Let $m$ be a non-prime positive integer. Under what conditions on $m$ and $f\in \Bbb Z_m[x]$ will $\Bbb Z_m[x]/(f)$ be a field/integral domain?

My obsevations:

1. When there exists some $a\in \Bbb Z$ such that f(a) (by viewing $f$ as polynomial in $\Bbb Z[x]$) is a multiple of a factor of m which are not equal to 1, then $\Bbb Z_m[x]/(f)$ cannot be a field.

2. Since any finite field must satisfy the condition that the number of elements is equal to a power of a prime number, if $f$ is a polynomial with leading coefficient being a unit in $\Bbb Z_m$ (so that division algorithm works) and m is not a power of prime, then $\Bbb Z_m[x]/(f)$ is a finite ring with $m^h$ elements, where $h=(deg(f))-1$. Then $\Bbb Z_m[x]/(f)$ cannot be a field.

3. Define the mutiplicity of a prime p in a unique factorization of an integer to be the power of p in the factorization. If m can be factored in product of prime with multiplicities 1, i.e.$$m=p_1 ...p_n$$

then $\Bbb Z_m[x]/(f)$ is isomorphic to $\Bbb Z_{p_1}[x]/(f)\times ...\times \Bbb Z_{p_n}[x]/(f)$, which is not a field.

Unfortunately I can't think of a way to deal with other cases. Any suggestions?

  • $\begingroup$ Your investigation in 3. is wrong. For instance $$\mathbb Z_6[X]/(4x-1) = \mathbb Z_6[x]/(4x-1,3) = \mathbb Z_3[x]/(x-1) = \mathbb Z_3$$ is a field. $\endgroup$ – MooS Apr 24 '17 at 8:47
  • $\begingroup$ Where does the first and second equality come from? $\endgroup$ – Jerry Apr 24 '17 at 9:12
  • $\begingroup$ $3(4x-1)=3 \pmod 6$, so $3$ is contained in the ideal. The second equality is trivial. $\endgroup$ – MooS Apr 24 '17 at 9:13
  • $\begingroup$ By CRT the question reduces to the study of $\mathbb Z_{p^k}[X]/(f)$, $k\ge1$. All these rings must be zero excepting one which has to be a field. For $k=1$ such a ring is zero iff $f$ is invertible, that is, all coefficients excepting $a_0$ are multiple of $p$ and $\gcd(a_0,p)=1$. It is a filed iff $f$ is irreducible modulo $p$. If $k\ge 2$ then our ring is never a field, and it must be zero. This happens when $f$ is invertible, and it is also equivalent to all coefficients excepting $a_0$ are multiple of $p$ and $\gcd(a_0,p)=1$. $\endgroup$ – user26857 Apr 24 '17 at 16:56
  • $\begingroup$ Now I think you can draw a conclusion. (If you can post an answer to your own question will be very nice.) $\endgroup$ – user26857 Apr 24 '17 at 16:58

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