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Given an arbitrary finite field $K$ (not necessarily $\mathbb{F}_p$ with $p \in \mathbb{P}$) with $|K| = q$ and an irreducible polynomial $f$ with $\alpha$ as root and degree of $n$. Is $|K(\alpha)| = q^n$ and why? Its clear to me for $K$ isomorphic to $\mathbb{Z}_p$ with $p$ is a prime number.

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I'm fairly sure that this has been answered somewhere here... – Jyrki Lahtonen Jul 4 '12 at 20:15
It's essentially the same as your other question…. – Cocopuffs Jul 4 '12 at 20:18
The simplest way to see this is that $K(\alpha)$ is a vector space over $K$, and so $|K(\alpha)| = |K|^{[K(\alpha) \, : \, K]} = q^n$ where $n = [K(\alpha) \, : \, K] = \deg f$. – Brandon Carter Jul 4 '12 at 20:21
@Cocopuffs: Does the euclidian algorithm work on things like $\mathbb{Z}_p/ \langle f \rangle [x]$ ? – joachim Jul 4 '12 at 20:25
@joachim: The Euclidean algorithm works in any ring $k[x]$, where $k$ is a field. – Brandon Carter Jul 4 '12 at 20:28
up vote 2 down vote accepted

The division algorithm holds in $k[x]$ for any field $k$. In fact, it holds in $R[x]$ for any commutative ring $R$ provided that the divisor has leading term a unit.

Theorem. Let $R$ be a commutative ring, and let $a(x)$ be a polynomial in $R[x]$ whose leading coefficient is a unit in $R$. Then for every $b(x)\in R[x]$ there exist unique $q(x),r(x)\in R[x]$ such that $b(x) = q(x)a(x)+r(x)$, where $r(x)=0$ or $\deg(r)\lt\deg(a)$.

The proof is exactly the same as the proof in any field, and can be done by induction on $\deg(b)$.

With the division algorithm in hand, it is immediate that if $f(x)$ is of degree $n$, then $k[x]/\langle f\rangle$ is a vector space over $k$ of dimension $n$, with basis $1+\langle f\rangle$, $x+\langle f\rangle,\ldots,x^{n-1}+\langle f\rangle$.

In particular, for $k=\mathbb{F}_q$, the resulting quotient is dimension $n$ over $\mathbb{F}_q$, and so has $q^n$ elements. This holds whether or not $f(x)$ is irreducible, though we need $f(x)$ irreducible in order to know that the quotient is a field.

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I think uniqueness of quotient and remainder requires $R$ to be a domain. – Mikko Korhonen Jul 4 '12 at 21:57
@m.k. Not if the leading coefficient of $a$ is a unit. From $q(x)a(x) +r(x) = q'(x)a(x)+r'(x)$ we get $(q(x)-q'(x))a(x) = r'(x)-r(x)$. Because the leading coefficient of $a$ is a unit, it is not a zero divisor, so if $q(x)-q'(x)\neq 0$ then $\deg((q-q')a) = \deg(q-q')+\deg(a)$, and if $r'-r\neq 0$ then $\deg(r'-r)\leq\max(\deg(r),\deg(r'))$. The usual argument now follows. You don't need $R$ to be a domain, you just need there to be no cancellation in the leading coefficient when multiplying by $a(x)$, and that follows from the fact that the leading coefficient is a unit. – Arturo Magidin Jul 4 '12 at 22:02
Thanks for setting me straight on this, I can see why it has to be true now. I got this false belief from some book (I think it was Galois Theory by J. Rotman) and I never looked into it.. – Mikko Korhonen Jul 4 '12 at 22:28
@m.k. In fact, uniqueness follows provided theleading coefficient of $a$ is not a zero divisor; but you need it to be a unit in order to guarantee existence. – Arturo Magidin Jul 4 '12 at 22:41

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