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Let \(D\) be an integral domain. A multiplicative norm \(N\) on \(D\) is a function \(N: D\to \Bbb Z\) such that
- \(N(\alpha)=0\) iff \(\alpha=0\),
- \(N(\alpha\beta)=N(\alpha)N(\beta)\) for all \(\alpha,\beta\in D\).
Let \(D\) be an integral domain with a multiplicative norm \(N\). Then
- \(N(1)=1\),
- if \(u\in D\) is a unit, then \(N(u)=\pm 1\),
- Assume that every \(\alpha\in D\) satisfying \(N(\alpha)=\pm 1\) is a unit. If \(\pi\) is an element such that \(N(\pi) =\pm p\) for some prime \(p\in \Bbb Z\), then \(\pi\) is an irreducible in \(D\).
For any multiplicative norm \(N(a)\), let \(\nu(a) := \vert N(a)\vert\). We can verify that EF2 is satisfied. It remains to show that \(\nu\) satisfies EF1. Consider the following example.
Let \(D=\Bbb Z[\sqrt{2}]\). Define a multiplicative norm on \(D\) that is a Euclidean function.
Let us regard \(D\) as a subring of \(\Bbb Q(\sqrt{2})\). The left multilplication by \(x=a+b\sqrt{2}\in\Bbb Q(\sqrt{2})\), denoted by \(L_x\), is a linear transformation on \(D\), so that
\[[L_x]_\alpha = \begin{pmatrix} a & 2b \\ b & a \end{pmatrix}.\]In this case, we can define the norm function as
\[N(x) = \det(L_x) = a^2 - 2b^2,\]which satisfies the property \(N(xy)=N(x)N(y)\) for all \(x,y\in Q(\sqrt{2})\). Moreover, since \(x\to L_x\) is a ring homomorphism and every nonzero element in \(\Bbb Q(\sqrt{2})\) is invertible (a field), \(\det(L_x)\not = 0\) when \(x\not = 0\). Therefore, \(N(x)=0\) iff \(x=0\).
\(\det\) 保持乘法！
Show that \(D\) is a Euclidean domain.
Solution
First, re-define the norm function as \(N(a+b\sqrt{2})=\vert a^2-2b^2\vert\).
Since \(D\) is a subring of \(\Bbb Q(\sqrt{2})\) with unity, \(D\) is an integral domain.
Let \(a, b\in D\). In \(\Bbb Q(\sqrt{2})\), write \(a/b = c_1+c_2\sqrt{2}\), where \(c_1, c_2 \in \Bbb Q\).
Choose \(q_1, q_2\in \Bbb Z\) be the cloest integers to \(c_1\) and \(c_2\) respectively, such that \(\vert q_i-c_i\vert \le {1\over 2}\) for \(i=1, 2\). Let \(q=q_1+q_2\sqrt{2} \in D\) and \(r=a-bq\).
We are to show that either \(r=0\) or \(N(r)<N(b)\). Suppose \(r\not=0\). Then
\[\begin{align*} N(r) &= N(a-bq) = N(b)N({a\over b}-q) \\ &= N(b)N\big((c_1+c_2\sqrt{2}) - (q_1+q_2\sqrt{2})\big) \\ &= N(b)\vert (c_1-q_1)^2 - 2(c_2-q_2)^2\vert \\ &\le N(b)\big(2\cdot ({1\over 2})^2\big) < N(b). \end{align*}\]Thus, for any \(a, b\in D\), there exist \(q, r\in D\) such that \(a=bq + r\) with \(r=0\) or \(N(r)<N(b)\). Since \(N\) is a multiplicative norm, which already satisfies EF2, we now have proved \(D\) is a Euclidean domain. ◼
\(N(x)\) 定義在 \(\Bbb Q(\sqrt{2})\) 上，因此 \(N({a\over b}-q)\) 是合法的。
Show that \(\alpha = 3+\sqrt{2}\) is a prime in \(D\).
We have
\[\big[D:\langle \alpha\rangle \big] = \big\vert\det(L_\alpha)\big\vert = N(\alpha) = 7,\]which is a prime. This implies that \(\langle \alpha \rangle\) is maximal.
Suppose there is an ideal \(I\) properly contains \(\langle \alpha \rangle\), i.e., \(\big[I:\langle \alpha\rangle\big]>1\). Since \(7\) is a prime number, we must have \([R:I] = 1\). Thus \(I=R\) and \(\langle \alpha \rangle\) is maximal.
By this corollary, \(\langle \alpha\rangle\) is also a prime ideal, which means that \(\alpha\) is a prime! ◼
Let \(\alpha\) be a zero of \(x^2+bx+c\). Then \(\Bbb Z[\alpha]\) is ED if \(\vert b\vert + \vert c\vert \le 2\).
用和 Example 一樣的方法，只是推廣 irreducible 而已！
高斯整數 \(\Bbb Z[i]\) 是 ED！
\(D = \Bbb Z[\sqrt{-n}]\) is not a UFD if \(n\ge 3\).
Proof
To prove \(D\) is not a UFD, we may find an irreducible which is not a prime. Here, we will show that \(2\) is the case.
First, define the multiplicative norm on \(D\) as
\[N(a+b\sqrt{-n}) = a^2 + nb^2, a, b\in \Bbb Z.\]Then, suppose \(2=z_1z_2, z_i\in D\), and we have \(4 = N(2) = N(z_1)N(z_2)\). Since \(n \ge 3\), one of \(N(z_1)\) and \(N(z_2)\) must be \(1\), which means that one of them is a unit. Thus \(2\) is irreducible in \(D\).
Suppose \(n\) is odd. Let \(\alpha=1+\sqrt{-n}\). We have \(2\vert {\alpha\overline{\alpha}} = (1+n^2)\), but \(2\) divides neither \(\alpha\) nor \(\overline\alpha\). This means that \(2\) is not a prime.
Now suppose \(n\) is even. Let \(\beta=\sqrt{-n}\). We have \(2\vert \beta\overline{\beta} = n^2\), but again \(2\) divides neither \(\beta\) nor \(\overline\beta\).
We have shown that a irreducible element, \(2\), is not a prime no matter what \(n\) is. Therefore, \(D\) is not a UFD. ◼
我們可以證明 \(3\) 在 \(\Bbb Z[i]\) 中是不可約的。而因為 \(\Bbb Z[i]\) 是 ED，代表他也是 PID；根據此引理，\(\langle 3\rangle\) 在 \(\Bbb Z[i]\) 中是極大的。再根據此定理，我們得知 \(\Bbb Z[i]/\langle 3\rangle\) 是 field！更進一步，
\[\begin{align*} \Bbb Z[i] &\cong \Bbb Z[x]/\langle x^2+1\rangle, \\ \Bbb Z[i]/\langle 3\rangle &\cong \Bbb Z[x]/\langle 3,x^2+1\rangle \\ &\cong \Bbb Z_3[x]/\langle x^2+1\rangle \\ &\cong \Bbb Z_3[i]. \end{align*}\]而 \(\Bbb Z_3[i]\) 是有 9 個元素的 Galois field！