Let $\phi$ be an irreducible polynomial and $q$ a prime number. Let $R=Z_q[X]/(\phi(X))$ be the ring of polynomials modulo $\phi$ with the coefficients in $Z/qZ$.
I wonder why $R$ is referred always as a ring. Isn't it a field? Every polynomial has an inverse since $q$ is prime, right?
If $\phi$ is irreducible modulo $q$ of degree $n$, then $\mathbb Z_q[X]/\langle \phi(X)\rangle$ is the field of cardinality $q^n$ -- and this is in fact the standard way of constructing fields of non-prime size.
Different irreducible polynomials give different representations of the same field -- that is, the quotient rings are always isomorphic when the polynomials have the same degree.
As Magdiragdag notes, it is not enough for $\phi$ to be irreducible over $\mathbb Z$ -- for example, $X^2+1$ is irreducible over $\mathbb Z$, but $\mathbb Z_5[X]/\langle X^2+1\rangle$ is not a field because $\mathbb Z_5$ already contains square roots of $-1$, namely $2$ and $3$ -- and in fact $X^2+1$ splits as $(X-2)(X-3)$ in $\mathbb Z_5[X]$. So $X-2$ and $X-3$ are zero divisors in the quotient ring.