4

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...

Rohan Marwah
  • 693

2 Answers2

6

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 $(x^{-1})^{-1}bx^{-1}=xbx^{-1}$?

rschwieb
  • 160,592
  • 4
    I'd really love to know how OP was helped by this answer. – Junglemath May 09 '20 at 08:35
  • @Junglemath what's funnier is that OP accepted it – Aram Nazaryan Nov 28 '23 at 15:47
  • @Junglemath I think framing the only hard part of a proof (that the center is closed under inverses) in such a way that the user only has to mechanically verify something (see if $x^{-1}$ commutes with everything, i.e. does $x^{-1}yx=y$ for all $y$?) is among the most straightforward help one can give for a problem without actually doing all the work. – rschwieb Nov 28 '23 at 16:13
  • @rschwieb Your edit makes it clear and can be considered a (good and) complete answer. However, your answer before the edit doesn't really help other than to reframe the question in different words. Thanks for the helpful edit. – Junglemath Nov 29 '23 at 02:11
0

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.

Rajdeep
  • 444
  • This could be a weird question, could we conclude $yr=ry$ when $x(yr)=x(ry)$ even if we were not on a division ring even though if $r\neq0$? Or can we do the cancellation only because we are on a division ring and $r\neq0$? – Zek Nov 11 '23 at 23:41
  • @Zek Yes, we could have concluded $x(yr) = x(ry) \implies yr = ry$ if $x \neq 0$ even if we weren't on a division ring (say, we were on an integral domain). In particular, if the ring has no zero divisors, then the multiplicative cancellation law holds (note that the converse is also true). – Rajdeep Nov 12 '23 at 04:55
  • @Zek What's important is that $x \neq 0$. It doesn't matter what $r$ is. – Junglemath Nov 29 '23 at 02:14