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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)?

user26857
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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 $\,p_i\mid a^n\,\Rightarrow\,p_i\mid a\,\Rightarrow\prod p_i \mid a\,\Rightarrow\, a\equiv 0$

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Bill Dubuque
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  • 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 $U({\bf Z}{2015})$ –  Mar 09 '15 at 21:02
  • @Jade What do you know about the relationship between the two unit groups from the above? – Bill Dubuque Mar 09 '15 at 21:06
  • Does $U({\bf Z}_{2015})$ not being cyclic imply that the corresponding polynomial ring can't be either? –  Mar 09 '15 at 21:09
  • @Jade The answer implies that the units of this polynomial ring are precisely the units of its coefficient ring. So how are the unit groups related? – Bill Dubuque Mar 09 '15 at 21:13
  • 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... –  Mar 09 '15 at 21:28
  • @Jade Yes, the unit groups are equal. Units remain units in any extension ring, and the above shows that there are no other units in this extension. – Bill Dubuque Mar 09 '15 at 21:30
  • Right. Does the fact that there's no element in $U({\bf Z}_{2015})$ which generates that group mean that there isn't one to generate all of the units in the poly. ring and so it isn't cyclic either? –  Mar 09 '15 at 21:39
  • The groups are equal (or isomorphic, depending on how on constructs polynomial rings; cyclicity is preserved by an isomorphism - think about it). – Bill Dubuque Mar 09 '15 at 21:51