Is every finite field a quotient ring of ${Z}[x]$?

Question

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 ${Z}[x]$?

Answer 1

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.

Answer 2

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: $$\mathbf F_{27}\simeq \mathbf Z/3\mathbf Z[x]/(x^3-x+1)\mathbf Z/3\mathbf Z[x]\simeq\mathbf Z[x]/(3\mathbf Z[x]+(x^3-x+1)\mathbf Z[x]).$$

Answer 3

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 $F=\Bbb{Z}[X]/(p, \tilde{f})$.