Lecture 25 - The Primitive Root Theorem

Summary

Today we proved the (majority of) the primitive root theorem, and so we are now able to determine exactly which moduli m have primitive roots. Indeed, this means we can count precisely how many primitive roots a number m has just by looking at a factorization of m.

Integers with Primitive Roots

Last class period we were able to show that there are a handful of integers which do not have primitive roots. Our two big classes of such moduli were

  • powers of 2 that are at least 8
  • products mn so that $(m,n) = 1$ and so that $m,n>2$.

Today we'll aim to show that the remaining integers — i.e., the integers $1,24$ together with integers of the form $p^e$ or $2p^e$, where p is an odd prime — all *do* have primitive roots. We already know that primitive roots exist modulo p, so we'll start our investigation by verifying that primitive roots exist modulo $p^2$.

Theorem: A modulus m has primitive roots if and only if $m = 1,2,4,p^e \mbox{ or } 2p^e$, where p is an odd prime.

Example: When do primitive roots exist?

From the previous theorem we know that there are no primitive roots modulo 65, since $65 = 3\cdot 13$ and hence doesn't match any of the "right" forms from our theorem. On the other hand, we do know that $242 = 2 \cdot 11^2$ does have primitive roots. $\square$

Proposition: If p is an odd prime, then there exists a primitive root mod $p^2$. In fact, if $r$ is a primitive root modulo p, then r is a primitive root modulo $p^2$ if and only if

$r^{p-1} \nequiv 1 \mod{p^2}$.

In the case that $r^{p-1} \equiv 1 \mod{p^2}$, then the number $r+p$ is a primitive root modulo $p^2$.

Notice that this proposition doesn't just say there are primitive roots mod $p^2$, but gives us a test we can run to actually product primitive roots modulo $p^2$ quickly.

Proof: Suppose that r is a primitive root mod p; we already know such an element exists. Now we know that if $n = \mbox{ord}_{p^2}(r)$, then we have $n \mid \phi(p^2) = p(p-1)$. We also know that $r^n \equiv 1 \mod{p^2}$ implies $r^n \equiv 1 \mod{p}$, and so we get $p-1 = \mbox{ord}_p(r) \mid n$. Together, this means that we have either $n = p-1$ or $n = p(p-1)$. If the latter case is true, then we're finished: such an r is a primitive root mod $p^2$.

Otherwise, consider the element $r+p$. Since it is congruence to r mod p, it is still a primitive root mod p. Hence its order is either $p-1$ or $p(p-1)$. We'll show that the former is impossible, and so we will be able to conclude that $r+p$ is a primitive root mod $p^2$. To see that $r+p$ can't have order $p-1$, we compute:

(1)
\begin{split} (r+p)^{p-1} = r^{p-1} + \left(\begin{array}{c}p-1\\1\end{array}\right) r^{p-2}p + \left(\begin{array}{c}p-1\\2\end{array}\right) r^{p-3}p^2 + \cdots. \end{split}

Now the "tail" of this expression is clearly divisible by $p^2$, and so modulo $p^2$ we have

(2)
\begin{split} (r+p)^{p-1} \equiv r^{p-1} + (p-1) r^{p-2}p \equiv r^{p-1} - pr^{p-2}.\mod{p^2} \end{split}

We are already assuming that the order of r mod $p^2$ is $p-1$, and hence we get $r^{p-1} \equiv 1 \mod{p^2}$. Also, we know that $pr^{p-2} \not\equiv 0 \mod{p^2}$ since $(r,p) = 1$. Therefore the previous equation says

(3)
\begin{align} (r+p)^{p-1} \equiv 1 - pr^{p-2} \not\equiv 1 \mod{p^2}. \end{align}

Hence we get that $\mbox{ord}_{p^2}(r+p) \neq p-1$, and so we must have that $r+p$ is a primitive root mod $p^2$. $\square$

Example: Finding a primitive root mod 121

From the book we know that 2 is a primitive root modulo 11 (note that we could prove that 2 is a primitive root by calculating $2^1,2^2$ and $2^5$ mod 11 and showing that none of these quantities is $1 \mod{11}$). Hence to find a primitive root modulo 121, we need to determine the value of $2^{10} \mod{121}$. From the result above, if this quantity isn't 1, then we'll know 2 is a primitive root; on the other hand, if $2^{10} \equiv 1 \mod{121}$, then we'll know that 2 isn't a primitive root mod 121, but that $2+11 = 13$ is. Using successive squaring we were able to see that

(4)
\begin{align} 2^{10} \equiv 56 \mod{121}, \end{align}

and so we know that 2 is a primitive root mod 121. $\square$

This might seem like we've only made slight progress, moving from primitive roots mod p to primitive roots mod $p^2$. As it turns out, though, this firs step is all one needs to do: primitive roots mod $p^2$ are always primitive roots mod $p^m$.

Proposition: Any primitive root mod $p^2$ is a primitive root mod $p^e$, where p is an odd prime.

Proof: Let r be a primitive root mod $p^2$, and write n for $\mbox{ord}_{p^e}(r)$. As before, we know that

(5)
\begin{align} r^{\phi(p^e)} \equiv r^{p^{e-1}(p-1}} \equiv 1 \mod{p^e} \Longrightarrow n \mid p^{e-1}(p-1). \end{align}

Moreover we have that

(6)
\begin{align} r^{n} \equiv 1 \mod{p^e} \Rightarrow r^n \equiv 1 \mod{p^2} \Rightarrow \mbox{ord}_{p^2}(r) \mid n. \end{align}

Since we know that $\mbox{ord}_{p^2}(r) = p(p-1)$ (since r is a primitive root mod $p^2$), Equations (5) and (6) together imply

(7)
\begin{align} n = \mbox{ord}_{p^e}(r) = p^k(p-1) \quad \mbox{for some }1 \leq k \leq e-1. \end{align}

Now we would like to show that $n = p^{e-1}(p-1)$, so we need to show that no choice of $k<e-1$ is possible in the above equation. To do this, we'll show that

(8)
\begin{align} r^{p^{e-2}(p-1)} \not\equiv 1 \mod{p^e} \end{align}

in the next lemma. In particular, this equation will force $\mbox{ord}_{p^e}(r) \neq p^k(p-1)$ for $k<e-1$, and hence we will be forced into the condition $\mbox{ord}_{p^e}(r) = p^{e-1}(p-1)$; i.e., r will be a primitive root. $\square$

Lemma: Suppose that r is a primitive root mod $p^2$. Then

$r^{p^{e-2}(p-1)} \not\equiv 1 \mod{p^e}$.

Proof: We'll prove this result by induction. The case $e=2$ is already taken care of since we know that

(9)
\begin{align} r^{p-1} \not\equiv 1 \mod{p^2} \end{align}

(since r is a primitive root, the smallest exponent which sends r to 1 is $\phi(p^2) = p(p-1)$). So assume that we know the result holds for $e-1$.

Now we know that

(10)
\begin{align} r^{\phi(p^{e-2})} \equiv 1 \mod{p^{e-2}}, \end{align}

and hence we have

(11)
\begin{equation} r^{p^{e-3}(p-1)} = 1 + cp^{e-2}; \end{equation}

notice that $p \nmid c$ since otherwise we'd have $r^{p^{e-3}(p-1)} \equiv 1 \mod{p^e-1}$, contrary to the induction hypothesis. Now we'll raise both sides of Equation (11) to the pth power. We get

(12)
\begin{split} r^{p^{e-2}(p-1)} &= (1+cp^{e-2})^p \\&= 1+\left(\begin{array}{c}p\\1\end{array}\right)cp^{e-2} + \left(\begin{array}{c}p\\2\end{array}\right)(cp^{e-2})^2 + \cdots \end{split}

Notice that all the first two terms are the only which survive when we consider this equation mod $p^e$. For instance, we know that $\left(\begin{array}{cc}p\\2\end{array}\right)(cp^{e-2})^2 = \frac{p(p-1)}{2}c^2p^{2e-4}$. Since p is odd, the term $\frac{p(p-1)}{2}$ is a multiple of p, and so this term is divisible by $p^{2e-3}$. Since we know $e \geq 3$, this means $2e-3 \geq e$. Hence the third term of this expansion is 0 modulo $p^e$. Any other term has a factor of the form $(p^{e-2})^k$ for $k \geq 3$, and since $ke-2k \geq e$ for such k, we know these terms are also congruent to 0 modulo $p^e$.

Hence we have

(13)
\begin{align} r^{p^{e-2}(p-1)} \equiv 1+pcp^{e-2} \equiv 1 + cp^{e-1} \mod{p^e}. \end{align}

Now since $p \nmid c$ we get $cp^{e-1} \not\equiv 0 \mod{p^e}$, and so we have

(14)
\begin{align} r^{p^{e-2}(p-1)}\equiv 1+cp^{e-1} \not\equiv 1 \mod{p^e} \end{align}

as desired. $\square$

Having proven this result, we then stated the following

Proposition: If r is an odd primitive root modulo $p^2$, then it is also a primitive root modulo $2p^e$ for every e.

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