# 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!

• Please share your thoughts so far. It's easier to help you then. – Shaun Mar 21 '17 at 17:42
• @Shaun, actually I don't manage to do much in this part... – Galc127 Mar 21 '17 at 18:08
• Perhaps your solutions to the first three parts might guide us as to what you know and how to help you. Sometimes the first few solutions are even used later on in multiple part questions. – Shaun Mar 21 '17 at 18:12
• @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