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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 ${Z}[x]$?

cps
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4 Answers4

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

Bernard
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  • Can it be generalized for $|F|=p^n$ as $$\mathbf F_{p^n}\simeq \mathbf Z/p\mathbf Z[x]/(x^{p^n}-x)\mathbf Z/p\mathbf Z[x]\simeq\mathbf Z[x]/(p\mathbf Z[x]+(x^{p^n}-x)\mathbf Z[x]).$$ – cps Jun 07 '15 at 16:00
  • @cps Yes, that is correct. – Thomas Poguntke Jun 07 '15 at 16:07
  • @cps: it doesn't work as is, because $x^{p^n}-x$ is not irreducible. If you factor $x$ out, the quotient is not necessarily irreducible, and if it is, it is not necessarily irreducible over $\mathbf F_p$. – Bernard Jun 07 '15 at 16:14
  • So what would be the possible generalization? – cps Jun 07 '15 at 16:35
  • Using the primitive element theorem, any finite extension of $\mathbf F_p$ is the quotient of $\mathbf F_p[x]$ by the minimal polynomial of a primitive element of $\mathbf F_{p^n}$. This polynomial is irreducible over $\mathbf F_p$, hence it is also irreducible over $\mathbf Z$. So $\mathbf F_{p^n}$ is the quotient of $\mathbf Z[x]$ by the ideal generated by $p$ and this polynomial. – Bernard Jun 07 '15 at 16:51
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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.

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

Crostul
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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 $F\cong \mathbf Z[x]/\ker \phi$.

tomasz
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