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While looking at proofs that every finite integral domain is also a field (or merely a division ring according to Jacobson), I notice the arguments all start with something like the following.

given the ring is finite, there can only be finitely many powers $a, a^2, a^3, \dots$ before they repeat

This is certainly true, but it apparently neglects the additive group of the ring; one can just as easily argue there are finite many factors $a, 2a ,3a, \dots$ before they repeat. So, what does the "typical" element of a finite ring look like?

My bet would be a linear combination like $n_1a^{m_1} + n_2a^{m_2} + \cdots + n_ka^{m_k}$. If this is the case, however, it's not clear to me how the above argument works. So I must misunderstand one of the two.

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    $\begingroup$ You can certainly also look at elements of the form $a,2a,3a,\ldots$. You can then conclude that there exists an integer $n$ such that $na=0$. So which way you go depends on what you want to prove. Your more general recipe works, if you look at elements of the subring generated by a single element $a$. That would be good for proving that the said subring is commutative. Again depending what you need to conclude. In a general finite ring that won't cover all, consider rings of matrices over a finite field for examples when you won't get the entire ring with a single $a$. $\endgroup$ Commented Apr 24, 2017 at 5:05
  • $\begingroup$ Yes. I should have made it more clear that this is really confined to a single generator. Yet, of course, the theorem (every finite id is a field) is not. $\endgroup$
    – JMJ
    Commented Apr 24, 2017 at 5:14
  • $\begingroup$ The claim is just that the list $a,a^2,a^3\ldots$ repeats; it's not claiming that all elements of the ring appear in that list. So the argument is not inconsistent with the existence of other elements of the ring that "look different". $\endgroup$
    – stewbasic
    Commented Apr 24, 2017 at 5:33
  • $\begingroup$ Yes, I understand now. $\endgroup$
    – JMJ
    Commented Apr 24, 2017 at 5:35

1 Answer 1

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Let's stick to commutative rings with unity.

The argument that the sequence $(na)$ repeats is not neglected. When $a=1_R$ we find that there is a least positive integer with $n1_R=0$. This is the characteristic of the ring, and we can regard $R$ as containing $\Bbb Z/n\Bbb Z$.

Your image of the typical ring as consisting of the polynomials in a fixed element $a$ is not accurate. There may be no such $a$ one can use. As an example consider the quotient ring $R=\Bbb Z/2\Bbb Z[X,Y]/(X^2,XY,Y^2)$. This consists of all elements $r+sx+ty$ where $r$, $s$, $t\in \Bbb Z/2\Bbb Z$ and $x^2=xy=y^2=0$. Then for any $a$ you take in $R$, there are at most $4$ elements of $R$ you can express as polynomials in $a$ with integer coefficients.

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