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I'm trying to understand the following explicit construction of a finite field with $p^n$ elements where $p$ is prime, $n \geq 2$:

Take any irreducible polynomial $f(X) \in \mathbb{F}_p$ of degree $n$. Then $\mathbb{F}_p[X] / (f(X))$ is a field with $p^n$ elements.

Now, I understand that if $f$ is irreducible, the ideal generated by $f$ is maximal, so $\mathbb{F}_p[X] / (f(X))$ is indeed a field.

I don't understand the following points:

  • Why does for every $n \geq 2$ an irreducible $f(X) \in \mathbb{F}_p[X]$ of degree $n$ exist?
  • Why does the field $\mathbb{F}_p[X] / (f(X))$ have $p^n$ elements?

Thanks in advance for any help!

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The existence of an irreducible polynomial can be shown by counting arguments. For example, a quadratic is either irreducible, or the product of two linear polynomials. We can restrict everything to be monic. We now how many monic linear polynomials there are, and we know how many monic quadratic polynomials there are. So we can count the product of two monic linear polynomials, and see that there are still more monic quadratics that must therefore be irreducible. Note that this even allows you to calculate the exact number of irreducible polynomials.

This can be generalized to higher degrees; for example if a quadratic is not irreducible, we can factor it into a product of irreducible cubics, quadratics, and linear factors in certain predictable ways. Lidl and Niederreiter use the Moebius function to give a formula for the number of irreducible polynomials of any given degree over a finite field.

For the second question, consider the elements of $\mathbb{F}_{q}[x]/(f(X))$; they have as representatives the polynomials of degree at most $n-1$, since a polynomial of degree $n$ or larger can have multiples of $f(x)$ subtracted from it to reduce the degree. Counting these polynomials gives the order of the new field.

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  • $\begingroup$ Thank you for your answer! I still don't understand though why any polynomial of degree bigger or equal to $n$ is divisible by $f(X)$. Could you elaborate on that? $\endgroup$ – Tom Bombadil Sep 13 '15 at 16:39
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    $\begingroup$ The nice thing is, it doesn't have to be divisible by $f(x)$ for it to be reduced. For example, in $\mathbb{F}_{3}[x]/(x^2+1)$; consider the polynomial $x^3$. We have that $x^3 = x(x^2+1) -x$, so we can reduce $x^3$ to $-x = 2x$ (a polynomial of degree less than 2). $\endgroup$ – Morgan Rodgers Sep 13 '15 at 17:21
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    $\begingroup$ (as a side note, if the polynomial is divisible by $f(X)$ it will reduce to $0$ in the new field; otherwise, you replace it with the remainder when you divide by $f(X)$, which always has degree less than $f(X)$.) $\endgroup$ – Morgan Rodgers Sep 13 '15 at 17:30

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