# Showing that for any prime $p$ and any $d\ge 1$, exists a field of size $p^d$

I'm trying to finish the following question:

Let $\mathbb{F}$ be a field and $f$ irreducible polynomial with degree $d$. Let $$\mathbb{K}=\{g\in\mathbb{F}[x]\mid \deg(g)<\deg(f)\}$$ Addition of elements of $\mathbb{K}$ defined the same as addition of polynomials. The multiplication is defined modulo $f$, i.e if $g_1\cdot g_2=s\cdot f+r$, then $g_1\cdot_\mathbb{K}g_2=r$.

1. Show that $\mathbb{K}$ is a field.
2. Show that if $f$ is reducible, then $\mathbb{K}$ is not a field.
3. Prove that the dimension of $\mathbb{K}$ as vector space over $\mathbb{F}$ is $d$.
4. Deduce that for any prime $p$ and positive integer $d$ exists some finite field of size $p^d$ (hint: you may use the fact that if $\mathbb{F}$ is finite, then for any $d\ge 1$ exists irreducible polynomial over $\mathbb{F}$).

I have managed to solve $1,2,3$, but I'm stuck with part $4$.

As suggested in comments, I'm editing the post, including my solutions for parts $2,3$ (the solution for part $1$ is only proving axioms of field).

Part $2$: If $f$ is reducible, then exist two polynomials $g,h$ such that $g\cdot h=f$ and $1\le \deg (g),\deg (h)<\deg (f)$, thus $g,h\in\mathbb{K}$ and $g\cdot_{\mathbb{K}}h=0$. Assume that $\mathbb{K}$ is a field. There are no zero divisors in field, thus $g=0$ or $h=0$, but this is a contradiction, hence $\mathbb{K}$ is not a field.

Part $3$: we can write $\displaystyle \mathbb{K}=\left\{\sum_{i=0}^{d-1}a_ix^i\mid a_i\in\mathbb{R}\right\}$, i.e the set of polynomials $\{x^0,x^1,\dots,x^{d-1}\}$ spans $\mathbb{K}$. Also, it is independent set of polynomials, hence it is a basis with dimension $d$, as required.

Based on the above, how should one show part $4$?

Thanks!

• @Shaun, thanks for your suggestion. I've edited the post, I would like to get feedback on my solution and some hints for part $4$. – Galc127 Mar 21 '17 at 18:56
• Do you know that there exists a field with $p$ elements? Take that field as ${\mathbb F}$. Then, using 3, how many elements does ${\mathbb K}$ have? – Magdiragdag Mar 21 '17 at 19:01
$\mathbb{Z}/p\mathbb{Z}$ is a (finite) field for prime values of $p$. Using this as the field $\mathbb{F}$ in the construction of $\mathbb{K}$, we obtain a construction of a field that we'll denote $\mathbb{K}_p^d$. Now the game is to count how many elements $\mathbb{K}_p^d$ has, and to see that that number is $p^d$.
To count the elements of $\mathbb{K}_p^n$, we will proceed combinatorially. For each coefficent, we have $p$ many options because the polynomials are in $\mathbb{F}_p[x]$, and there are $d$ different exponents because the polynomials have degree bounded by $d$. Thus $p^d$ is an upper bound on the size of $\mathbb{K}_p^d$. To finish the proof, you need to show that every two functions with coefficents in $\mathbb{F}_p$ and degree less than $d$ are different if they have different coefficients which can be easily done by checking the values of the functions.
• Nice one :) we actually have proved in lecture that if $V$ is a vector space with dimension $d$ over $\mathbb{Z}_p$, then $|V|=p^d$. – Galc127 Mar 21 '17 at 19:27