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Suppose we have a ring $R$ such that the Grothendieck group $K_{0}(R)=0$.

Question 1: Does it follow that there exists two positive natural numbers $n\neq m$ such that $R^{m}$ is isomorphic to $ R^{n}$ as $R$-module, where $R^{k}$ is the free $R$-module of rank $k$ ?

Question 2 (added later, in comment thread): Does it follow that $R$ is isomorphic to $R^2$?

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2 Answers 2

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Question 1:

Does $K_0(R)=0$ imply that $R^m\approx R^n$ for some $m\neq n$?

Answer to Question 1:

Yes.

Choose any distinct natural numbers $m$ and $n$. Using square brackets to denote $K_0$ classes, $K_0(R)=0$ implies in particular that $[R^m]=[R^n]$. It follows that there is some projective module $Q$ such that $$R^m\oplus Q\approx R^n\oplus Q$$ Adding a projective complement to $Q$ on both sides lets us assume $Q=R^d$ for some $d$. So $R^{m+d}\approx R^{n+d}$

Question 2:

Does $K_0(R)=0$ imply $R\approx R^2$?

Answer to Question 2:

No.

William Leavitt, in his paper Modules Without Invariant Basis Number constructs, for any $N>0$, a ring with these properties:

a) All of the modules $R^1,R^2,R^3,\ldots R^N$ are pairwise non-isomorphic.

b) $R^N\approx R^{N+1}\approx R^{N+2}\approx\cdots$.

Now take Leavitt's ring for $N=2$.

Claim 1. $K_0(R)=0$.

Proof. By b) above (with $N=2$) $R^2\approx R^3$.

Now let $P$ be a projective module. Then $$P\oplus P\approx P\otimes R^2\approx P\otimes R^3\approx P\oplus P\oplus P$$ so in $K$-theory, $2[P]=3[P]$ and $[P]=0$.

Claim 2. $R$ is not isomorphic to $R^2$.

Proof. This is a) above, with $N=2$.

Remark. Benjamin Steinberg's comment refers to a later paper of Leavitt that constructs (in the notation of that comment) the rings $L_{m,n}$ for all $m,n$; the paper I cited constructs only the rings $L_{m,m+1}$, which is all that is needed here.

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  • $\begingroup$ Thanks, so as far as I understand your argument, in general we can not deduce that if $K_{0}(R)=0$ then $R\simeq R^2$, right ? $\endgroup$
    – cellular
    Commented Apr 24 at 21:55
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    $\begingroup$ How would you deduce $R=R^2$ ? $\endgroup$
    – Steven Landsburg
    Commented Apr 24 at 22:02
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    $\begingroup$ That was my question :) because I do not see how to prove it, I suspect there is examples such that K_0(R)=0 but R^2 not isomorphic to R^. $\endgroup$
    – cellular
    Commented Apr 24 at 22:08
  • $\begingroup$ Oh, sorry. I missed the word "not" in your comment, and thought you'd written that we CAN deduce $R\approx R^2$. I do not think we can deduce this. $\endgroup$
    – Steven Landsburg
    Commented Apr 24 at 22:10
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    $\begingroup$ Leavitt constructed rings call Leavitt algebras $L_{m,n}$ where $R^m\cong R^n$ but this does not hold for smaller m,n. So take m=2 and n=3. Then K_0 is 0 and but R^2 is not isomorphic to R $\endgroup$
    – Benjamin Steinberg
    Commented Apr 25 at 11:49
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This answer only adds more references and remarks, serving as a complement to the accepted answer of Steven Landsburg.

Rings are supposed to be associative and unital, but not necessarily commutative.

Definition 1. A ring $R$ has *Invariant Basis Number (IBN), if the following condition holds: if $R^m \simeq R^n, $ as $R$-modules for positive $m$ and $n$, then $m = n$.

It is well known that a non-trivial commutative ring has IBN and so has any left Noetherian ring [2, Theorem 1.3.9]. The ring $A(R)$ of column-finite matrices over a non-trivial ring $R$, with rows and columns indexed by $\mathbb{N}$, hasn't IBN [2, Example 1.37] and satisfies moreover $K_0(R) = 0$ [2, Exercise 3C.1].

W. G. Leawitt has classified the rings without IBN [1, 2]. The proof of this classification has been revisited by P. M. Cohn [3]. More on this below.

A positive answer to the original question follows immediately from this general fact:

[4, Proposition 4.5]. Let $R$ be a ring. The map $n \mapsto n[R]$ from $\mathbb{Z}$ to $K_0(R)$ is injective if and only if $R$ has IBN.

The following definition helped Leawitt and Cohn classify rings without IBN.

Definition 2. A ring $R$ is said to be of type $(h, k)$ in the sense of P. M. Cohn, for some positive integers $h$ and $k$, if the following are equivalent:

  • $R^m \simeq R^n$ as $R$-module with $m,n > 0$.
  • Either $m = n$, or else $m, n \ge h$ and $m \equiv n \mod (k)$.

The next proposition is an easy exercise:

[4, Exercise 1C.2]. Let $R$ be a non-trivial ring. Then the following are equivalent:

  • The ring $R$ hasn't IBN.
  • The ring $R$ is of type $(h, k)$ for some positive integers $h$ and $k$.

Remark.

  • A ring $R$ is of type $(1, 1)$ if and only if $R \simeq R^2$ as $R$-modules.
  • The trivial ring and the ring $A(R)$ are both of type $(1, 1)$.
  • A $R$ satisfies $K_0(R) = 0$ if and only if $R$ is of type $(h, 1)$ for some positive integer $h$; this follows from the above exercise and from Steven Landsburg's answers to Questions 1 and 2 (for the "only if" part, you may also check [4, Proposition 4.3]).

W. G. Leavitt has shown that rings of all types exist, by proving $(a)$ rings of type $(h, 1)$ exist for all $h > 0$ [1] and $(b)$ rings of type $(1, k)$ exist for all $k > 0$ [2] and by observing that if $R$ has type $(h, 1)$ and $S$ has type $(1, k)$ then the direct product $R \times S$ has type $(h, k)$. As already observed by Benjamin Steinberg and Steven Landsburg, the second question can be answered in the negative thanks to these constructions.

  • [1] W. G. Leawitt, "Modules without invariant basis number", 1957.
  • [2] W. G. Leawitt, "The module type of a ring", 1962.
  • [3] P. M. Cohn, "Some remarks on the invariant basis property", 1966.
  • [4] B. A. Magurn, "An algebraic introduction to $K$-theory", 2002.
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  • $\begingroup$ Thank you for this excellent answer ! $\endgroup$
    – cellular
    Commented Apr 27 at 8:19

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