# Order of group of units of ${\bf Z}_{2015}[X]$

I know that the order of the group of units $U({\bf Z}_{2015})$ is 1440 and so the order of the group of units of the polynomial ring ${\bf Z}_{2015}[X]$ must be at least that because we can view each unit from the former group as a constant polynomial.

I have been lead to believe that the order of the polynomial ring above is also 1440. Is this true and if so, why?

Finally, how do I figure out if $U({\bf Z}_{2015}[X])$ is cyclic without going through all of its elements to see if there is one with order 1440 (or other if this is not the group's order)?

• See here for when $\,U(\Bbb Z_n)$ is cyclic. – Gone Mar 9 '15 at 20:39

Recall that a polynomial $\,f(x)\,$ is a unit iff $\,f(0)\,$ is a unit and all other coefficients are nilpotent. But $\,2015 = 5\cdot 13\cdot 31 \,$ is a product of distinct primes, so there are no nontrivial niltpotents mod $2015$ since $\,p_i\mid a^n\,\Rightarrow\,p_i\mid a\,\Rightarrow\prod p_i \mid a\,\Rightarrow\, a\equiv 0$
• Thanks for the help. I've changed the final question about being cyclic: I meant to enquire about the polynomial ring $U({\bf Z}_{2015}[X])$, not just $U({\bf Z}_{2015})$ – user220680 Mar 9 '15 at 21:02
• Does $U({\bf Z}_{2015})$ not being cyclic imply that the corresponding polynomial ring can't be either? – user220680 Mar 9 '15 at 21:09
• So elements of $U({\bf Z}_{2015})$ are the relatively prime congruence classes and the units of the polynomial ring exactly are the same ones. I have to admit I'm struggling to come up with what you're hinting at... – user220680 Mar 9 '15 at 21:28