# Find an infinity ring with a finite characteristic p

Find an infinity ring with finite characteristic p, which has following properties:

(i) is not commutative

(ii) is a field

(iii) does not have unit

I think that the (ii) can be field of fractions using integral domain ${\displaystyle \mathbb {Z} }_p[x]$ set of all polynomials over the field ${\displaystyle \mathbb {Z} }_p$

But I cannot find other examples. Probbaly some sequences with finite non-zero elements can work but I have no idea how to finish it.

Thank you very much for any help.

• $X(\mathbb{Z}/p\mathbb{Z})\langle X,Y\rangle$ works for (i) and (iii), where $(\mathbb{Z}/p\mathbb{Z})\langle X,Y\rangle$ is the free algebra generated by $X$ and $Y$, or non-commutative polynomials, if you want to call it that way. – user553213 Apr 25 '18 at 11:44

## 1 Answer

For (i), take the ring of $n\times n$ matrices for some $n>1$, with entries in $\Bbb Z_p[x]$. For (iii), take $p\Bbb Z_{p^2}[x]$.

• I have one more question. Is there any "simpler" example for (ii)? Thanks! – Waney Apr 25 '18 at 12:10
• @Waney No, I think the field of rational functions in one variable with coefficients from $\Bbb Z_p$ is about as simple as it gets. – Arthur Apr 25 '18 at 12:12
• If I'm not mistaken, $2\mathbb Z_{2p}$ actually does have a unit, namely $p+1$. It might be better to take $p\mathbb Z_{p^2}$. – Marc Paul Apr 25 '18 at 16:50
• @MarcPaul Chinese remainder theorem says you're right. Modulo $p^2$ it is. – Arthur Apr 25 '18 at 17:35
• Now I am little bit confused. I am trying to do it on example. First of all I take p=3 and use the orginal form $pZ_2p$ it means that $3Z_6$ is set {0,3,6,9,12,15} how I know that the characteristic is 3, and why it has unit, since 3+1=4 is not in set. Now I am trying to do same with $3Z_9$ is set {0,3,6,9,12,15,18,21,24} how I know that the characteristic is 3. Probbaly I do not understand the notation or somethin other. I am sorry. Thank you very much. – Waney Apr 25 '18 at 18:38