I have to show the following:

Let $p$ be a prime and $r \in \mathbb{N}$ with $\gcd(r,p)=1$. Prove the existence of a field extension $E$ of $\mathbb{F}_p$ which contains an $r$-th primitive root of unity.

I don't know how to start here.

Any help is appreciated.


  • 3
    $\begingroup$ Have you considered $\mathbb{Z}_p[x]/(1-x^r)$? $\endgroup$ – Alex Becker Jan 6 '12 at 20:29
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    $\begingroup$ @AlexBecker - That's not a field. $\endgroup$ – Thomas Andrews Jan 6 '12 at 21:04
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    $\begingroup$ Do you mean $\mathbb{Z}_p$ the p-adic integers, or integers modulo $p$ ? $\endgroup$ – Joel Cohen Jan 6 '12 at 22:13

Consider the residues modulo $r$ of $p, p^2, p^3, \dots, p^{r-1}$.

Claim: $p^m \equiv 1 \bmod r$ for some $m, 1 \leq m \leq r-1$.

The proof is by contradiction. All $r-1$ residues belong to the set $\{1,2,\ldots, r-1\}$, and so if none of them equals $1$, then two residues must have the same value and thus $r$ is a divisor of $p^i - p^j = p^j(p^{i-j}-1)$. Since $\gcd(r, p^j) = 1$, $r$ must be a divisor of $(p^{i-j}-1)$. that is, $p^{i-j} \equiv 1 \bmod r$ in contradiction of the assumption that none of the residues equals $1$. $\quad\qquad\Box$

Let $m$ denote the smallest positive integer such that $r$ divides $p^m-1$. There exist irreducible polynomials of degree $m$ in $\mathbb F_p[x]$ (for the exact number, see, for example, here), and if $g(x)$ is such a polynomial, then $E = \mathbb F_p[x]/g(x)$ is an extension field of $\mathbb F_p$. $|E| =p^m$ and its nonzero elements constitute a cyclic group of order $p^m-1$ under multiplication. This group has elements of all orders $n$ that divide $p^m-1$. Since $r$ is a divisor of $p^m-1$, $E$ contains a primitive $r$-th root of unity, which is what you have to show. $\quad\qquad\Box$


Note that $X^r-1$ is separable over $\mathbb F_p$, and let $E$ be a splitting field for $X^r-1$ over $\mathbb F_p$.

EDIT A. If you don't know the notion of splitting field, you can proceed as follows. Let $E/\mathbb F_p$ be an extension containing $m$ roots of $X^r-1$, with $m$ maximum. Assume by contradiction $m < r$. Then $X^r-1$ has a factor $f$ of degree at least $2$ which is irreducible over $E$. Form the extension $E[X]/(f)$ to get a contradiction.

Of course, the key point is that the roots of $X^r-1$ in any extension of $\mathbb F_p$ are simple. (Do you see why?)

EDIT B. The following fact has been implicitly used: a finite subgroup of the multiplicative group of a field is cyclic. So, once you have $r$ distinct roots of $X^r-1$ in $E$, you know that one of them (at least) is primitive.

EDIT C. The argument in Edit A shows this: Let $K$ be a field and $f\in K[X]$ a polynomial of degree $n$. Then there is an extension $L/K$ which contains $n$ roots of $f$, counted with their multiplicities. It's a very basic fact.


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