Table of Content
Do \(ab=0\) always imply \(a=0\) or \(b=0\)?
A non-zero element \(a\) of a ring \(R\) is a zero divisor if there exists a non-zero element \(b\) such that \(ab = 0\).
把 \(0\) 因數分解成 \(a\) 和 \(b\) 的感覺。
\(M_n(\mathbb{R})\) 中有很多 zero divisor！
The zero divisors of \(\mathbb{Z}_n\) are precisely the nonzero elements whose representatives are not relatively prime to \(n\).
If \(p\) is a prime, then \(\mathbb{Z}_p\) has no zero divisors.
The cancellation law holds for a ring \(R\) iff \(R\) has no zero divisors.
Proof
\((\Rightarrow)\) Assume that the cancellation law holds, but \(ab=0\) for some \(a, b \not = 0\). Then we have \(ab = 0 = a0\), but \(b \not = 0\), which is a contradiction.
\((\Leftarrow)\) Assume that \(R\) has no zero divisors. If \(ab=ac\) and \(a\not = 0\), then \(ab-ac=0=a(b-c)\), by distributive law. Since \(R\) has no zero divisors and \(a\) is assumed to be nonzero, we have \(b-c=0\) and thus \(b=c\). ◼
就算 \(R\) 有 zero divisors，只要 \(a^{-1}\) 存在，\(ab=ac\) 依舊 implies \(b=c\)。
如果 \(R\) 沒有 zero divisor，則當 \(a\not = 0\)， \(ax=b\) 在 \(R\) 中至多有一解（\(a(x_1-x_2) = 0 \Rightarrow x_1=x_2\)；可能無解）。
If \(R\) is a commutative ring with unity and \(R\) has no zero divisors, for convenience, \(a^{-1}b\) can be denoted as \(b/a\). In this case, \(a^{-1}\) is denoted by \(1/a\).
A commutative ring \(R\) with unity \(1\not=0\) and has no zero divisors is an integral domain.
The ring of integers \(\mathbb{Z}\) is an integral domain.
如果 \(R\) 是 integral domain，則 \(R\) 的結構和 \(\mathbb{Z}\) 便有些類似！
第二項 remark 說明了，我們難以用 direct product 的方式建構出性質良好的 ring，不像 groups。
Every subring \(R\) of a field containing the unity is an integral domain.
Every field \(F\) is an integral domain.
Proof
Show that for any \(a, b \in F\) with \(ab=0\), if \(a\not = 0\), then \(b=0\).
Every finite integral domain \(D\) is a field.
From this corollary, \(\mathbb{Z}_p\) is a field when \(p\) is a prime.
Wedderburn’s theorem: every finite division ring is a field.
Proof
We have to show that every nonzero element \(a\) of \(D\) has a multiplicative inverse. Let \(0, 1, a_1,\cdots ,a_n\) be all elements of the finite integral domain \(D\).
Let \(\phi: D\to D\) given by \(\phi_a(x) = ax\), for any \(a,x\in D\). Then we can see that \(\phi\) is a permutation since all the products
\[a0, a1, aa_1,\cdots,aa_n\]are distinct, since \(ab=ac\) implies \(b=c\) by Cancellation law. Thus,
\[0, a, aa_1, \cdots,aa_n\]must be all the elements of \(D\), and one of them must be \(1\), i.e. \(aa_i = 1\) for some \(a_i\). This proves the theorem. ◼
Remark
如果 \(D\) is infinite 呢？設 \(D=\mathbb{Z}, a=2\)， 將 \(\mathbb{Z}\) 中的全部元素乘上 \(2\) 之後會得到 \(2\mathbb{Z}\)，但 \(\mathbb{Z}\not = 2\mathbb{Z}\)！
Let \(R\) be a ring. Suppose that there is a positive integer \(n\) such that \(n\cdot a = 0\) for all \(a\in R\). The least \(n\) is the characteristic of the ring \(R\). If no such positive integer exists, then \(R\) is of characteristic \(0\).
像是 cyclic group 的 order！
Example
Let \(R\) be a ring with unity. If \(n\cdot 1 \not = 0\) for all \(n\in\mathbb{Z}^+\), then \(R\) has characteristic \(0\). If \(n\cdot 1=0\) for some \(n\in \mathbb{Z}^+\), then the smallest such integer \(n\) is the characteristic of \(R\).
判斷 unity 就夠了！
Proof
If \(n\cdot 1\not = 0\) for all \(n \in\mathbb{Z}^+\), then \(R\) is clearly of characteristic \(0\), by definition.
If \(n\cdot 1=0\) for some \(n\in \mathbb{Z}^+\), then for all \(a\in R\), we have
\[n\cdot a = (a+\cdots+a) = a(1+\cdots+1) = a(n\cdot 1) = a0 = 0.\]Then the theorem follows. ◼
Let \(R\) be a ring with unity of characteristic \(n\) (which can be zero). Then \(\langle 1_R \rangle \cong \mathbb{Z}_n\).
Proof
We know that \(\langle 1_R \rangle = \{0, 1_R, 2\cdot 1_R,\cdots ,(n-1)1_R\}\). Then we can define \(\phi: \langle 1_R \rangle \to \mathbb{Z}_n\) by
\[\phi(a \cdot 1_R) = a,\ a \in \mathbb{Z}_n,\]which can be shown to be an isomorphism.
應該將 \(\mathbb{Z}_n\) 視為 quotient ring \(\mathbb{Z}/n\mathbb{Z}\)。因此，當 \(n=0\)，\(\mathbb{Z}/0\mathbb{Z} \cong \mathbb{Z}\)。這就是將 characteristic 定義為 \(0\) 而非 infinity 的原因之一。
Every ring with unity contains \(\mathbb{Z}\) or \(\mathbb{Z}_n\) as its subring.