Is every finite field a quotient ring of ${Z}[x]$? Is every finite field  a quotient ring of ${Z}[x]$? For example, how a field with 27 elements can be written as a quotient ring of ${Z}[x]$?
 A: It is the splitting field of an irreducible polynomial $f(x)$ of degree $3$ over $\mathbf F_3$, for instance $x^3-x+1$. As $\mathbf F_3=\mathbf Z/3\mathbf Z$, we have:
$$\mathbf F_{27}\simeq \mathbf Z/3\mathbf Z[x]/(x^3-x+1)\mathbf Z/3\mathbf Z[x]\simeq\mathbf Z[x]/(3\mathbf Z[x]+(x^3-x+1)\mathbf Z[x]).$$
A: Every finite field has an order which is the power of a prime.
Every finite field of order $p$ is isomorphic to integers modulo $p$. Every finite field of order $p^k$ is isomorphic to polynomials over the field with p elements; modulo an irreducible polynomial of degree $k$. There are no other fields.
So yes. Taking $\mathbb Z[x]$ through first a quotient on $p\mathbb Z$, then on $d\mathbb Z[x]$ with $p, d$ prime/irreducible will yield a field.
A: Every finite field $F$ (of characteristic $p$) is isomorphic to a suitable quotient $F=\Bbb{F}_p[X]/(f)$ where $f \in \Bbb{F}_p[X]$ is an irreducible polynomial. Let $\tilde{f} \in \Bbb{Z}[X]$ be any lift of $f$.
Then $F=\Bbb{Z}[X]/(p, \tilde{f})$.
A: Here's another proof, which does not presuppose any knowledge of finite fields, only elementary group and field theory. Fix a finite field $F$ with $n$ elements.
Write $F^\times$ for the multiplicative group of nonzero elements of $F$. I claim that this group is cyclic (this is essentially Artin's theorem, but I will include a proof here for completeness, and to keep things elementary).

Note that $F^\times$ has $n-1$ elements, so the orders of its elements divide $n-1$. Now, let $m$ be the least number such that for all $x\in F^\times$ we have $x^m=1$. Now, since $F$ is a field, the polyniomial $x^m-1=0$ has at most $m$ distinct roots, so $m\geq n-1$, so $m=n-1$. Finally, note that orders of elements of a finite abelian group are closed under least common multiple (if $x$ has order $m_x$ and $y$ has order $m_y$, then $xy$ has order $\operatorname{lcm}(m_x,m_y)$). It follows that there is some $x\in F^\times$ of order $m=n-1$, so $F^\times$ is cyclic.

Now, given that $F^\times$ is cyclic, let $\alpha$ be its generator. Observe that there is a ring homomorphism $\phi\colon \mathbf Z[x]\to F$ such that $\phi(1)=1_F$ and $\phi(x)=\alpha$. Since $\alpha$ generates $F^\times$, it is easy to check that $\phi$ is onto, so $F\cong \mathbf Z[x]/\ker \phi$.
