Table of Content
Let \(R\) be a commutative ring with unity. Let \(a, b\in R\).
If there exists \(c\in R\) such that \(b=ac\), then \(a\) divides \(b\), i.e., \(a\) is a factor of \(b\), denoted by \(a\mid b\).
An element \(u\) is an unit if \(u\) divides \(1\).
\(a\) and \(b\) are associates if \(a=ub\) for some unit \(u\in R\).
A nonzero non-unit element \(p\) is an irreducible if whenever \(p=ab\), either \(a\) or \(b\) is a unit.
A nonzero non-unit element \(p\) of an integral domain is a prime if \(p\vert ab\) implies \(p\vert a\) or \(p\vert b\).
In \(\Bbb Z\), \(-n\) and \(n\) are associates.
Let \(D\) be an integral domain. Let \(a, b\in D\).
- \(a\vert b\) iff \(\langle b\rangle\subset \langle a\rangle\).
- \(a\) and \(b\) are associates iff \(\langle a\rangle = \langle b\rangle\).
- \(a\) and \(1\) associates iff \(a\) is an unit.
- \(a\) is irreducible iff \(\langle a\rangle\) is maximal among all proper principal ideals.
- A prime element \(a\) is always irreducible, while the converse is not necessarily true.
- An arbitrary integral domain need not be atomic (factorization domain).
A subring with unity of a field is an integral domain!
Proof of 4.
Let \(p=ab\). Since \(p\) is a prime and \(p\vert ab\), we have \(a=cp\) or \(b=dp\) for some \(c, d\in D\). WLOG, suppose \(a=cp\), and then
\[p=ab=cpb=pcb \implies p(1-cb) = 0,\]which means that \(cb=1\). Thus \(b\) is a unit.
On the other hand, \(2\) is irreducible in \(\Bbb Z[\sqrt{-5}]\), but \(2\vert (1+\sqrt{-5})(1-\sqrt{-5})\) doesn’t imply \(2\) divides either of them. ◼
An integral domain \(D\) is an unique factorization domain (UFD) if
Every nonzero non-unit element of \(D\) can be factored into a product of finite number of irreducibles.
If \(a\in D\) has two factorizations \(p_1\cdots p_r\) and \(q_1\cdots q_s\) into products of irreducibles, then \(r=s\) and \(q_j\) can be renumbered so that \(p_i\) and \(q_i\) are associates.
In a UFD, an element is irreducible iff it is prime.
Proof (irreduciblity implies prime)
Let \(p\) be an irreducible in a UFD \(D\). We shall prove that for some \(q, x, y\in D\), whenever \(pq = xy\), \(p\vert x\) or \(p\vert y\).
Case 1: \(xy=0\).
Since \(D\) is an integral domain, either \(x\) or \(y\) is zero, and we have \(p\vert 0\).
Case 2: \(x\) or \(y\) is a unit.
Without loss of generality, suppose \(x\) is a unit. Then, \((x^{-1}q)p = y\), which implies that \(p\vert y\).
Case 3: \(x\) and \(y\) are non-zero, non-unit.
First, suppose \(q\) is a unit. Then \(p = q^{-1}xy = (q^{-1}x)y\). Since \(p\) is a prime, it is also irreducible, which implies that either \((q^{-1}x)\) or \(y\) is a unit. By assumption, \(y\) is not a unit. Suppose \((q^{-1}x)\) is a unit, let \(r\) be its inverse. Then
\[rq^{-1}x = 1 \implies x = r^{-1}q,\]which means that \(x\) is as well a unit. A contradiction. Thus \(q\) is not a unit.
Since \(pq=xy\) and all of them are non-zero and non-unit, there must be a one-to-one correspondence between the irreducibles on both sides (unique factorization). This means that \(p\) must correspond to an irreducible factor of \(x\) or that of \(y\), i.e., \(p\) divides either \(x\) or \(y\). ◼
要先確認大家都 non-zero、non-unit，才可以分解！
To show an integral domain is not a UFD, one may show that there exists some irreducible element which is not a prime.
Show that \(R=\Bbb Z[\sqrt{-6}]\) is not a UFD.
First note that \(D = \Bbb Z[\sqrt{-6}]\) is an integral domain since it is a subring with unity of \(\Bbb C\).
Let \(N(z) = z\bar z\). Then, for all \(z_1, z_2\in D\), \(N(z_1)N(z_2) = N(z_1)N(z_2)\). Let \(z=3+2\sqrt{-6}\), then \(N(z) = 3^2 + 6\cdot 2^2 = 33\). Suppose \(z=z_1z_2\), then
\[33 = N(z) = N(z_1)N(z_2).\]Since \(N(z_i) = 3\) or \(11\) has no solutions, we must have one of \(N(z_i) = 1\). Therefore, \(z_1\) or \(z_2\) is a unit and \(z\) is irreducible.
The same argument implies that \((3-2\sqrt{-6}), 3\), and \(11\) are all irreducible. Therefore,
\[33 = 3 \cdot 11 = (3+2\sqrt{-6})(3-2\sqrt{-6}).\]Since \((3-2\sqrt{-6})\) is not associated to \(3\) and \(11\), we obtain two non-equivalent factorization of \(6\). That is to say, \(D\) is not a UFD. ◼
Show that every non-zero prime ideal in \(R\) is maximal.
Solution
Let \(I\) be a prime ideal in \(R\). To show \(I\) is maxiaml, we may prove that \(R/I\) is a field, and it suffices to show that \([R:I]\) is finite.
Since \(R\) is a Dedekind’s domain, every non-zero proper ideal in \(R\) can be factored into a product of prime ideals. Choose a principal ideal, \(R\alpha\), such that \(R\alpha\) is divided by \(I\) as ideal, which means that \(R\alpha\) is a subset of \(I\). Moreover, we can verify that \(R\alpha\) is an ideal of \(I\).
Suppose \(\alpha=a+b\sqrt{-6}\). This implies that
\[\begin{align*} R\alpha &= \langle a+b\sqrt{-6}\rangle \\ &= (a+b\sqrt{-6})\Bbb Z + \sqrt{-6}(a+b\sqrt{-6})\Bbb Z \\ &= (a+b\sqrt{-6})\Bbb Z + (-6b + a\sqrt{-6})\Bbb Z. \end{align*}\]Let \(M = \Bigg\langle\begin{pmatrix}a\\ b\end{pmatrix}, \begin{pmatrix}-6b \\ a\end{pmatrix}\Bigg\rangle\). If we map \(R\) to \(\Bbb Z^2\) (which is bijective), we have
\[[R:R\alpha] = [\Bbb{Z}^2:M] = \det\begin{pmatrix} a&-6b\\ b&a \end{pmatrix} = a^2 + 6b^2,\]which is indeed finite. Since \(I\le R\), \([I:R\alpha]\) must also be finite. Therefore,
\[[R:I] = [R:R\alpha]/[I:R\alpha]\]is finite and \(I\) is maximal. ◼
Maybe we need some isomorphism theorem for rings to justify the last equation.
試著證明 \([\Bbb{Z}^2:M] = \det\begin{pmatrix}a&-6b\\ b&a\end{pmatrix}\)！可以方格圖觀察看看。
An integral domain \(D\) is a principal ideal domain if every ideal in \(D\) is principal.
Let \(D\) be a PID and \(a, b\in D\). Then \(a\) and \(b\) are called coprime if \(c\vert a\) and \(c\vert b\), \(c\in D^\times\).
For \(a, b\in D\) where \(D\) is a PID, \(a\) and \(b\) are coprime iff \(\langle a, b\rangle=D\).
An ideal \(\langle p\rangle\) in a PID is maximal iff \(p\) is irreducible.
Proof idea
If \(p\) is not irreducible, there exist \(a\mid p\) for some non-unit \(a\), which means that \(\langle p\rangle \subset \langle a\rangle\).
In a PID \(D\), if an irreducible \(p\) divides \(ab\), then either \(p\mid a\) or \(p\mid b\).
Suppose \(p\mid ab\), then \(ab=pr\) for some \(r\in D\). This implies that \(ab\equiv 0\) in \(D/\langle p\rangle\). Since \(\langle p\rangle\) is maximal (by the previous lemma), \(D/\langle p\rangle\) is a field; we conclude that \(a\equiv 0\) or \(b\equiv 0\) (no zero divisor), i.e., \(a\in \langle p\rangle\) or \(b\in \langle p\rangle\). ◼
Alternative proof
Suppose \(p\vert ab\), then \(ab=pr\) for some \(r\in D\). This implies that \(ab\in \langle p\rangle\). Since \(p\) is irreducible, \(\langle p\rangle\) is maximal, which implies that \(\langle p \rangle\) is prime. Thus, either \(a\in \langle p\rangle\) or \(b\in \langle p\rangle\). ◼
\(x\vert n\iff n\in \langle x\rangle\).
小就是大！
In a PID, if an irreducible \(p\) divides \(a_1\cdots a_n\), then \(p\) divides \(a_i\) for some \(i\).
想想 \(\Bbb F[x]\)！
Let \(R\) be a commutative ring. Suppose that \(I_1\subseteq I_2\subseteq\cdots\) is an ascending chain of ideals in \(R\). Then \(I=\bigcup_i I_i\) is an ideal of \(R\).
Let \(D\) be a PID. Let \(I_1\subseteq I_2\subseteq\cdots\) be an ascending chain of ideals. Then there is a positive number \(N\) such that \(I_n = I_N\) for all \(n\ge N\).
當 \(D\) 是 infinite 才有討論價值！
Proof
Since \(D\) is a PID, \(I=\langle a\rangle\) for some \(a\in I\) (Because we have unity in \(D\), \(a\) is guaranteed to stay in \(I\).). Then there exist some \(N\) such that \(a\in I_N\), which implies that \(I=\langle a\rangle = Ra\subseteq I_N\). We conclude that \(I=I_N\). ◼
If \(D\) is a PID, then it is also a UFD.
Let \(D\) be a PID. Then every non-zero, non-unit element of \(D\) is a product of irreducibles. Such product is unique upto order and associates.
Proof
Let’s first show that every nonzero non-unit element \(a\) has an irreducible factor, with an algorithmic way.
\[\begin{align*} &\textbf{procedure}: \text{FindIrrFactor}(a \in D) \\ &\textbf{1. }\textbf{if } a \text{ is irreducible}: \textbf{return } a. \\ &\textbf{2. }\text{Factorize } a \text{ into } a=a_1b_1, \text{ where neither is a unit}. \text{Thus } \langle a\rangle \subset \langle a_1\rangle. \\ &\textbf{3. }\textbf{return }\text{FindIrrFactor}(a_1). \end{align*}\]With this algorithm, we obtain a strictly ascending chain of ideals \(\langle a\rangle \subset\langle a_1\rangle \subset \cdots\). By ACC, this process must terminate at some point, and thus there exists an irreducible factor \(a_n\) of \(a\).
Now, we shall show that \(a\) is a product of irreducibles.
\[\begin{align*} &\textbf{procedure}: \text{Factorization}(a \in D) \\ &\textbf{1. }\textbf{if } a \text{ is irreducible}: \textbf{return } a. \\ &\textbf{2. }\text{Factorize } a \text{ into } a=p_1a_1, \text{ where } p_1 \text{ is irreducible and } a_1\text{ is not a unit}. \text{Thus } \langle a\rangle \subset \langle a_1\rangle. \\ &\textbf{3. }\textbf{return }\text{Factorization}(a_1). \end{align*}\]With the same argument, we have \(a=p_1p_2\cdots p_r\) when the process terminates, where \(p_i\) are all irreducibles.
Finally, the proof of uniqueness of factorization resembles the proof here. Hence it is omitted here. ◼
The polynomial ring over a UFD is still a UFD.
Let \(\Bbb F\) be a field. Then \(\Bbb F[x_1, \cdots, x_n]\) is a UFD, but not a PID when \(n\ge2\).
Proof (not a PID)
Let \(I=\langle x_1, x_2\rangle = \{fx_1 + gx_2\mid f, g\in \Bbb F[x_1, x_2] \}\).
Suppose \(I\) is principal: \(I=\langle f(x_1, x_2)\rangle\). There exist \(g_1, g_2\in \Bbb F[x_1, x_2]\) such that
\[\begin{align*} x_1 = f\cdot g_1 &\implies \deg_{x_2}f = 0, \\ x_2 = f\cdot g_2 &\implies \deg_{x_1}f = 0. \end{align*}\]This means that \(f\) is a unit. However, \(\langle x_1, x_2\rangle\) doesn’t contain any unit, which is a contradiction. Therefore, \(I\) is not principal. ◼
A Euclidean function on an integral domain \(D\) is a function \(\nu: D\backslash \{0\} \to \Bbb N\cup \{0\}\) such that:
(EF1) For all \(a, b\in D\) with \(b\not =0\), there exist \(q\) and \(r\) in \(D\) such that
\[a=bq+r, \text{ with either } r=0 \text{ or } \nu(r)<\nu(b).\](EF2) For all \(a, b\in D\) with \(ab\not = 0\), \(\nu(a)\le \nu(ab)\).
An integral domain \(D\) is a Euclidean domain if there exists a Euclidean function on \(D\).
Examples of \(\nu\)
Every Euclidean domain \(D\) is a PID.
Every Euclidean domain is a UFD.
Proof hint
Claim that \(I=\langle b\rangle\), where \(b\) has the minimum \(\nu\) value in \(I\).
Let \(D\) be a Euclidean domain with Euclidean function \(\nu\). Then
- \(\nu(1)\) is minimal among all \(\nu(a)\) for \(a\in D\backslash\{0\}\).
- \(u\in D\) is a unit iff \(\nu(u)=\nu(1)\).
Proof
By EF2.
設 \(D\) 為一 integral domain。
證明 \(D\) 不是 UFD / PID 簡單（找反例：prime, irreducible）、證明 \(D\) 是 ED 也簡單（找 function \(\nu\)）；但是反之卻很困難！這就是定義這麼多 domain 的原因。
具體來說，如果找不到 Euclidean function，就可以試著證明 \(D\) 不是 UFD。