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I am attempting to prove what the title says, that $\mathbb{F}_p[x]/\langle f(x)\rangle$ with $f(x)$ irreducible of degree $n$ is a field with $p^n$ elements.

I have already proven that for any field $K$ and polynomial $f(x)$ in $K[x]$, $K[x]/\langle f(x)\rangle$ is a field if and only if $f(x)$ is irreducible in $K[x]$. So I know that $\mathbb{F}_p[x]/\langle f(x)\rangle$ is a field for sure. Where I'm lost is figuring out how to know for sure that there are $p^n$ elements. It's fairly clear if $n\in\{ 0 ,1\}$, but I don't think induction is going to help.

I know from a proposition in class that since $F_p = \mathbb{Z}/\langle p\rangle$ and $F_p = \mathbb{Z}/\langle p\rangle$ is an integral domain, $F_p[x] = (\mathbb{Z}/\langle p\rangle)[x] = (\mathbb{Z}[x])/(\langle p\rangle[x])$, but I'm not sure whether or not that's helpful information.

KCd
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Xindaris
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3 Answers3

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Isn't this enough? If $g\in\mathbb{F}_p[x]/f(x)$, then $\max(\deg g) = \deg(f)-1=n-1$. Thus, all elements in $\mathbb{F}_p[x]/f(x)$ are of the form $\sum_{i=0}^{n-1}a_ix^i$, for $a_i\in\mathbb{F}_p$ for all $1\le i\le n-1$. Since $|\mathbb{F}_p|=p$, and there are $n$ coefficients, there are $p^n$ ways to construct a polynomial in $\mathbb{F}_p[x]/f(x)$, so $|\mathbb{F}_p[x]/f(x)|=p^n$.

tc1729
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  • It sounds reasonable when I read it, but then I wonder if there's some way to get a polynomial with degree >= n if n is big enough and f(x) has strange enough coefficients. It seems reasonable to say that there isn't, but I don't know how to be rigorously sure. – Xindaris Oct 28 '13 at 00:53
  • See the comment by Zev Chonoles at http://math.stackexchange.com/questions/83265/prove-that-fx-x21-has-121-elements – tc1729 Nov 03 '13 at 17:18
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Well, you can always suppose $f$ to be a monic polynomial, $f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0$. But then, if $\alpha \in \mathbf F_p[x]/(f(x))$ is the image of $x$ in the quotient, the relation $\alpha^n + a_{n-1}\alpha^{n-1} + \dots + a_0$ shows that $\alpha^n, \dots, \alpha, 1$ are linearly dependent over $\mathbf F_p$. Moreover, it is easy to see by induction on $m$ that $\alpha^m$ is a linear combination of $\alpha^{n-1}, \dots, \alpha, 1$ with coefficients in $\mathbf F_p$, for every $m\geq 0$. Since the elements $\alpha^m$ span $\mathbf F_p[x]/(f(x))$ as an $\mathbf F_p$-vector space, it follows that the elements $\{\alpha^{n-1}, \dots, \alpha, 1\}$ also span it. Moreover, $\{\alpha^{n-1}, \dots, \alpha, 1\}$ are linearly independent by the irreducibility of $f$.

Alternatively, use the division algorithm to write any element of $\mathbf F_p[x]$ as $p(x)f(x) + r(x)$, where the degree of $r(x)$ is strictly less than the degree of $f$.

This implies that $\mathbf F_p[x]/(f(x))$ is a vector space of dimension exactly $n$ over $\mathbf F_p$, so it has $p^n$ elements.

Bruno Joyal
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The possible remainders upon division by $f(x)$ (where $\deg(f(x)) = n$) are those polynomials of the form $$ r(x) = a_0 + a_1x + \dotsb + a_{n-1}x^{n-1}, $$ where $a_i \in \mathbb{F}_p$. There are $n$ such $a_i$'s, and for each of the $a_i$'s there are $p$ possible choices from $\mathbb{F}_p$. Thus, there are $p^n$ elements in $\mathbb{F}_p / (f(x))$.

tylerc0816
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