168 III More on Divisibility
Proposition 39For any m×n matrix A, the following conditions are equivalent:
(i)for some, and hence every, partition A=(A 1 A 2 ), the submatrices A 1 and A 2
are left coprime;
(ii)there exists an n×mmatrixA†such that A A†=Im;
(iii)there exists an(n−m)×nmatrixAcsuch that
(
A
Ac
)
is invertible;
(iv)there exists an invertible n×n matrix V such that AV=(ImO).
Proof IfA=(A 1 A 2 )for some left coprime matricesA 1 ,A 2 , then there existX 1 ,
X 2 such thatA 1 X 1 +A 2 X 2 =Imand hence (ii) holds. On the other hand, if (ii) holds
then, for any partitionA=(A 1 A 2 ),thereexistX 1 ,X 2 such thatA 1 X 1 +A 2 X 2 =Im
and henceA 1 ,A 2 are left coprime.
Thus(i)⇔(ii). Suppose now that (ii) holds. ThenAhas rankmand hence there
exists an invertiblen×nmatrixUsuch thatA=(DO)U,wherethem×mmatrix
Dis non-singular. In factDis invertible, sinceAA†=Imimplies thatDis a left
divisor ofIm. Consequently, by changingU, we may assumeD=Im.Ifwenowtake
Ac=(OIn−m)U, we see that(ii)⇒(iii).
It is obvious that(iii)⇒(iv)and that(iv)⇒(ii).
We now consider the extension of these results to other rings besidesZ.LetRbe
an arbitrary ring. A nonempty setM⊆Rmis said to be anR-moduleifa,b∈M
andx,y∈Rimplyxa+yb∈M. The moduleMisfinitely generatedif it contains
elementsa 1 ,...,ansuch that every element ofMhas the formx 1 a 1 +···+xnanfor
somex 1 ,...,xn∈R.
It is easily seen that ifRis aBezout domain ́ , then the whole of the preceding dis-
cussion in this section remains valid if ‘module’ is interpreted to mean ‘R-module’ and
‘matrix’ to mean ‘matrix with entries fromR’. In particular, we may takeR=K[t]to
be the ring of all polynomials in one indeterminate with coefficients from an arbitrary
fieldK. However, bothZandK[t] are principal ideal domains. In this case further
results may be obtained.
Proposition 40If R is a principal ideal domain and M a finitely generated
R-module, then any submoduleLofMis also finitely generated. Moreover, ifMis
generated by n elements, so also isL.
Proof SupposeMis generated bya 1 ,...,an.Ifn=1, then anyb∈Lhas the form
b=xa 1 for somex∈Rand the set of allxwhich appear in this way is an ideal ofR.
SinceRis a principal ideal domain, it follows thatLis generated by a single element
b 1 ,whereb 1 =x′a 1 for somex′∈R.
Suppose now thatn>1 and that, for eachm<n, any submodule of a module
generated bymelements is also generated bymelements. Anyb∈Lhas the form
b=x 1 a 1 +···+xnanfor somex 1 ,...,xn∈ Rand the set of allx 1 which appear in this way is an ideal
ofR.SinceRis a principal ideal domain, it follows that there is a fixedb 1 ∈Lsuch