# How existence of an irreducible polynomial in $F_p(x)$ of degree $n$ guarantees existence of finite field of order $p^n$

I read somewhere that if $\pi$ is an irreducible polynomial of degree $m$ then $F_p(x)\ \backslash \left< \pi \right>$ is a finite field of order $p$. What is $F_p(x)\ \backslash \left< \pi \right>$? How does it guarantee existence of a finite field of order $p^m$?

Here is my understanding of the answer:

Say $\pi$ is an irreducible polynomial of degree $m$. Consider any two polynomials $p_1$ and $p_2$ each of degree less than $m$. Say $p_1(x) = a_0 + a_1x +...+a_{m-1}x^m$ and $p_2(x) = b_0 + b_1x + ...+ b_{m-1}x^{m-1}$. Since polynomials uniquely factor into irreducible polynomials $\pi$ can't divide the product of these two polynomials since the factorization of the product contains irreducible polynomials of degree at most $m-1$. Therefore $(p_1\times p_2)/\pi(x)$ is a polynomial of degree less than $m$.

Therefore for polynomials $p_1, p_2$ of degree less than $m$, we define $p_1\cdot p_2 = (p_1\times p_2)/\pi$ and for non-zero polynomials it is guaranteed that this $p_1\cdot p_2$ is not equal to zero.

This also means that for every non-zero polynomial $p_1$ of degree less than $m$, $p_1\cdot p_2$ is not equal to $p_1 \cdot p_3$ if $p_2$ is not the same as $p_3$. Otherwise $(p_1\times p_2)/\pi-(p_1\times p_3)/\pi=((p_1\times p_2)-(p_1\times p_3))/\pi=(p_1\times (p_2-p_3))/\pi = 0$ implying that $p_2 = p_3$, a contradiction.

This also guarantees existence of multiplicative inverse since for exactly one polynomial $u$ we must have $p_1\cdot u=1$.

Therefore if multiplication is defined this way, then all axioms of the field are satisfied and hence we've obtained a field.

If $p$ is irreducible in $F[X]$ ($F$ being any field), the quotient ring $F[X]/(p)$ is a field because an irreducible polynomial generates a maximal ideal of $F[X]$, and an $F$-vector space of dimension $n=\deg p$.

If $F=\mathbf F_p$ is the prime field with characteristic $p$, a vector space of dimension $n$ has exactly $p^n$ elements, and this one is a field. If we denote $x=X+(p)$,this field is denoted $\mathbf F_p(x)$.

• Sorry I'm not familiar with rings and groups. What is a quotient ring? Is there an easier way to explain this, i.e. by explaining the idea and avoiding terminologies. You could provide a link where I could read this up – Siddharth Joshi Jul 4 '17 at 12:37
• Do you integers modulo $n$ in number theory? – Bernard Jul 4 '17 at 12:44
• Is quotient ring $F_p(x)$ modulo $\pi$, I.e. the remainder of all polynomials in $F_p(x)$ with $\pi$. Then it would be the set of all polynomials of degree less than $m$. How do we know if it is a field. – Siddharth Joshi Jul 4 '17 at 12:50
• This is because of the irreducibility of $p$: for any non-zero polynomial $f$ of degree $<\deg p$, we have a Bézout's relation between $f$ and $p$: $\;uf+vp=1$, which becomes mod. $p$: $\; uf\equiv 1$, so $u$ is the modular inverse of $f$. – Bernard Jul 4 '17 at 12:58
• Is this explanation also correct: $\pi$ of degree $d$ doesn't divide any polynomial of the form $(a_0 + a_1x+...+a_{d-1}x^{d-1})(b_0+b_1x+...+b_{d-1}x^{d-1}) : a_0,...,a_{d-1}, b_0,...,b_{d-1}\in F_p,$ which guarantees existence of multiplicative inverse. – Siddharth Joshi Jul 4 '17 at 13:08

Are you at all familiar with the idea of constructing field as a quotient ring of a polynomial ring? Basically it amounts to imagining a zero of the polynomial generating the ideal you "mod out".

A most familiar example is probably $$\Bbb{C}\simeq \Bbb{R}[x]/\langle x^2+1\rangle.$$ The idea is basically the following. The polynomial $p(x)=x^2+1$ has no real zeros. Because it is quadratic, this implies that it is also irreducible. Therefore we can "imagine" it has a zero, call it $i$, and the we construct a bigger field that contains this element $i$. Not as such, but as a coset $u=x+\langle x^2+1\rangle$. Then it follows that $u^2+1$ is the coset $(x^2+1)+\langle x^2+1\rangle$. But, because $x^2+1$ is in the ideal we mod out, this is the zero element in the quotient ring. As $p(u)=0$ we can decide to call that coset $i$ instead of $u$, and voila! We have the complex numbers.

We can do the same to a finite field $\Bbb{Z}_3$. The polynomial $x^2+1$ has no zeros there either, so we can construct a field of nine elements as $\Bbb{Z}_3[x]/\langle x^2+1\rangle$. Again $i=x+\langle x^2+1\rangle$ is a solution of $x^2+1=0$.

Observe that to get a field of $25$ elements we need to use a different polynomial. This is because $x^2+1$ has a zero in $\Bbb{Z}_5$, namely $p(2)=2^2+1=5=0$ (modulo five). This implies that $p(x)$ is not irreducible, and that spells trouble. Because this time $p(x)=x^2+1=x^2-4=(x-2)(x+2)$ it follows that the product of the cosets of $x-2$ and $x+2$ is zero. Therefore the quotient ring $\Bbb{Z}_5[x]/\langle p(x)\rangle$ is not a field (it has zero divisors).

This generalizes to the quotient ring of the polynomial ring by any ideal generated by an irreducible polynomial. You can find the details in basic abstract algebra textbooks. Given a polynomial $f(x)$, irreducible over $K$, we can, following this recipe, construct a bigger field $L$ containing a zero of $f(x)$.

• Sorry, I'm not familiar with quotient and polynomial rings. I'm editing the question to explain my understanding of the answer. Please see if the explanation is correct? – Siddharth Joshi Jul 4 '17 at 13:30

If $F$ is a field and $\pi$ is an irreducible polynomial of degree $m$ over $F$, then $K=F[X]/(\pi)$ is a field of degree $m$ over $F$.

In particular, $K \cong F^m$ as vector spaces over $F$.

Therefore, $K$ has $p^m$ elements when $F$ is a finite field of order $p$.