appendix

DEFINITION 1.10 (temporary)   Let $R$ be a commutative ring. A finitely generated $R$ -module $M$ is said to be projective if there exists an $R$ -module $M'$ such that

$$M\oplus M' \cong R^{\oplus n}$$

for some $n$ .

We shall argue homological algebra in much more detail later. So the definition here is meant to be minimum. See a book on homological algebra for ``correct definition''.

DEFINITION 1.11   Let $R$ be a domain. The generic rank of $M$ is the rank of $Q(R)\otimes_R M$ over the quotient field $Q(R)$ .

LEMMA 1.12   Let $R$ be a commutative unique factorization domain(UFD). Suppose an $R$ module $M$ satisfies the following conditions.

1. $M$ is of generic rank $1$ .
2. $M$ is finitely generated.
3. $M$ is projective.

Then $M$ is $R$ -free of rank $1$ .

PROOF.. (Essentially borrowed from [1]) Let $K=Q(R)$ be the quotient field of $R$ .

Since $M$ is projective and $R$ is an integral domain, $M$ is torsion free. So

$$\iota: M\to M\otimes_R K$$

is injective. Since $M$ is of generic rank $1$ , $M\otimes_R K$ is isomorphic to $K$ . as a $K$ -module. We may thus assume that $M\subset K$ . Since $M$ is finitely generated, we may further assume that $M\subset R$ .

Now, Let us paraphrase the condition that $M$ being projective. First of all, the condition is equivalent to an existence of $R$ -module homomorphisms

$$f:M \to R^n , g:R^n \to M$$

such that $g\circ f=\operatorname{id}$ . Secondly, we may then represent $f,g$ in matrix form.

$$f(m)= \begin{pmatrix} f_1(m)\\ f_2(m) \\ \vdots\\ f_n(m) \end{... ..._2, g_3, \dots ,g_n) \begin{pmatrix} r_1\\ r_2\\ \vdots \\ r_n \end{pmatrix}$$

$$f:M \to R^n , g:R^n \to M$$

such that $g\circ f=\operatorname{id}$ .

Thirdly, each $f_i,g_i$ is represented by a linear map from $K$ to $K$ . That means, by an element of $K$ .

$$f= \begin{pmatrix} a_1\\ a_2 \\ \vdots\\ a_n \end{pmatrix}, \qquad g= (b_1, b_2, b_3, \dots ,b_n)$$

We may obtain several properties of $\{a_i\}, \{b_j\}$ :

1. $g(R^n)\subset M$ implies $b_i \in M \subset R (\forall i)$ .
2. $g\circ f: R^n \to R^n$ implies $a_i b_j\in R (\forall i\forall j)$ .
3. $g(R^n)\subset M$ implies $b_i \in M (\forall i)$ .
4. $g(R^n)= M$ implies $\sum_i R b_i = M$ .
5. $g\circ f=\operatorname{id}$ implies $\sum_i b_i a_i=1$ .

Let us write

$$a_i=\frac{l_i}{m_i} \qquad (\gcd(l_i,m_i)=1).$$

$$R\ni a_i b_j=\frac{l_i b_j}{m_i}$$

Since $l_i$ and $m_i$ are coprime, we conclude that $b_j$ is divisible by $m_i$ . Let us denote by $l$ the largest common multiple of $m_1,m_2,\dots,m_n$ .

$$l \vert b_j$$

By (iv) we see $M \subset l R$ .

By (v) we see

$$l=l\sum_i b_i a_i=\sum_i b_i l_i \frac{l}{m_i} \in \sum_i b_i R \subset M.$$

Thus $M\supset l R$ . So $M=l R$ . $\qedsymbol$

LEMMA 1.13   Let $R$ be a commutative UFD. Then for any $R$ -algebra homomorphism

$$\rho: M_n(R)\to M_n(R),$$

there exits an element $G\in \operatorname{GL}_n(R)$ such that

$$\rho(x)=G x G^{-1}$$

holds for any $x\in M_n(R)$ .

PROOF.. Let us denote by $e_{ij}\in M_n(R)$ the matrix element. Let us consider $R$ -modules

$$M_i=\rho(e_{ii}) R^n \qquad(i=1,2,3,\dots,n).$$

Then by an argument similar to that in (I,Lemma 7.9), we see that

$$R^n=\bigoplus_{i=1}^n M_i.$$

and that the multiplication by $\rho(e_{ji})$

$$M_i \overset{\rho(e_{ji}).}{\to} M_j$$

give isomorphisms between the modules. Hence we see easily that $M_0$ satisfy the assumptions of the previous lemma. We conclude that $M_1$ is freely generated by single element $v_1$ . Then we put

$$v_j=\rho(e_{j1}) v_1.$$

and

$$G=(v_1 v_2 v_3\dots v_n) .$$

We may easily see that $G$ plays the roles as expected. $\qedsymbol$