\;\;\;\; (as \space x \neq 0)$$ $$\implies y \in Z(D)$$ Hence, every non-zero element in $Z(D)$ is a unit, which completes the proof.