1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ $X(\mathbb{Z}/p\mathbb{Z})\langle X,Y\rangle$ 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. $\endgroup$ – user553213 Apr 25 '18 at 11:44
2
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ I have one more question. Is there any "simpler" example for (ii)? Thanks! $\endgroup$ – Waney Apr 25 '18 at 12:10
  • $\begingroup$ @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. $\endgroup$ – Arthur Apr 25 '18 at 12:12
  • $\begingroup$ If I'm not mistaken, $2\mathbb Z_{2p}$ actually does have a unit, namely $p+1$. It might be better to take $p\mathbb Z_{p^2}$. $\endgroup$ – Marc Paul Apr 25 '18 at 16:50
  • $\begingroup$ @MarcPaul Chinese remainder theorem says you're right. Modulo $p^2$ it is. $\endgroup$ – Arthur Apr 25 '18 at 17:35
  • $\begingroup$ 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. $\endgroup$ – Waney Apr 25 '18 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.