Let $p_i : M \to M$ denote the homothety $m \mapsto e_i m$; the map $p_i$ is $A$-linear since $e_i$ is central; since $e_i^2 = e_i$, $\sum_{i \in I} e_i = 1$ and $e_i e_j = 0$ for $i \neq j$, we have

and the $p_i$ form an orthogonal family of projectors whose sum is the identity (II, p. 209, Def. 7). By loc. cit., Prop. 12, the module M is the direct sum of the submodules $p_i(M) = e_iM$. In addition $e_iM$ is annihilated by $b_i = A(1 - e_i)$; if $i \neq j$ and $m \in M$ then $(1 - e_i)e_jm = e_jm$, whence no nonzero element of $e_jM$ is annihilated by $1 - e_i$, and so a fortiori by $b_i$. It follows that $e_iM = M_i$, and the proposition is proved.

Clearly, if $\pi$ and $\pi'$ are two associate irreducible elements, then $M_\pi = M_{\pi'}$; thus, for a given module M, the submodule $M_\pi$ depends only on the ideal $(\pi)$ of A; it is called the rr-primary component of the module M, and the decomposition of M as a direct sum of the $M_\pi$ is called the canonical decomposition of M as a direct sum of its rr-primary components.

This follows from the fact that N ∩ M, is the π-primary component N, of N.

Indeed, if N and N' are two submodules of M, then M = N ⊕ N' if and only if M, − N, ⊕ N'_π for every irreducible element π of $\mathscr{A}$ (Cor. 1).

Indeed, the condition $(M/N)_\pi = 0$ implies $N_\pi = M_\pi$ and Cor. 2 applies.

An A-module M is called semi-simple if every submodule of M is a direct factor (cf. A, VIII, § 3).

First suppose that M is semi-simple; let x ∈ M and let π be an irreducible element of $\mathscr{A}$. If N is a complement of $Annx$ in M, then we can write $x = \alpha \pi x + y$, with $\alpha \in \mathscr{A}$ and $y \in N$; but that implies $y = (1 - \alpha n)x$, so

Conversely, by Cor. 2 it is enough to prove that an A-module M annihilated by an irreducible element π is semi-simple; but that is clear, since M then has a natural structure of a vector space over the field $\mathscr{A}/(\pi)$, and the submodules of M are precisely the vector subspaces under this structure.

This follows immediately from Lemma 1, for by Lemma 2 there exists $\alpha \neq 0$ in $\mathscr{A}$ such that $M = M(\alpha)$.

where $H$ and the $s(p)$ are uniquely determined by these conditions.

Moreover, if $R_p$ denotes a subset of $A$ containing precisely one element of each residue class mod $p$ in $A (p \in P)$, then each $x \in K$ can be uniquely expressed in the form

where $a \in A$ and $r_{ph} \in R_p$ for all $h$ and $p$, and all but finitely many of the $r_{ph}$ are zero.

Consider $K$ as an $A$-module; then $A$ is the submodule generated by 1. The quotient module $K/A$ is the quotient of $K$ by the equivalence relation $x' - x \in A$, which is also written, in the notation of VI, p. 6, as $x \equiv x'$ (mod 1); let $f$ denote the natural homomorphism from $K$ onto $M = K/A$.

The quotient module $M$ is a torsion module, for every element of $M$ has the form $f(a/b)$ ($a \in A, b \in A, b \neq 0$), whence $b f(a/b) = f(a) = 0$. Hence Th. 1 of VII, p. 8 applies. Let $M_p$ ($p \in P$) be the submodule of elements of $M$ annihilated by powers of $p$; then $f^{-1}(M_p)$ is the subring $A_p$ of elements of $K$ of the form $a p^{-n}$ where $a \in A$ and $n \geq 0$ is an integer. The module $M$ is the direct sum of the $M_p$, so every $x \in K$ is congruent mod 1 to an element in the sum of the $A_p$; in other words, formula (1) holds, with the $s(p)$ integers $> 0$, and the $a$, finitely many elements of $A$ such that $a$, is not a multiple of $p$.

We now show that these conditions on the $s(p)$ and the $a$, completely determine $H$ and the $s(p)$. Indeed $H$ is then the set of $p \in P$ such that the component of $f(x)$ in $M_p$ is $\neq 0$. On the other hand if $s$ and $s'$ are two integers $> 0$ such that s \geq s' and if $a$ and $a'$ are elements of $A$ not divisible by $p$ such that $ap^{-s} \equiv a'p^{-s'} \pmod{1}$, then we deduce that $a \equiv a'p^{s-s'} \pmod{p^s}$; if $s > s'$ then $a \equiv 0 \pmod{p}$, contradicting the hypotheses. This argument also shows that each $a_i$ is well defined mod $p^{s(p)}$.

To complete the proof, we note first of all that in each residue class mod $p^i$ in $A$ there exists a unique element of the form $\sum_{h=0}^{s-1} r_h p^h$ with $r_h \in R_p$ for $0 \leq h \leq s-1$. In fact, we proceed by induction on $s$ (the property follows from the definition of $R_p$ for $s = 1$) : let $x \in A$; by hypothesis there is a unique element of the form $\sum_{h=0}^{s-2} r_h p^h$ ($r_h \in R_p$) in the residue class of $x$ mod $p^{s-1}$; then $x - \sum_{h=0}^{s-2} r_h p^h$ is a multiple $ap^{s-1}$ of $p^{s-1}$; now there exists a unique element $r_s$, of $R_p$ such that $a \equiv r_s \pmod{p}$; whence $x \equiv \sum_{h=0}^{s-1} r_h p^h \pmod{p^s}$. To obtain formula (2) it is now enough to apply this fact to each $a_p$ in formula (1). The uniqueness is clear in view of the above.

The following are the most important applications of Th. 2 :

I. The ring A is the ring Z of rational integers, and K = Q. Let P be the set of prime numbers > 0, and for each $p \in P$ let $R_p$ be the interval $(0, p-1)$ in $\mathbf{Z}$. Thus we have the canonical decomposition

with $a \in \mathbf{Z}$, $e_{ph} \in \mathbf{Z}$ and $0 \leq e_{ph} \leq p-1$.

II. The ring A is the ring E[X] of polynomials in one indeterminate over a commutative field E, and K = E(X). Let P be the set of monic irreducible polynomials in $E[X]$ (VII, p. 5). For $p \in P$ we can, by virtue of euclidean division of polynomials (IV, p. 10), take $R_p$ to be the set of polynomials of degree strictly less than that of $p$. Thus we have the decomposition (called canonical) of a rational function $r(X) \in E(X)$:

where $a(X)$ is a polynomial and $v_{ph}(X)$ is a polynomial of degree strictly less than that of $p(X)$, for all $p$ and $h$. In particular, if the field E is algebraically closed, then the $p(X)$ have the form $X - a$ with $a \in E$, and the $v_{ph}(X)$ are thus constants.

Hence we can say that the vector space $E(X)$ over the field E has a basis consisting of the monomials $X^n$ ($n \geq 0$ an integer) and the rational functions of the form $X^m / (p(X))^h$, where $p \in P$ and $h$ and $m$ are integers with $h \geq 1$ and $0 \leq rn < \deg(p)$.

To prove the first assertion, it is enough to show that, if $k \geq 1$ and $x \equiv 1 + p^ky \mod p^{k+1}$, then $x^p \equiv 1 + p^{k+1}y \mod p^{k+2}$, and then to argue by induction on the integer k. For all $a \in \mathbf{Z}$ and $k \geq 1$, it is immediate that

Since $U(p^n)$ has order $p^{n-1}$, the order of every element $u$ of $U(p^n)$ is a power of $p$, and $u$ is a generator of $U(p^n)$ if and only if $u^{p^{n-2}} \neq 1$. Now if $u$ is the class of $x = 1 + py$, then $u^{p^{n-2}}$ is the class of $1 + p^{n-1}y$, by Lemma 3, whence $u$ generates $U(p^n)$ if and only if $y \not\equiv 0 \mod p$, in other words $x \not\equiv 1 \mod p^2$. For example, the class $1 + p$ generates $U(p^n)$. Similarly, the class $u$ of $x$ mod $2^n$ generates $U(2^n)$ if and only if $u^{2^{n-3}} \neq 1$, which means that $x$ is not congruent to 1 mod 8, by Lemma 3; this is satisfied by $x = 5$.

The module $M$ is torsion, since $abM = 0$. The first assertion follows from Cor. 3 of VII, p. 9. If $N$ and $P$ are cyclic, then they are finitely generated, and hence so is $M$ (II, p. 17, Cor. 5); since each $p$-primary component of $M$ is isomorphic to a $p$-primary component either of $N$ or of $P$, it follows from Remark 2 of VII, p. 10, that $M$ is cyclic.

The orders $p^n \ ^1$ of $U(p^n)$ and $p - 1$ of $(\mathbf{Z}/p\mathbf{Z})^*$ are coprime; since $U(p^n)$ and $(\mathbf{Z}/p\mathbf{Z})^*$ are cyclic (Prop. 3 and V, p. 78, Lemma 1), the group $(\mathbf{Z}/p^n\mathbf{Z})^*$ is cyclic (apply Lemma 4 to the exact sequence (3)). If $n \geq 2$ then the restriction of the homomorphism $v : (\mathbf{Z}/2^n\mathbf{Z})^* \to (\mathbf{Z}/4\mathbf{Z})^*$ to the subgroup $(1, -1)$ is bijective; the group $(\mathbf{Z}/2^n\mathbf{Z})^*$ is thus the direct product of this subgroup and the kernel $U(2^n)$ of $v$; the result follows from Prop. 3.