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The group of units of the rings $\text{GF}(p^r)$ and $\mathbb{Z}/p^r\mathbb{Z}$ are both cyclic (except for the exception of prime powers are not cyclic when $p=2$ and $r\ge 3$). This is a strong result which I have been using a lot but I don't understand it properly.

I would like to know more alternative proofs of these results. If there are any generalizations (e.g. can both cases be treated together?) and specific cases (e.g. is there something special about the intersection $\text{GF}(p) = \mathbb{Z}/p\mathbb{Z}$.

I hope that question doesn't seem vague, it's more like a couple of questions together - I will greatly appreciate anything on this topic so thanks very much!


Update

Inspired by lhf's answer to a question about Wilson's theorem is there any way to prove that if the product of all elements of a group are $-1$ then the units group is cyclic?

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    $\begingroup$ any finite multiplicative subgroup of a field is cyclic $\endgroup$ – yoyo Mar 8 '11 at 23:19
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    $\begingroup$ @quanta: The group of units of $\mathbb{Z}/p^r\mathbb{Z}$ is not cyclic for $p=2$, $r \ge 3$. $\endgroup$ – Chris Eagle Mar 8 '11 at 23:29
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    $\begingroup$ @quanta: The fact that the group of units of $\mathbb{Z}/p^r\mathbb{Z}$ is cyclic is a consequence of the existence of primitive roots, which in turn can be seen as a consequence of Hensel's Lemma; would that be what the kind of observation you are hoping for? $\endgroup$ – Arturo Magidin Mar 8 '11 at 23:37
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    $\begingroup$ @Arturo Magidin, Neat! I can use the fact that Z/pZ is cyclic to prove Z/p^rZ using Hensel. That's really a slick proof that I didn't know! $\endgroup$ – quanta Mar 8 '11 at 23:46
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    $\begingroup$ Sometimes negative results are helpful: lots of rings similar to these rings do not have cyclic groups of units. For instance, replace Z by R=Z[√2], and take p=3 (which is still prime), then R/p^2R is a finite local ring whose group of units is 3×3×8 and so not cyclic. $\endgroup$ – Jack Schmidt Mar 9 '11 at 4:58
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For what it's worth, here is my writeup of the theorem on non/existence of primitive roots modulo $N$, used for a course in number theory at the advanced undergraduate level. (It does not use Hensel's Lemma. As others have pointed out, although there is definitely something Henselesque going on here, it is not so straightforward to use HL to give a proof.)

The basic idea for showing that $U(p^n) = (\mathbb{Z}/p^n \mathbb{Z})^{\times}$ is cyclic for any odd prime $p$ was communicated to me by David Savitt:

As usual, we use the fact that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic, which I prove here in the usual way (i.e., in essentially the same way as in Qiaochu's answer to this question). So we know there exists an element of order $p-1$ (any lift of a generator mod $p$ will have order divisible by $p-1$, so take a suitable power of that element). Since

$\# U(p^n) = \varphi(p^n) = p^{n-1}(p-1)$

and

$\operatorname{gcd}(p^{n-1},p-1) = 1$,

it is enough to find an element of order $p^{n-1}$. In fact the element $1+p \pmod{p^n}$ has order $p^{n-1}$, which is proved using elementary and relatively painless binomial coefficient considerations.

This proof is about 50% shorter than most others I have seen. It does use a little familiarity with finite commutative groups, so would not be appropriate for the most general possible number-theoretic audience.

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    $\begingroup$ I seem to remember that if $g$ is a primitive root modulo $p$, and $g+tp$ is a primitive root modulo $p^2$ (and the existence of such a $t$ can be obtained with HL), then $g+tp$ will also be a primitive root modulo $p^n$ for all $n\geq 2$ (assuming $p$ is odd of course); am I misremembering something? $\endgroup$ – Arturo Magidin Mar 9 '11 at 6:46
  • $\begingroup$ @Arturo: no, everything you've said sounds right to me. It does take some time to establish these facts, though... $\endgroup$ – Pete L. Clark Mar 9 '11 at 7:06
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Once you show that $\mathbb{Z}/p\mathbb{Z}$ has a primitive root, my favorite proof of the "ramping up" that proves the existence of primitive roots in $\mathbb{Z}/p^n \mathbb{Z}$ (for $p$ odd) is in LeVeque "Fundamentals of Number Theory" Theorem 4.4 and the comment following it. This proof also makes clear that if an integer $g$ is a primitive root in $\mathbb{Z}/p\mathbb{Z}$, then either $g$ or $g+p$ is a primitive root in $\mathbb{Z}/p^n \mathbb{Z}$ for all $n$.

He proves this as a ready consequence of the following:

Suppose $p$ is a prime relatively prime to $a$. Let $t_n$ be the order of $a \mod p^n$ and assume that $p^z$ exactly divides $a^{t_1}-1$. Then if $p>2$ or $z>1$,

$$ t_n = \begin{cases} t_1, \quad &\text{for $n \leq z$}\\\\ t_1 p^{n-z}, \quad &\text{for $n > z$.}\end{cases}$$

I like this approach because I find the above formula more useful than it is usually given credit for. I also find elegance in the following proof of the above formula, which employs

  1. the characterization of the order of a prime $a$ modulo $p^r$ as the inertial degree of the prime $a$ in the cyclotomic extension $\mathbb{Q}(\zeta_{p^r})$
  2. the fact that in a cyclic extension of number fields of degree $p^n$, a prime that is inert at the bottom degree $p$ extension must stay inert through the whole extension.

First, find a primitive root $g$ in $\mathbb{Z}/p \mathbb{Z}$. Employ Dirichlet's Theorem on primes in arithmetic progressions to find a prime $a$ in the same class as $g$. In the cyclotomic $\mathbb{Z}_p$ extension of $\mathbb{Q}$, $a$ must have inertial degree $p$ between the $z$th and $z+1$st layers (that is, between $\mathbb{Q} (\zeta_{p^z})$ and $\mathbb{Q} (\zeta_{p^{z+1}}))$. Thus, $a$ has inertial degree $p$ between all of the higher layers of the cyclotomic $\mathbb{Z}_p$ extension, giving LeVeque's formula for the orders $t_n$. The proof of the existence of primitive roots in $\mathbb{Z}/p^n\mathbb{Z}$ is very short once this is established.

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It seems like you aren't aware of the result yoyo cites in the comments, so I think it's worth pointing out as a start.

Theorem: A finite subgroup of the multiplicative group of any field is cyclic.

Proof. Let $G$ be a finite subgroup of $F^{\ast}$, $F$ a field, and $n = |G|$. Then the elements of $G$ must be precisely the roots of the polynomial $x^n - 1$. We have the factorization

$$x^n - 1 = \prod_{d | n} \Phi_d(x)$$

where $\Phi_d$ is the $d^{th}$ cyclotomic polynomial. It is not hard to see that the roots of $\Phi_d$ over $F$ must in fact be precisely the set of elements of order exactly $d$ in $G$; in particular, there exists an element of order exactly $n$.

I've been told that the result for $\mathbb{Z}/p^n\mathbb{Z}$ is clearer if one first proves the corresponding result for the $p$-adic integers, but I'm not sure if this actually saves you any work.

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The p-adic exponential from $p\cdot \mathbb Z_p$ (with $p>2$) to $1+p\mathbb Z_p\subset \mathbb Z_p^\times$ is an isomorphism. The former has (pro-) generator $p\cdot 1$, in the sense that any quotient $p\mathbb Z_p/p^{n+1}\mathbb Z_p\approx p\mathbb Z/p^{n+1}\mathbb Z$ is cyclic, generated by the image of $p\cdot 1$. Thus, the index-$p$ part of $(\mathbb Z/p^n)^\times$ is cyclic.

Combining this with the cyclic-ness of $\mathbb Z/p^\times$ is easy.

The exponential also does show why/how having "primitive roots" becomes progressively more difficult.

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