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Let $D$ be a division ring/algebra. I'm confused about what are the generators of $D$.

On one hand, the multiplicative group $D^\times$ is generated by some generators $S$ as a group. Most algebra textbooks carefully defines this process. But at this point, we have all the elements of $D$ other than zero.

On the other hand, there is a general categorical result asserting every ring is the quotient of a free ring, which is the $\mathbb{Z}$-polynomials of words over some set, with juxtaposition as the multiplication. Say the generating set is $S$ and the free ring is $FR(S)$. Then $D\simeq FR(S)/I$, where $I$ is some ideal.

The difference of these two generating processes occurs at the inverses. In the group generating process, $S^{-1}$ is simply a set with the same cardinality as $S$. Whereas, in the quotient of free ring construction, I'm not sure how the inverses are defined from $S$.

Can someone please guide me through the quotient ring method, and show how the inverses are created? Book reference is welcome too. Thanks in advance.

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When defining a group presentation, you work in the category of groups, so you want to force your construction to have inverses.

When defining an algebra presentation, you work in the category of algebras, where elements usually don't have inverses. So they are not hardwired in the construction. You have to check yourself, for a given presentation, if the resulting algebra is a division algebra or not (it might not be easy to do in practice).

A closer parallel would be to talk about a presentation of monoid (where a monoid is given as a quotient of a free monoid by some congruence), and then check whether the given presentation defines a group (and not just a monoid).

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  • $\begingroup$ Thanks for the answer. After some thought, is it correct to say for $D\simeq FR(S)/I$, it is equivalent to finding a maximal ideal $I$ that is also a maximal left/right ideal? $\endgroup$
    – user760
    May 15 at 13:22
  • $\begingroup$ And, in general, if $R\simeq FR(S)/I$, where $\pi: FR(S)\to R$ is the canonical projection, then $u\in R^\times$ is a unit iff for every $u^*\in \pi^{-1}(u)$, the subring of the free ring $\langle\{1\}\cup \{u^*\}\cup I\rangle$ has no proper one-sided ideals containing $I$ $\endgroup$
    – user760
    May 15 at 17:17

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