# Form of a typical element in a finite ring

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.

• 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 $a$. Commented Apr 24, 2017 at 5:05
• 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.
– JMJ
Commented Apr 24, 2017 at 5:14
• 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". Commented Apr 24, 2017 at 5:33
• Yes, I understand now.
– JMJ
Commented Apr 24, 2017 at 5:35

## 1 Answer

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.