# Every finite field can be obtained by a quotient of the ring $\Bbb Z[x]$.

Every finite field can be obtained by a quotient of the ring $\Bbb Z[x]$.

My try:

Any finite field $F$ is of the order $p^n$ where $p$ is a prime and $n\in \Bbb N$ .

If we want to make a field of order say $p^n$ then take the quotient ring $\Bbb Z[x]/\langle p \rangle$ and then form the field $\Bbb Z_p[x]$ .

Then choose an irreducible polynomial of degree $n$ in $\Bbb Z_p[x]$ and form the quotient $\Bbb Z_p[x]/\langle f(x)\rangle$ and we get a field of order $p^n$ elements.

Two questions:

1. Is the result correct?

2. Is my proof correct?

• you should probably prove that there is such an irreducible polynomial of degree $n$ in $\mathbb{Z}_p[x]$ – reuns May 7 '16 at 1:26

Hint: Let $F$ be a finite field, $(F-\{0\},\times)$ is cyclic. Let $u$ a generator. Consider the map $g:Z[X]\rightarrow F$ defined by $g(X)=u$.
For your proof, you have to show that there exists $f$ such that $F=Z_p[X]/(f)$
• I think mathoverflow.net/a/54741/84768 is the most convincing. and how do you know that there is only one field with $p^k$ elements ? – reuns May 7 '16 at 1:35
• I don't know if every finite field can be written as a quotient $Z_p[X]/(f)$ this may be true if $F$ is a separable extension of $Z/p$ since you may apply the primitive element theorem. – Tsemo Aristide May 7 '16 at 1:46
• @user1952009, there is no need of knowing that there is only pne field with $p^k$ elements to prove that a finite field has cyclic multiplicative group. – Mariano Suárez-Álvarez May 7 '16 at 2:19