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We saw from Proposition 2.13 and Corollary 2.14 that the direct sums of stably free modules are stably free. Another question to ask is whether the direct sum of stably free modules is free. Obviously, if the sum contains enough free summands, then the sum will be free. But how about K ⊕K? We saw that this module is stably free but is it free? If K ⊕K is not free, then how about K3? In particular, does there exists a natural numbern, such thatKn _{is free but for alli < n, K}i _{is not free? The} following two examples show that either case can occur.

Example 3.1. In a paper by Evans [7], he constructs a ring R and a sequence of modules {Mn} with the properties thatMn is stably free for each n, Mn has rank n, and Mn has no free direct summands. This last property shows that Mn is not free. Thus, in his ring R, there exist finitely generated, non-free, stably free modules of arbitrary rank.

Example 3.2. On the other hand, a paper by Brown, Lenagan, and Stafford [3] shows an opposite result. LetGbe the fundamental group of the Klein bottle. Then,

by application of Brown, Lenagan, and Stafford’s Theorems 4.9, 4.10, and 4.11,1 _it

can be shown that any stably free ZG-module with rank larger than or equal to 4 is necessarily free.

Thus, for the integral group ring over the Klein bottle, we have an explicit bound on the rank of non-free, stably free modules. For ZG, we know that K4 _{is free from}

above. What about K2 _{or K}3_{? The conclusion of this chapter will be to show that}

for the Klein bottle, K3 _{is free, but like the previous chapter, we prefer to work in a}

more general setting and mention the Klein bottle as a corollary. We take this portion now to develop a few criteria that would guarantee the possible freedom of K⊕K.

Now, ifK⊕K is free, there exists some positive integer`, such thatK⊕K ∼=kG`. We claim that ` = 2, for kG3 ∼=K⊕K ⊕kG∼= (kG`)⊕kG=kG`+1. Then, by the IBN, `+ 1 = 3. A simple generalization of this argument tells us that for a ring R satisfying the IBN, if a stably free R-module P is free, then the notions of rank for stably free modules and rank for free modules coincide.

In particular, if K ⊕K is free, we see that K ⊕K ∼= K ⊕kG. Thus, to show the freedom of K ⊕K, it suffices to prove the last isomorphism. We actually will consider a slightly more general isomorphism. Namely, is it possible that there exists a module L, such that K⊕K ∼=L⊕kG, and if so, what does this tell us about the freedom of K⊕K?

Throughout the rest of this section, we assume R to be a right Noetherian ring. In addition, we assume all R-modules to be right modules unless otherwise stated.2

1_{Brown, Lenagan, and Stafford don’t work with the Klein bottle but instead with polycyclic-by-} finite group rings. The Klein bottle is just a special case.

2_{We assume thatRis right Noetherian simply because in terms of free modules,Ris well behaved.} That is, every right Noetherian ring is stably finite [12, p.7], which means that ifN⊕Rn∼_=Rn_{, then}

Proposition 3.3. Let P be a stably free R-module. If P has a free summand, that

is, P ∼=Q⊕R for some R-module Q, then Q is stably free and there exists an integer

`, such that L

`P is free.

Proof. SinceP is stably free, there existm, n∈Z, such thatP⊕Rm ∼_=Rn_{. Therefore,}

Q⊕Rm+1 ∼_{= (Q⊕R)⊕R}m ∼_{= (P)⊕R}m ∼_=Rn_{. Thus, Qis stably free.} Now we shall prove the second statement,

Pm+1 ∼= (Q⊕R)m+1 = Qm+1⊕Rm+1

∼_{= R}(m+1)(n−(m+1))+(m+1) _{by Prop. 2.13}

= R(m+1)(n−(m+1)+1)

= R(m+1)(n−m) (3.1)

Example 3.4. Sincekis a commutative Noetherian ring,kGis a Noetherian ring [14, Corollary 10.2.8]. According to equation (3.1) in Proposition 3.3, ifK⊕K ∼=L⊕kG for somekG-moduleL, then (K⊕K)2 =K4 is free. The details here are not difficult to work out. IfL⊕kG2 ∼=kG3, then (K⊕K)⊕(K⊕K)∼= (L⊕kG)⊕(L⊕kG) = L⊕(L ⊕kG⊕kG) ∼= L⊕(kG⊕kG⊕kG) ∼= kG4_{. Thus, K}4 _{is free given the}

assumption.

We can do better than the last statement. K ⊕K ⊕ K ∼= K ⊕(L ⊕kG) = L⊕(K ⊕kG) ∼= L⊕(kG⊕kG) ∼= kG3. Thus, K ⊕K ⊕K would be free if the

without mention. Stable finiteness also implies the Invariant Basis Number. Proofs of all these statements can be found in Lam [12,§1].

assumption holds. Thus, the integer mentioned in the proof of Proposition 3.1 might not necessarily be the smallest number for which the sum is free.

Once again, let us assume that K ⊕K has a free factor. As the above example shows, K⊕K⊕K is free of rank 3. From this it is clear to see that all higher sums of K must also be free. Note for all ` <sub>></sub> 4 that `−3 _> 1 and K` = K3 ⊕K`−3 ∼= kG3_⊕K`−3 ∼

=kG`_{. This unfortunately does not give enough information to conclude} whether K⊕K is free or not. If another appropriate assumption is made on L, we can prove the freedom of K⊕K.

Proposition 3.5. LetP be anR-module, such thatP⊕R∼=R2_{. IfP⊕P} ∼_=Q⊕Rfor

someR-moduleQand there exists an epimorphism q:Q→P, then the epimorphism

is an isomorphism and P ⊕P ∼=R⊕R.

Proof. By our assumption, there is an exact sequence

0−→kerq−→Q−→q P −→0.

But P is a projective module, which implies that this sequence splits. Thus, we have that Q∼= kerq⊕P. Hence,

P ⊕P ∼= Q⊕R

∼

= (kerq⊕P)⊕R

∼

= kerq⊕(R2)

Remember that P ⊕P is stably free. Then,

∼

= (kerq⊕R2)⊕R

∼

= kerq⊕R3.

Thus, R3 ∼= kerq⊕R3. ButR is Noetherian, and hence kerq = 0.3 This gives us immediately thatP ∼=Qand more importantly P⊕P ∼=Q⊕R∼=P⊕R∼=R⊕R. We next provide another property L might exhibit that will imply thatK⊕K is free.

Proposition 3.6. If P is a one-relator, projective R-module, then P ⊕R is free.

Proof. LetP =he1, e2, . . . , en |Peimii with m1 6= 0. The presentation ofP gives

rise to the following exact sequence,

0−→kerq−→Rn−→q P −→0.

Since, P is projective, the sequence is split and Rn ∼_{= P ⊕ kerq. Since m}

1 6= 0

by assumption, kerq 6= 0. Also, kerq is generated by a single element. Now, say kerq =aR. Then, multiplication by a for R →aR is an epimorphism and is not an isomorphism only if a is a torsion element. But free modules and hence projective modules are torsion-free. Thus, a projective cyclic module must be free. Therefore, P ⊕R∼=Rn_.

Corollary 3.7. If P ⊕ P ∼= Q ⊕ R and Q = he1, e2 |e1m1+e2m2i, then

P ⊕P ∼=R⊕R.

Proof. By Proposition 3.6, Q⊕R∼=R2_{. But P ⊕P} ∼_{=Q⊕R by assumption.}

3_{If a ring is stably finite andP is a stably free (or a projective) module, then (N⊕P} ∼_=P)⇒ (N = 0).