GTM84 - Ch4 - The structure of U(Z/nZ)

The discussion is followed by the congruence and its property, applications(CRT, \(ax\equiv b(m)\)) we have discussed before.

\[m=\prod_{i=1}^{n}m_i \]

Using CRT, we have got:

\[\phi: \mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}/m_1\mathbb{Z} \oplus \mathbb{Z}/m_2\mathbb{Z} \cdots \oplus \mathbb{Z}/m_n\mathbb{Z} \ \ where\ (m_i,m_j)=1 \]

Because of isomorphism, it also maps the unit group to unit group, so:

\[\phi: U(\mathbb{Z}/m\mathbb{Z})\cong U(\mathbb{Z}/m_1\mathbb{Z}) \oplus U(\mathbb{Z}/m_2\mathbb{Z}) \cdots \oplus U(\mathbb{Z}/m_n\mathbb{Z}) \ \ where\ (m_i,m_j)=1 \]

1 Primitive Roots and the Group Structure of \(U(\mathbb{Z}/n\mathbb{Z})\)

Lemma 1 The number of polynomials' roots on a field

Let \(f(x)\in \mathbb{K}[x]\), \(\mathbb{K}\) is a field. Suppose that \(deg\ f(x)=n\). Then \(f\) has at most \(n\) distinct roots.

Using Euclidean Division, it is easy to see.

Proposition 1 \(x^{p-1}-1\equiv (x-1)(x-2)\cdots(x-p+1)(p)\)

Proposition 2 if \(d|p-1\), then \(x^d\equiv 1(p)\) has exactly \(d\) solutions

Consider the distribution of roots to prove them.

Theorem 1 \(U(\mathbb{Z}/p\mathbb{Z})\) is cyclic group

Theorem 1' Finite subgroups of the multiplicative group of a field are cyclic

Proof: See Section 3

Lemma 2 For \(l\ge 3\), \(U(\mathbb{Z}/2^l\mathbb{Z})\) is the direct product of two cyclic groups, one of order \(2\), the other of order \(2^{l-1}\)

Theorem 2 Let \(n=2^ap_1^{a_1}\cdots p_m^{a_m}\) be the prime decomposition of \(n\). Then

\[\phi: U(\mathbb{Z}/n\mathbb{Z})\cong U(\mathbb{Z}/2^a\mathbb{Z}) \oplus U(\mathbb{Z}/p_1^{a_1}\mathbb{Z}) \cdots \oplus U(\mathbb{Z}/p_n^{a_n}\mathbb{Z}) \]

where:

\(U(\mathbb{Z}/p^{a}\mathbb{Z})\) is cyclic group of order \(p^{a-1}(p-1)\). \(U(\mathbb{Z}/2^a\mathbb{Z})\) is cyclic group when \(a=1, 2\), For \(l\ge 3\), \(U(\mathbb{Z}/2^l\mathbb{Z})\) is the direct product of two cyclic groups, one of order 2, the other of order \(2^{l-1}\)

Then:

n posseses primitive roots iff. n is of the form 2, 4, \(p^a\), \(2p^a\)

Proof: Construction(consider the distribution of factors) and Use the property of isomorphism.

2 \(n\)th Power Residues

Question: when \(x^n\equiv a(m), \ (a,m)=1\) is solvable

Use CRT, consider \(x^n\equiv a(p^\alpha)\), where \(p\) is an odd prime.

Theorem 3 If \(p\not | a,p\not | n\), then if \(x^n\equiv a(p^\alpha),\ \alpha\ge1\) has the same solutions with \(x^n\equiv a(p)\)

Always in number theory, \(2^\beta\) is considered seperately.

Theorem 4 \(a\) is odd, \(e\ge 3\), consider the congruence $x^n \equiv a(2^e) $

If \(n\) is even, a solution always exists and it is unique.

If \(n\) is odd, a solution exists iff. \(a\equiv 1(4)\), \(a^{2^{e-2}/d}\equiv 1(2^e)\), where \(d=(n,2^{e-2})\), and there are \(2d\) solutions.

3 Some Proofs and Exercises

本节主要是考察了U(Z/nZ)的结构，主要是利用CRT做一个同构，然后分开利用原根(当然也求解了原根存在的n的形式，也即U(Z/nZ)为循环群n的取值)或者其他工具进行讨论，涉及到一些同余式的求解，我们可以注意到一些初步的同余式求解的根的分布情况(结合原根和CRT很容易求解n次剩余)