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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

$\displaystyle 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

$\displaystyle \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

$\displaystyle 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.

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

$\displaystyle 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$ .

$\displaystyle 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

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

$\displaystyle 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$ .

$\displaystyle l \vert b_j
$

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

By (v) we see

$\displaystyle 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

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

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

$\displaystyle \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

$\displaystyle 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

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

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

$\displaystyle 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

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

and

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

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


next up previous
Next: appendix 2 Up: A review Previous: Algebra endomorphisms and centers
2008-03-15