math0412133.tex
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\begin{document}
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{\footnotesize For the last version of this text, type {\em didrygaillard} on Google. Date of this version: Tue Dec 23 09:13:38 CET 2008. Jean-Marie Didry and Pierre-Yves Gaillard} \vfill

\cl{\LARGE Around the Chinese Remainder Theorem}\vfill

\tableofcontents\vfill\newpage

\section{Introduction}
%
The purpose of this text is to provide a convenient formulation of some folklore results. The five main statements are 
Theorem \ref{rp1} page \pageref{rp1}, 
Theorem \ref{rp2} page \pageref{rp2}, 
Theorem \ref{rp3} page \pageref{rp3}, 
Theorem \ref{rp4} page \pageref{rp4}, and 
Theorem \ref{rp5} page \pageref{rp5}. \bb

\ii By ``\p'' we mean ``complex coefficient \p\ in the indeterminate $X$''. \bb

\ii We show that

\begin{itemize}
\item the computation of the quotient and the remainder of the \de\ of a \p\ by a nonzero \p,
\item the partial fraction decomposition of a \fr, 
\item the computation of an inductive sequence,
\item the exponentiation of a matrix,
\item the integration of an order $n$ constant coefficient linear \edo,
\end{itemize}

follow from a unique, simple and obvious formula: Formula (\ref{tg}) page~\pageref{tg}, which we call {\em Taylor-Gauss Formula}. \bb

\ii The field $\C(X)$ of \frs\ and the ring $E$ of entire functions are two important examples of differential rings containing $\C[X]$. The subring $\C[X]$ ``controls'' $E$ in the sense that an entire function can be euclideanly divided by a nonzero \p, the remainder being a \p\ of degree strictly less than that of the divisor. Among the rings having this property, one contains all the others: it is the differential ring 
\begin{equation}\label{ce}
\prod_{a\in\C}\C[[X-a]].
\end{equation}
Any torsion $\C[X]$-module is a module over the ring (\ref{ce}), and this ring is universal for this property. In particular any element $f$ of (\ref{ce}) can be evaluated on a square matrix $A$, the matrix $f(A)$ being by definition $R(A)$, where $R$ is the remainder of the \de\ of $f$ by a nonzero \p\ annihilating $A$. Since there is an obvious formula for this remainder (the Taylor-Gauss Formula), we are done. An important example consists in taking for $f$ the exponential function, viewed as  the element of (\ref{ce}) whose $a$-th component is the Taylor series of $e^X$ at $a$. We recover of course the usual notion of exponential of a matrix, but cleared form its artificial complications. \bb

\ii It is handy to embed $\C(X)$ and $E$ into the differential ring $$\prod_{a\in\C}\C((X-a)),$$ which we use as a huge container. A side advantage of this ring is that it shortcuts the usual (particularly unilluminating) construction of the field of \frs\ (as a differential field) from the ring of \p s. \bb

%\ii The main idea used in this text shows up in {\em Local Class Field Theory} by Serre (see [1]). 
%
\section{Laurent Series}
%
Let $a$ be a \nc. A {\bf\sel} is an expression of the form 
%
$$f=f(X)=\sum_{n\in\Z}f_{a,n}\ (X-a)^n,$$
%
where $(f_{a,n})_{n\in\Z}$ is a family of complex numbers for which there is an integer $n_a$ such that $n<n_a$ implies $f_{a,n}=0$. \bb

\ii We define the operations of addition, multiplication and differentiation on the \sels\ by
%
$$(f+g)_{a,n}=f_{a,n}+g_{a,n},$$
%
$$(f\,g)_{a,n}=\sum_{p+q=n}f_{a,p}\ g_{a,q},$$
%%
$$(f')_{a,n}=(n+1)\ f_{a,n+1},$$
%
and we check that these operations have the same properties as on \p s.
%
\att Let $f$ be a \sel. If $f\not=0$, then there is a unique Laurent series $g$ in $X-a$ such that $f\,g=1$.\ztt
%
{\bf Proof.} Exercise.\bb

\ii Put $g=1/f=\frac{1}{f}$ and $h\,g=h/f=\frac{h}{f}$ if $h$ is a \sel.
%
\section{Generalized Rational Fractions}
%
A {\bf \frg} is a family $f=(f_a)_{a\in\C}$ each of whose members $f_a$ is a \sel. The complex numbers $f_{a,n}$ are called the {\bf coefficients} of $f$. The \frgs\ are added, multiplied and differentiated componentwise. \bb

\ii To each \p\ $P$ is attached the \frg
%
\begin{equation}\label{P}
%
\left(\sum_{n=0}^\infty\frac{P^{(n)}(a)}{n!}\ 
%
(X-a)^n\right)_{a\in\C}.
%
\end{equation}
%
As the formal sum in the parenthesis contains only a finite number of nonzero terms, it can be viewed as a \p. As such, it is of course equal to the \p\ $P$. Hence the \spd\ of \p s as \p s coincide with their \spd\ as \frgs. These facts prompt us to designate again by $P$ the \frg\ (\ref{P}).\bb

\ii We can now define a {\bf \fr} as being a \frg\ obtained by dividing a \p\ by a nonzero \p. The \spd\ of \frs\ are \frs. If a nonzero \frg\ $f=(f_a)_{a\in\C}$ is a \fr, then $f_a$ is nonzero for all $a$. \bb

\ii For all \frg\ $f$ and all \nc\ $a$ put
%
$$\mu(a,f):=\inf\ \{n\in\Z\ |\ f_{a,n}\not=0\}$$
%
with the convention $\inf\ \varnothing=+\infty$, and say that $\mu(a,f)$ is the {\bf multiplicity} of $a$ as zero, or root, of $f$. We have 
%
$$\mu(a,f g)=\mu(a,f)+\mu(a,g).$$ 
%
If $\mu(a,f)\ge0$ say that $f$ is {\bf defined at} $a$, and denote $f_{a,0}$ by $f(a)$. If a \frg\ $f$ is defined at a \nc\ $a$, then we have
%
$$f_a=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}\ (X-a)^n.$$
%
Any sum, product or derivative of \frgs\ defined at $a$ is a \frg\ defined at $a$. \bb

\ii For all \frg\ $f$, all \nc\ $a$, and all integer $\mu$ set 
%
$$\DL_a^\mu(f):=\sum_{n\le\mu}f_{a,n}\ (X-a)^n,$$
%
and say that this \sel\ is the {\bf order} $\mu$ {\bf \dl\ of $f$ at $a$}. If $f$ and $g$ are \frgs\ defined at $a$, we have
%
$$\DL_a^\mu(f+g)=\DL_a^\mu(f)+\DL_a^\mu(g),\quad
%
\DL_a^\mu(f\,g)
%
=\DL_a^\mu\Big(\DL_a^\mu(f)\DL_a^\mu(g)\Big).$$

\ii Let $a$ be a \nc, let $\mu$ be an integer, and let $f$ and $g$ be two \frgs. 
%
\ax\label{mod} Show that the following conditions are equivalent
\begin{enumerate}
\item $\mu(a,f-g)\ge\mu$,
\item $(X-a)^{-\mu}\ (f-g)$ is defined at $a$,
\item $\DL_a^{\mu-1}(f)=\DL_a^{\mu-1}(g)$.
\end{enumerate}
\zx 

\ii If these conditions are satisfied and if $f$ and $g$ are defined at $a$, then we say that $f$ and $g$ are {\bf congruent modulo} $(X-a)^\mu$ and we write $$f\e g\bmod(X-a)^\mu.$$
%
We thus have
%
$$\DL_a^{\mu-1}(f)\e f\bmod(X-a)^\mu,$$
%
as well as
%
$$\left.
\begin{array}{c}
 f_1\e g_1\bmod (X-a)^\mu\\ [1em]
 f_2\e g_2\bmod (X-a)^\mu
\end{array}\right\}\then\left\{
\begin{array}{c}
 f_1+f_2\e g_1+g_2\bmod (X-a)^\mu\\  [1em]
 f_1\,f_2\e g_1\,g_2\bmod (X-a)^\mu.
\end{array}\right.$$
%
\ax\label{pr} Assume $f$ is defined at $a$ and $\mu\ge0$. Let $R$ be a degree $<\mu$ \p. Show that the following conditions are equivalent. 
\begin{enumerate}
\item $R=\DL_a^{\mu-1}(f)$,
\item $R\e f\bmod(X-a)^\mu$,
\item $(X-a)^{-\mu}\ (f-R)$ is defined at $a$. 
\end{enumerate}
In other words 
%
$$\left(\frac{f-\DL_a^{\mu-1}(f)}{(X-a)^\mu}\ ,\ 
%
\DL_a^{\mu-1}(f)\right)$$ 
%
is the unique pair $(q,R)$ such that
%
\begin{enumerate}
\item $q$ is a \frg\ defined at $a$, 
\item $R$ is a \p\ of degree $<\mu$, 
\item $f(X)=(X-a)^\mu\,q(X)+R(X)$.
\end{enumerate}\zx

\ii If we wish to extend the result of Exercise \ref{pr} to the division by an arbitrary \p\ $D$, it is natural to restrict to the \frgs\ defined at all point of $\C$. 
We thus define a {\bf \pg} as being a \frg\ defined at all point of $\C$. Any sum, product or derivative of \pgs\ is a \pg. 
%
\ans Fix a nonconstant \p\ $D$ \label{D} and, for all complex number $a$, let $\mu_a$ be the multiplicity of $a$ as a root of $D$. 
\zns

\ax\label{pg}Let $f$ be a \pg. Show that the \frg\ $f/D$ is a \pg\ if and only if $f\e0\bmod(X-a)^{\mu_a}$ for all $a$. 
\zx

\ii We admit 
%
\att[Fundamental Theorem of Algebra] There a nonzero  complex number $c$ satisfying
%
$$D(X)=c\prod_{D(a)=0}(X-a)^{\mu_a}.$$
%
\ztt
%
\ax\label{p}Let $P$ be a \p. Show that the following conditions are equivalent
\begin{enumerate}
\item $P\e0\bmod(X-a)^{\mu_a}$ for all $a$, 
\item $P/D$ is a \pg,
\item $P/D$ is a \p,
\item $P/D$ is a degree $\deg P-\deg D$ \p.
\end{enumerate} 
\zx
%
\section{Chinese Remainder Theorem}
%
\att[Chinese Remainder Theorem] For all \pg\ $f$ and all nonconstant \p\ $D$ there is a unique couple $(q,R)$ such that\label{tc}
%
\begin{enumerate}
\item $q$ is a \pg, 
\item $R$ is a \p\ of degree $<\deg D$, 
\item $f=D\,q+R$.
\end{enumerate}
In Notation \ref{D} page \pageref{D} we have the {\bf Taylor-Gauss Formula} (see [1])
\aeq\label{tg}\boxed{\boxed{
%
R(X)=\sum_{D(a)=0}\ \DL_a^{\mu_a-1}\!\!\left(f(X)\
%
\frac{(X-a)^{\mu_a}}{D(X)}\right)
%
\frac{D(X)}{(X-a)^{\mu_a}}}}\zeq\ztt

\ii Say that $R$ is the {\bf remainder of the \de} of $f$ by $D$, and that $q$ is its {\bf quotient}. \bb

{\bf Proof.} Uniqueness. Assume
%
$$f=D\ q_1+R_1=D\ q_2+R_2$$
%
(obvious notation). We get
%
$$\frac{R_2-R_1}{D}=q_1-q_2$$
%
and Exercise \ref{p} (2. $\then$ 4.) implies $R_1=R_2$ and thus $q_1=q_2$. \bb

\ii Existence. Put
%
$$s_{(a)}(X):=\DL_a^{\mu_a-1}\!\!\left(f(X)\ 
%
\frac{(X-a)^{\mu_a}}{D(X)}\right)
%
\frac{1}{(X-a)^{\mu_a}}\quad,\quad s:=\sum\ s_{(a)},$$
%
and note that $\frac{f}{D}-s_{(a)}$ is defined at $a$ by Exercise~\ref{pr}, that $\frac{f}{D}-s$ is a \pg\ $q$, and that $s D$ is a degree $<\deg D$ \p\ $R$. QED\bb

{\bf Example.} Let $a$ and $b$ be two complex numbers. For $n\ge0$ write $s_n$ for the sum of the degree $n$ monomials in $a$ and $b$. The remainder of the \de\ of the \p\ $\sum_{n\ge0}a_n\ X^n$ by $(X-a)(X-b)$ is
%
$$\sum_{n\ge1}\ a_n\ s_{n-1}\ X+a_0
%
-a\ b\ \sum_{n\ge2}\ a_n\ s_{n-2}.$$
%
\acc {\em (First Serret Formula --- see [1].)} The principal part of $g:=f/D$ at $a$ is \label{s1}
%
$$\boxed{\boxed{
%
\DL_a^{\mu_a-1}\!\!\Big(g(X)(X-a)^{\mu_a}\Big)
%
(X-a)^{-\mu_a}}}
%
$$\zcc

\ii There are partial analogues to Theorem \ref{tc} and Corollary \ref{s1} over an arbitrary commutative ring (see Section~\ref{aq}), but the uniqueness of the partial fraction decomposition disappears. For instance we have, in a product of two nonzero rings, 
%
$$\frac{1}{X-1}-\frac{1}{X}
%
=\frac{(1,-1)}{X-(1,0)}+\frac{(-1,1)}{X-(0,1)}\quad.$$
%
\att Let $P$ be a \p, let $f$ be the \fr\ $P/D$, let $Q$ be the quotient, which we assume to be nonzero, of the \de\ of $P$ by $D$, and let $q$ be the degree of $Q$. Then $Q$ is given by the {\bf Second Serret Formula} (see [1])~: 
%
$$\boxed{Q(X^{-1})
%
=\DL_0^q(f(X^{-1})\ X^q)\ X^{-q}}$$
%
\ztt
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{\bf Proof.} If $d$ is the degree of $D$ and if $R$ is the remainder of the \de\ of $P$ by $D$, then the \fr
%
$$g(X):=X^\inv\Big(f(X^\inv)-Q(X^\inv)\Big)
%
=\frac{X^{d-1}\ R(X^\inv)}{X^d\ D(X^\inv)}$$
%
is defined at 0, and we have
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$$X^q\ f(X^\inv)-X^q\ Q(X^\inv)=X^{q+1}\ g(X)
%
\e0\bmod X^{q+1},$$
%
which, as $X^q\,Q(X^\inv)$ is a \p\ of degree at most $q$, implies
%
$$\DL_0^q\Big(X^q\ f(X^\inv)\Big)-X^q\ Q(X^\inv)=0.$$
%
QED\bb

\ii In Notation \ref{D} page \pageref{D} fix a root $a$ of $D$, denote by $B$ the set of all other roots, and, for any map $u:B\to\N,b\mapsto u_b$, denote by $|u|$ the sum of the $u_b$. 
%
\att The coefficient $c_{a,k}$ of $(X-a)^k$ in 
$\DL_a^{\mu_a-1}(\frac{(X-a)^{\mu_a}}{D(X)})$ is \label{fe}
%
\begin{equation}\label{efe}
c_{a,k}=(-1)^k\sum_{\deux{u\in\N^{B}}{|u|=k}}\ \prod_{b\in B}\ 
%
\binom{\mu_b-1+u_b}{\mu_b-1}\ \frac{1}{(a-b)^{\mu_b+u_b}}\quad.
\end{equation}
\ztt
%
{\bf Proof.} It suffices to multiply the \dls
%
$$\DL_a^{\mu_a-1}\!\!\left(\frac{1}{(X-b)^{\mu_b}}\right)
%
=\sum_{n=0}^{\mu_a-1}\ \binom{\mu_b-1+n}{\mu_b-1}\ 
%
\frac{(-1)^n\ (X-a)^n}{(a-b)^{\mu_b+n}}\quad.$$
%
QED
%
\acc The \p\ $R(X)$ in the Chinese Remainder Theorem (Theorem~\ref{tc} page \pageref{tc}) is \label{e}
%
$$R(X)=\ds\sum_{\deux{D(a)=0}{k+n<\mu_a}}\ c_{a,k}\ 
%
\frac{f^{(n)}(a)}{n!}\ (X-a)^{k+n},$$
%
where $c_{a,k}$ is given by (\ref{efe}). 
%
\zcc
%
\acc The coefficients of the partial fraction decomposition of a \fr\ are integral coefficient \p s in \label{cu} \bb
 
\ii $\bullet$ the coefficients of the numerator, 

\ii $\bullet$ the roots of the denominator, 
 
\ii $\bullet$ the inverses of the differences of the roots of the denominator. \bb

These \p s are homogeneous of degree one in the coefficients of the numerator and depend only on the multiplicities of the roots of the denominator. (We assume the denominator is monic.) 
\zcc 
%
\section{Exponential}
% 
The most important example of \pg\ is perhaps the exponential
%
$$e^X:=\left(e^a\sum_{n=0}^\infty\frac{(X-a)^n}{n!}
%
\right)_{a\in\C},$$
%
which satisfies
%
$$\frac{d}{d X}\ e^X=e^X$$
%
as a \pg. \bb

\ii More generally we can, for any real number $t$, define the \pg
%
\begin{equation}\label{etx}
e^{t X}
%
:=\left(e^{a t}\sum_{n=0}^\infty\frac{t^n\ (X-a)^n}{n!}
%
\right)_{a\in\C},
\end{equation}
%
and observe the identity between \pgs
%
$$e^{t X}\ e^{u X}=e^{(t+u) X}.$$

\ii We would like to differentiate $e^{t X}$ not only with respect to $X$ but also with respect to $t$. To do this properly we must allow the coefficients of a \pg\ to be not only complex constants, but $C^\infty$ functions from $\R$ to $\C$. Let us call {\bf \pgcv} a \pg\ whose coefficients are $C^\infty$ functions from $\R$ to $\C$. We then have the following identity between \pgcvs
%
$$\frac{\partial e^{t X}}{\partial t}=X\,e^{t X}.$$

\ii One is often lead to compute \dls\ of the form $$\DL_a^\mu(e^{t X}f(X)),$$ where $f(X)$ is a \frd. This \dl\ is the unique degree $\le \mu$ \p\ satisfying
%
\begin{equation}\label{w1}
%
\DL_a^\mu(e^{t X}f(X))\e e^{a t}\ \DL_a^\mu (f(X))\
%
\sum_{n=0}^\mu\frac{t^n\ (X-a)^n}{n!}\ \bmod(X-a)^{\mu+1}.
%
\end{equation}

\ii Let us sum this up by 
%
\att In Notation \ref{D} page \pageref{D} the remainder of the \de\ of $e^{t X}$ by $D(X)$ is given by the {\bf Wedderburn Formula} (see [1])
%
\begin{equation}\label{w2}
%
\boxed{\sum_{D(a)=0}\ \DL_a^{\mu_a-1}\!\!\left(e^{t X}\
%
\frac{(X-a)^{\mu_a}}{D(X)}\right)
%
\frac{D(X)}{(X-a)^{\mu_a}}}
%
\end{equation}
%
the \dl\ being given by (\ref{w1}). 
%
\ztt
%
\section{Matrices}
%
Let $A$ a complex square matrix. 
%
\ax\label{pm} Show that there is a unique monic \p\ $D$ which annihilates $A$ and which divides any \p\ annihilating $A$. 
\zx 

\ii Say that $D$ is the {\bf minimal \p} of $A$. 
%
\ax Let $f$ be a \pg, $D$ a \p\ annihilating $A$, and $R$ the remainder of the \de\ of $f$ by $D$. Show that the matrix $R(A)$ does not depend on the choice of the annihilating \p\ $D$. [Suggestion: use Exercises \ref{p} and \ref{pm}.]
\zx 

\ii It is then natural to set $$f(A):=R(A).$$
%
\ax Let $f,g$ be two \pgs\ and $D$ the minimal \p\ of $A$. Show $f(A)=g(A)\iff D$ divides $f-g$.
\zx 
%
\ax Show $(f\,g)(A)=f(A)\,g(A)$. \zx 
%
\att The matrix $e^{t A}:=f(A)$, where $f(X)=e^{t X}$ (see (\ref{etx})), depends differentiably on $t$ and satisfies $\frac{d}{d t}\ e^{t A}=A\ e^{t A},e^{0A}=1$.  
%
\ztt
%
\ax Prove the above statement.\zx 
%
\ax Let $$D(X)=X^q+a_{q-1}\,X^{q-1}+\dots+a_0$$ be a monic \p, $(e_j)$ the canonical basis of $\C^q$, and $A$ the $q$ by $q$ matrix characterised \label{co} by
%
$$j<q\then A\,e_j=e_{j+1},\quad 
%
A\,e_q=-a_0\,e_1-\dots-a_{q-1}\,e_q.$$
%
Show that $D$ is the minimal \p\ of $A$.\zx 

\ii Let $D$ be a degree $q\ge1$ polynomial, let $f$ be a \pg, and $b_{q-1}\,X^{q-1}+\dots+b_0$ the remainder of the \de\ of $f$ by $D$. 
%
\ax Show $f(A)\,e_1=b_0\,e_1+\dots+b_{q-1}\,e_q$. \label{fA}
\zx  
%
\section{Inductive Sequences}
%
Let $D$ be a degree $q\ge1$ monic polynomial; let $\C^\N$ be the set of complex number sequences; let $\DD$ be the shift operator mapping the sequence $u$ in $\C^\N$ to the sequence $\DD u$ in $\C^\N$ defined by $(\DD u)_t=u_{t+1}$; let $f$ be in $\C^\N$; let $c_0,\dots,c_{q-1}$ be in $\C$; let $y$ be the unique element of $\C^\N$ satisfying 
%
$$D(\DD)\ y=f,\quad y_n=c_n\mbox{ for all } n<q\ ;$$
%
for $(n,t)$ in $\N^2$ denote by $g_n(t)$ the coefficient of $X^n$ in the remainder of the \de\ of $X^t$ by $D$. 
%
\att If $t\ge q$ then $\ds y_t=\sum_{n<q}\ c_n\ g_n(t)
%
+\sum_{k<t}\ g_{q-1}(t-1-k)\ f_k.$
%
\ztt
%
{\bf Proof.} Introduce the sequence of vectors $x_t:=(y_t,\dots,y_{t+q-1})$. We have 
%
$$x_{t+1}=B\,x_t+f_t\,e_q,\quad x_0=c,$$
%
where $e_q$ is the last vector of the canonical basis of $\C^q$, and $B$ is the transpose of the matrix $A$ in Exercise \ref{co}. Hence
%
$$x_t=B^t c+f_0\,B^{t-1}\,e_q+f_1\,B^{t-2}\,e_q+\dots +f_{t-1}\,e_q.$$
% 
It suffices then to take the first component of the left and right hand sides, and to invoke Exercise \ref{fA}. QED
%
\section{Differential Equations}
%
Let $D$ be a degree $q\ge1$ monic polynomial and $y$ the unique solution to the \edo

\aeq\label{edo}D\left(\frac{d}{d t}\right)y=f(t),\quad
 %
 y^{(n)}(0)=y_n\ \forall\ n<q:=\deg D,\zeq

where $f:\R\to\C$ is a continuous function.
%
\att We have the {\bf Collet Formula} (see [1])
%
$$\boxed{y(t)=\sum_{n<q}y_n\ g_n(t)
%
+\int_0^t g_{q-1}(t-x)\ f(x)\ d x}$$
%
where $g_n(t)$ is the coefficient of $X^n$ in the remainder of the \de\ of $e^{t X}$ by $D$. 
\ztt

\ii [For $D(X)=X^q$ we recover the Taylor Formula with integral remainder.] \bb 

{\bf Proof.} Putting $v_n:=y^{(n-1)}$, $v_{0n} :=y_{n-1}$ for $1\le n\le q$, and denoting by $e_q$ the last vector of the canonical basis of $\C^q$, Equation (\ref{edo}) takes the form
%
$$v'(t)-B\ v(t)=f(t)\ e_q,\quad v(0)=v_0,$$
%
where $B$ is the transpose of the matrix $A$ in Exercise \ref{co}. Applying $e^{-t B}$ we get
%
$$\frac{d}{d t}\ e^{-t B}\ v(t)=e^{-t B}\ f(t)\ e_q,\quad v(0)=v_0,$$
%
whence
%
$$v(t)=e^{t B}\ v_0+\int_0^t f(x)\ e^{(t-x)B}\ e_q\ d x.$$
%
In view of the Wedderburn Formula (\ref{w2}), it suffices then to take the first component of the left and right hand sides, and to invoke Exercise~\ref{fA}. QED\bb

\ii Let $h$ be a \pgcv, let $f$ and $y$ be two continuous functions from $\R$ to $\C^q$, let $y_0$ be a vector in $\C^q$, and $A$ a $q$ by $q$ complex matrix. If $y$ is differentiable and satisfies
%
\begin{equation}\label{a}
y'(t)+h(t,A)\ y(t)=f(t),\quad y(0)=y_0,
\end{equation}
%
then
%
$$H(t,A):=\int_0^t h(u,A)\ d u\quad\then\quad
%
\frac{d}{d t}\ e^{H(t,A)}\ y(t)=e^{H(t,A)}\ f(t),$$
%
whence 
%
\att The unique solution to (\ref{a}) is given by the {\bf Euler Formula} (see [1])

$$\boxed{y(t)=\exp\left(\int_t^0h(u,A)\ d u\right)y_0
%
+\int_0^t\exp\left(\int_t^v h(u,A)\ d u\right) f(v)\ d v}$$
%
\ztt
%
\section{Euclid}
%
By ``ring'' we mean in this text ``commutative ring with 1''. For the reader's convenience we prove the Chinese Remainder Theorem. 
%
\att [Chinese Remainder Theorem] Let $A$ be a ring and $I_1,\dots,I_n$ ideals such that $I_p+I_q=A$ for $p\not=q$. Then the natural morphism \label{tcc} from $A$ to the product of the $A/I_p$ is surjective. Moreover the intersection of the $I_p$ coincides with their product. 
\ztt
%
{\bf Proof.} Multiplying the equalities $A=I_1+I_p$ for $p=2,\dots,n$ we get
%
\aeq\label{12n}A=I_1+I_2\cdots I_n.\zeq
%
In particular there is an $a_1$ in $A$ such that
%
$$a_1\equiv1\bmod I_1,\quad a_1\equiv0\bmod I_p\ \forall\ p>1.$$
%
Similarly we can find elements $a_p$ in $A$ such that
$a_p\equiv\delta_{p q}\bmod I_q$ (Kronecker delta). This proves the first claim. Let $I$ be the intersection of the $I_p$. Multiplying (\ref{12n}) by $I$ we get 
%
$$I=I_1I+II_2\cdots I_n\subset I_1\ (I_2\cap\cdots \cap I_n)\subset I.$$
%
This gives the second claim, directly for $n=2$, by induction for $n>2$. QED \bb

\ii Let $A$ be a PID (principal ideal domain). Construct the ring $A^\w$ as follows. A family $(a_d)_{d\not=0}$ of \els\ of $A$ indexed by the nonzero \els\ $d$ of $A$ represents an \el\ of $A^\w$ if it satisfies
%
$$d\ |\ e\then a_d\e a_e\bmod d$$
%
(where $d\ |\ e$ means ``$d$ divides $e$'') for all pair $(d,e)$ nonzero \els\ of $A$. Two such families $(a_d)_{d\not=0}$ and $(b_d)_{d\not=0}$ represent the same \el\ of $A^\w$ if and only if $$a_d\e b_d\bmod d\quad\forall\ d\not=0.$$ The ring structure is defined in the obvious way. Embed $A$ into $A^\w$ by mapping $a$ in $A$ to the constant family equal to $a$. By abuse of notation we often designate by the same symbol an element of $A^\w$ and one of its representatives. Let $P$ be a representative system of the association classes of prime \els\ of $A$.
%
\al Let $a=(a_b)_{b\not=0}$ be in $A^\w$ and $d$ a nonzero \el\ of $A$ such that $a_d\e 0\bmod d$. \label{27} Then there is a $q$ in $A^\w$ such that  $a=d\,q$.
\zl
%
{\bf Proof.} Let $p$ be in $P$ and $i$ the largest integer $j$ such that $p^j$ divides $d$. In other words there is an \el\ $d'$ of $A$ which is prime to $p$ and satisfies $d=p^i d'$. For all nonnegative integer $j$ we have
%
$$a_{p^{i+j}}\e0\bmod p^i.$$ 
%
As a result there is an \el\ $a'_j$ of $A$ such that $a_{p^{i+j}}=p^i\,a'_j$. We have
%
$$p^i\,a'_{j+1}\e a_{p^{i+j+1}}\e a_{p^{i+j}}\e 
%
p^i\,a'_j\bmod p^{i+j}$$
%
and thus 
%
$$a'_{j+1}\e a'_j\bmod p^j.$$
%
For all $j$ choose $q_{p^j}$ such that 
%
$$d'\,q_{p^j}\e a'_j\bmod p^j$$
%
and thus
%
$$d\,q_{p^j}\e a_{p^j}\bmod p^j.$$
%
We get
%
$$d'\,q_{p^{j+1}}\e a'_{j+1}\e a'_j\e d'\,q_{p^j}
%
\bmod p^j$$
%
and thus
%
$$q_{p^{j+1}}\e q_{p^j}\bmod p^j.$$
%
Let $b$ be in $A$, $b\not=0$, and $P_b$ the (finite) set of those \els\ of $P$ which divide $b$. For $p$ in $P_b$ denote by $i(p)$ the largest integer $j$ such that $p^j$ divides $b$ and choose a solution $q_b$ in $A$ to the congruence system
%
$$q_b\e q_{p^{i(p)}}\bmod p^{i(p)},\quad p\in P_b,$$
%
solution which exists by the Chinese Remainder Theorem (Theorem \ref{tcc} page \pageref{tcc}). One then checks that the family $q:=(q_b)_{b\not=0}$ is in $A^\w$ and that we do have $d\,q=a$. QED\bb

\ii See Theorem \ref{rp4} page \pageref{rp4} for a generalization. \bb

\ii Let $a$ be in $A^\w$, let $d$ be a nonzero \el\ of $A$, let $\mu(p)$ be the multiplicity of $p$ in $P$ as a factor of $d$, let 
%
$$\DL_p^{\mu(p)-1}\!\!\left(a\
%
\frac{p^{\mu(p)}}{d}\right)$$
%
be an \el\ of $A$ satisfying
%
$$\DL_p^{\mu(p)-1}\!\!
%
\left(a\ \frac{p^{\mu(p)}}{d}\right)
%
\frac{d}{p^{\mu(p)}}\e a_{p^{\mu(p)}}\bmod p^{\mu(p)},$$
%
and put 
%
$$\rho:=\sum_{p\ |\ d}\ \DL_p^{\mu(p)-1}\!\!\left(a\
%
\frac{p^{\mu(p)}}{d}\right)\frac{d}{p^{\mu(p)}}\in A.$$

\ii Let $B$ be a ring and $A$ an integral domain contained in $B$. Say that $A$ is {\bf principal in} $B$ if for all $b$ in $B$ and all nonzero $d$ in $A$ there is a $q$ in $B$ and an $r$ in $A$ such that $b=d q+r$. \bb
%
\att[First Main Statement] Let $A$ be a PID, let $a$ be an element of $A^\w$, and let $d$ be a nonzero element of $A$. In the above notation, \label{rp1} the elements $a$, $a_d$ and $\rho$ of $A^\w$ are congruent modulo $d$. Moreover, if $A$ is principal in $B$, then there is a unique $A$-algebra morphism from $B$ to $A^\w$. 
\ztt
%
{\bf Proof.} Modulo $d$ we have $a\e a_d$ by the Lemma, and $a_d\e\rho$ by the Chinese Remainder Theorem (Theorem \ref{tcc} page \pageref{tcc}). This proves the first claim. Let us check the second one, starting with the uniqueness. Let $f$ be an $A$-linear map from $B$ to $A^\w$. If $b$ and $q$ are in $B$, and if $d\not=0$ and $r$ are in $A$, then the equality $b=d\,q+r$ implies $f(b)=d\,f(q)+r$, and thus $f(b)_d\e r\bmod d$. Consequently there is at most one such map. The existence is proved by putting $f(b)_d:=r$ in the above  notation, and by checking that this formula does define an $A$-linear map from $B$ to $A^\w$. QED \bb

\ii An $A$-module is {\bf torsion} if each of its vectors is annihilated by some nonzero scalar. 
%
\att If $A$ is principal in $B$, then any torsion $A$-module admits a unique $B$-module structure which extends its \label{w} $A$-module structure. Moreover any $A$-linear map between torsion $A$-modules is $B$-linear. 
\ztt

{\bf Proof.} Let us check the uniqueness. Let $b$ be in $B$~; let $v$ be in our torsion module $V$~; let $d\not=0$ in $A$ satisfy $d v=0$~; let $q$ in $B$ and $r$ in $A$ verify $b=d q+r$. We then get $b v=r v$, which proves the uniqueness. The existence is obtained by setting $b v:=r v$ in the above notation, and by checking that this formula does define a $B$-module structure on $V$ which extends the $A$-module structure. The last assertion is clear. QED \bb

\ii [Let $\oo{Hom}_{c t}(G,H)$ be the group of continuous morphisms from the topological group $G$ to the abelian topological group $H$, and $\widehat{\Z}$ the profinite completion of $\Z$. Equip $\Q/\Z$ with the discrete topology or with the topology induced by that of $\R/\Z$. Equip also $\Q/\Z$ with the $\widehat{\Z}$-module structure provided by Theorem~\ref{w}. We then have  
%
$$\oo{Hom}_{c t}(\widehat{\Z},\R/\Z)
%
=\oo{Hom}_{c t}(\widehat{\Z},\Q/\Z)
%
=\oo{Hom}_{\widehat{\Z}}(\widehat{\Z},\Q/\Z)
%
=\Q/\Z.$$
%
We recover the well known fact that $\Q/\Z$ is the dual of $\widehat{\Z}$ in the category of locally compact abelian groups. See Weil's {\em L'int\'egration dans les groupes topologiques}, pp. 108-109.] \bb

\ii Let $L$ be a sublattice of the lattice of ideals of an arbitrary ring $A$. Here are some examples: 
%
\begin{enumerate}
\item $L$ is the set of nonzero ideals of an integral domain, 
\item $L$ is the set of powers of a fixed ideal of an arbitrary ring, 
\item let $Y$ be a set of prime ideals of $A$, and $L$ the set of those \label{supp} ideals $I$ of $A$ such that any prime ideal containing $I$ is in $Y$, 
\item $L$ is the set of open ideals of a topological ring.
\end{enumerate}
%
Construct the ring $A^\w$ as follows. An \el\ of $A^\w$ is represented by a family $(a_I)_{I\in L}$ of \els\ of $A$ satisfying
%
$$I\subset J\then a_I\e a_J\bmod J$$
%
for all $I,J$ in $L$. Two such families $(a_I)_{I\in L}$ and $(b_I)_{I\in L}$ represent the same \el\ of $A^\w$ if and only if 
%
$$a_I\e b_I\bmod I\quad\forall\ I\in L.$$ 
%
The ring structure is defined in the obvious way. By abuse of notation we often designate by the same symbol an element of $A^\w$ and one of its representatives. Map $A$ to $A^\w$ by sending $a$ in $A$ to the constant family equal to $a$. \bb

\ii Call {\bf torsion module} any $A$-module each of whose vector is annihilated by some \el\ of $L$. [In the case of Example \ref{supp} above, an $A$-module is torsion if and only if its support is contained in $Y$ --- Bourbaki, {\em Alg. Com.} II.4.4.] Assume to simplify that the intersection of the \els\ of $L$ reduces to zero, and consider $A$ as a subring of $A^\w$. 
%
\att[Second Main Statement] Any torsion $A$-module $V$ admits a unique $A^\w$-module structure such that \label{rp2} 
%
$$a\in A^\w,v\in V,I\in L,Iv=0\then a v=a_I\,v.$$
%
This $A^\w$-module structure on $V$ extends the $A$-module structure.
\ztt
%
{\bf Proof.} The uniqueness being obvious, lets us check the existence. For $v$ in $V$ and $I,J$ in $L$ such that $I v=0=J v$, we have $a_I\,v= a_{I+J}\ v$; as this vector depends only on $a$ and $v$, it can be denoted $a v$. Let us show $a\,(v+w)=a v+a w$. If $a$ is in $A^\w$, and if $I,J$ are in $L$ and verify $I v=0=J w$, we have
%
$$a\ (v+w)=a_{I\cap J}\ (v+w)=a_{I\cap J}\ v+a_{I\cap J}\ w=a\,v+a\,w.$$
%
QED 
%
\att[Third Main Statement] Let $B$ a ring containing $A$. Assume that each torsion $A$-module is equipped with a $B$-module structure which extends its $A$-module structure, and that each $A$-linear map between torsion $A$-modules is $B$-linear. \label{rp3} Then there is a unique $A$-algebra morphism $f$ from $B$ to $A^\w$ which satisfies $b\,v=f(b)\,v$ for all vector $v$ of any torsion $A$-module, and all $b$ in $B$. 
\ztt
%
{\bf Proof.} For all $I$ in $L$ denote by $1_I$ the element 1 of $A/I$. Let us verify the uniqueness. Let $b$ be in $B$ and $I$ in $L$. There is an $a$ in $A$ satisfying
%
$$a\cdot1_I=b\cdot1_I=f(b)\cdot1_I=f(b)_I\cdot1_I,$$
%
and thus $f(b)_I\e a\bmod I$. To check the existence, we put $f(b)_I:=a$ in the above notation, and verify that this formula does define an $A$-linear map $f$ from $B$ to $A^\w$ which satisfies $b\,v=f(b)\,v$ for all vector $v$ of any torsion $A$-module, and all $b$ in $B$. [To check that $f(b)_I$ and $f(b)_J$ are congruent modulo $J$ for $I\subset J$ in $L$, we use the assumption that the canonical projection from $A/I$ to $A/J$ is $B$-linear.] QED \bb 

\ii Let $A$ be a Dedekind domain; let $L$ be the set of nonzero ideals of $A$; and let $M$ be the set of maximal ideals of $A$. For any $A$-module $V$, let $V^\w$ be the projective limit of the $V/I V$ with $I$ in $L$; and, for any $P$ in $M$, let $V^\w_P$ be the projective limit of the $V/P^n V$. Then $A^\w$ and $A^\w_P$ are $A$-algebras; by the Chinese Remainder Theorem (Theorem \ref{tcc} page \pageref{tcc}) $A^\w$ is the direct product of the $A^\w_P$; the $A$-module $V^\w$ is an $A^\w$-module; the $A$-module $V^\w_P$ is an $A^\w_P$-module; $V^\w$ is the direct product of the $V^\w_P$. Here is a generalization of Lemma \ref{27} page \pageref{27}. 
%
\att[Fourth Main Statement] The ring $A^\w$ is $A$-flat. If $V$ is a finitely generated $A$-module, then the natural morphism from $A^\w\otimes_A V$ to $V^\w$ is an isomorphism. \label{rp4} In particular, if $I$ is in $L$, then $I A^\w$ is the kernel of the canonical projection of $A^\w$ onto $A/I$. 
\ztt
%
{\bf Proof} \ \ The first claim follows from Proposition VII.4.2 of {\em Homological Algebra} by Cartan and Eilenberg (Princeton University Press, 1956). The second claim follows from Exercise II.2 of the same book in conjunction with Proposition 10.13 of {\em Introduction to Commutative Algebra} by Atiyah and Macdonald (Addison-Wesley, 1969). The third claim is obtained by tensoring the exact sequence 
%
$$0\to I\to A\to A/I\to0$$
%
with $A^\w$ over $A$. QED
%
\section{The Case of an Arbitrary Ring} \label{aq}
%
(Reminder: by ``ring'' we mean ``commutative ring with 1''.) Let $A$ be a ring, $X$ an indeterminate, and $a$ an element of $A$. For any formal power series 
$$f=\sum_{n\ge0}\ a_n\ (X-a)^n\in A[[X-a]]$$ 
and any nonnegative integer $k$ put 
%
$$\frac{f^{(k)}}{k!}:=\sum_{n\ge k}\ \bi{n}{k}\ a_n\
%
(X-a)^{n-k}\ \in A[[X-a]].$$ 
%
We then get the {\bf Taylor Formula}
%
$$f=\sum_{n\ge0}\ \frac{f^{(n)}(a)}{n!}\ (X-a)^n.$$

\ii We regard the $A$-algebra $A[[X-a]]$ as an $A[X]$-algebra via the morphism from $A[X]$ to $A[[X-a]]$ which maps $P$ to 
%
$$\sum_{n\ge0}\ \frac{P^{(n)}(a)}{n!}\ (X-a)^n.$$

\ii For any $f$ in $A[[X-a]]$ and any nonnegative integer $k$ we call $k$-{\bf jet} of $f$ at $a$ the polynomial 
%
$$\DL_a^k(f)
%
:=\sum_{n\le k}\ \frac{f^{(n)}(a)}{n!}\ (X-a)^n.$$
%
Then $\DL_a^k$ induces a ring morphism 
%
$$\DL_a^k:A[[X-a]]\longrightarrow
%
\frac{A[X]}{(X-a)^k}\quad.$$

\ii Let $a_1,\dots,a_r$ be in $A$; let $m_1,\dots,m_r$ be positive integers; let $D$ be the product of the $(X-a_i)^{m_i}$; and let $u$ be the natural morphism from $A[X]/(D)$ to the product $B$ of the $A[X]/(X-a_i)^{m_i}$:
%
$$u:\frac{A[X]}{(D)}\longrightarrow 
%
B:=\prod_{i=1}^r\ \frac{A[X]}{(X-a_i)^{m_i}}\quad.$$
%
Assume that $a_i-a_j$ is invertible for all $i\not=j$. 
In particular 
%
$$D_i(X):=\frac{D(X)}{(X-a_i)^{m_i}}$$
%
is invertible in $A[[X-a_i]]$. Let $v$ be the morphism 
%
$$v:\Big(P_i\bmod(X-a_i)^{m_i}\Big)_{i=1}^r\ \mapsto\  
%
\sum_{i=1}^r\ 
%
\DL_{a_i}^{m_i-1}\!\!\left(\frac{P_i}{D_i}\right)D_i
%
\bmod D$$
%
from $B$ to $A[X]/(D)$. 
%
\att[Fifth Main Statement] The maps $u$ and $v$ are inverse ring isomorphisms. \label{rp5} Moreover Corollary \ref{e} page \pageref{e} remains true (with the obvious changes of notation). 
\ztt 
%
{\bf Proof}\ \ The composition $u\circ v$ is obviously the identity of $B$. Since both rings are rank $\deg D$ free $A$-modules, this proves the statement. QED
%
\section{Universal Remainder}
%
In this Section we work over some unnamed ring 
(reminder: by ``ring'' we mean ``commutative ring with 1''). The remainder of the euclidean division of $X^r$ by $$(X-a_1)\cdots(X-a_k)$$ is a universal polynomial in $a_1,\dots,a_k,X$ with integral coefficients. To compute this polynomial first recall Newton's interpolation. \bb 

\ii As a general notation set
%
$$f(u,v):=\frac{f(v)-f(u)}{v-u}\quad,$$
%
define the {\em Newton interpolation polynomial}  
%
$$N(f(X);a_1,\dots,a_k;X)$$ 
%
of an arbitrary polynomial $f(X)$ at $a_1,\dots,a_k$ by

$$N(f(X);a_1,\dots,a_k;X):=f(a_1)+f(a_1,a_2)(X-a_1)$$
%
$$+f(a_1,a_2,a_3)(X-a_1)(X-a_2)+\cdots$$
%
$$+f(a_1,\dots,a_k)(X-a_1)\cdots(X-a_{k-1}),$$

and put 
%
$$g(X):=N(f(X);a_1,\dots,a_k;X),$$
 
$$h(X):=N(f(a_1,X);a_2,\dots,a_k;X).$$ 

\att[Newton Interpolation Theorem] We have $g(a_i)=f(a_i)$ for $i=1,\dots,k$. In particular 
$g(X)$ is the remainder of the euclidean division of $f(X)$ by $(X-a_1)\cdots(X-a_k).$ Moreover $f(a_1,\dots,a_i)$ is symmetric in $a_1,\dots,a_i$.\ztt 
%
{\bf Proof.} Argue by induction on $k$, the case $k=1$ being easy. Note $$g(X)=f(a_1)+(X-a_1)\ h(X).$$ The equality $g(a_1)=f(a_1)$ is clear. Assume $2\le i\le k$. By induction hypothesis we have $h(a_i)=f(a_1,a_i)$ and thus 
%
$$g(a_i)=f(a_1)+(a_i-a_1)\ f(a_1,a_i)=f(a_i).$$ 
%
Since $f(a_1,\dots,a_k)$ is the leading coefficient of $g(X)$, it is symmetric in $a_1,\dots,a_k$. QED \bb

\ii Assume that $$b_i:=\prod_{j\not=i}\ (a_i-a_j)$$ is invertible for all $i$. By Lagrange interpolation we have, for $f(X)=X^r=f_r(X)$,  
%
$$f_r(a_1,\dots,a_k)=\sum_i\ \frac{a_i^r}{b_i}\quad.$$
%
The generating function of the sum $s_n=s_n(a_1,\dots,a_k)$ of the degree $n$ monomials in $a_1$, \dots, $a_k$ being 
%
$$\frac{1}{(1-a_1\,X)\cdots (1-a_k\,X)}
%
=\sum_i\ \frac{a_i^{k-1}\ b_i^{-1}}{1-a_i\,X}
%
=\sum_{n,i}\ \frac{a_i^{k-1+n}}{b_i}\ X^n,$$
%
we get 
%
$$s_n=\sum_i\ \frac{a_i^{k-1+n}}{b_i}\quad,
%
\quad f_r(a_1,\dots,a_k)=s_{r-k+1}$$
%
with the convention $s_n=0$ for $n<0$, and the remainder of the euclidean division of $X^r$ by $(X-a_1)\cdots (X-a_k)$ is 
%
$$\sum_{i=1}^k\ s_{r-i+1}(a_1,\dots,a_i)\ 
%
(X-a_1)\cdots(X-a_{i-1})$$
%
(even when some of the $b_i$ are not invertible). 
%
\section{An Adjunction} 
%
Recall that a $\Q${\bf -algebra} is a ring containing $\Q$ (``ring'' meaning ``commutative ring with 1''), and that a {\bf derivation} on a $\Q$-algebra $A$ is a $\Q$-vector space endomorphism $a\mapsto a'$ satisfying $(a b)'=a' b+a b'$ for all $a,b$ in $A$. A {\bf differential} $\Q$-algebra is a $\Q$-algebra equipped with a derivation. We leave it to the reader to define the notions of $\Q$-algebra and of differential $\Q$-algebra {\bf morphism}. To each $\Q$-algebra $A$ is attached the $\Q$-algebra $A[[X]]$ (where $X$ is an indeterminate) equipped with the derivation $\frac{d}{d X}\ $. Let $A$ be a $\Q$-algebra and $B$ a differential $\Q$-algebra. Denote respectively by
%
$$\mathcal{A}(B,A)\quad\mbox{and}\quad
%
\mathcal{D}(B,A[[X]])$$
%
the  $\Q$-vector space of $\Q$-algebra morphisms from $B$ to $A$ and of differential $\Q$-algebra morphisms  from $B$ to $A[[X]]$. The formulas 
%
$$f(b)=F(b)_0,\quad
% 
F(b)=\sum_{n=0}^\infty\frac{f(b^{(n)})}{n!}\ X^n,$$
%
where $F(b)_0$ designates the constant term of $F(b)$, set up a bijective $\Q$-linear correspondence between vectors $f$ of $\mathcal{A}(B,A)$ and vectors $F$ of $\mathcal{D}(B,A[[X]])$. Let us check for instance the following point. Let $F$ be in $\mathcal{D}(B,A[[X]])$, let $f$ in $\mathcal{A}(B,A)$ be the constant term of $F$, and let $G$ in $\mathcal{D}(B,A[[X]])$ be the extension of $f$. Let us show $G=F$. We have

$$\begin{array}{lcll}
G(b)&=&\ds\sum_{n=0}^\infty\ \frac{G(b)^{(n)}(0)}{n!}\ X^n&\mbox{by Taylor}\\[2em]
%
&=&\ds\sum_{n=0}^\infty\ \frac{G(b^{(n)})(0)}{n!}\ X^n&\mbox{because $G(x)'=G(x')$}\\[2em]
%
&=&\ds\sum_{n=0}^\infty\ \frac{f(b^{(n)})}{n!}\ X^n&\mbox{by definition of $G$}\\[2em]
%
&=&\ds\sum_{n=0}^\infty\ \frac{F(b^{(n)})(0)}{n!}\ X^n&\mbox{by definition of $f$}\\[2em]
%
&=&\ds\sum_{n=0}^\infty\ \frac{F(b)^{(n)}(0)}{n!}\ X^n&\mbox{because $F(x')=F(x)'$}\\[2em]
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&=&F(b)&\mbox{by Taylor.}
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\end{array}$$ 
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\section{Wronski} 
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Here is a brief summary of Section II.2 in H. Cartan's book {\em Calcul diff\'erentiel}. If $E$ and $F$ are Banach spaces, let $\mathcal{L}(E,F)$ be the Banach space of continuous linear maps from $E$ to $F$. Let $E$ be a Banach space, $I$ a nonempty open interval, and $A$ a continuous map from $I$ to $\mathcal{L}(E,E)$. Consider the ODE's 

\begin{equation}\label{wr1}x'=A(t)\,x,\end{equation} 

where the unknown is a map $x$ from $I$ to $E$, and
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\begin{equation}\label{wr2}X'=A(t)\,X,\end{equation}
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where the unknown is a map $X$ from $I$ to $F:=\mathcal{L}(E,E)$. Let $t_0$ be in $I$. The solution to (\ref{wr2}) satisfying $X(t_0)=Id_E$ is called the {\bf resolvent} of (\ref{wr1}) and is denoted $t\mapsto R(t,t_0)$.\bb
 
\ii Let $B$ be a continuous map from $I$ to $E$ and $x_0$ a vector of $E$. The solution to 
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$$x'=A(t)\,x+B(t),\quad x(t_0)=x_0$$
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is 
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$$x(t):=R(t,t_0)\,x_0
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+\int_{t_0}^t R(t,\tau)\,B(\tau)\ d\tau.$$

\ii Here is an additional comment. For $t$ in $I$ let $\mathcal{A}(t)$ be the left multiplication by $A(t)$. In particular $\mathcal{A}$ is a continuous map from $I$ to $\mathcal{L}(F,F)$. Let $\mathcal{R}$ be its resolvent. Then $\mathcal{R}(t,t_0)$ is the left multiplication by $R(t,t_0)$.\bb

\ii Corollary. If $W$ is a differentiable map from $I$ to $F$ satisfying $W'=A\,W$, then 
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$$W(t)=R(t,t_0)\, W(t_0).$$

\ii Corollary to the Corollary. If in addition $W(t_0)$ is invertible, then 
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$$R(t,t_0)=W(t)\, W(t_0)^{-1}.$$

\centerline{*}\bb 

[1] For terminological justifications, type {\em didrygaillard} on Google. 
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\end{document}
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