A ring with IBN which admits a free module with a generator with less elements than a basis This is a follow-up to this question of mine: Cardinality of a minimal generating set is the cardinality of a basis
I observed in a comment to that question that it was, in fact, a duplicate of this question: Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$? (1)
The proof uses heavily that the ring is commutative.
I am wondering now of an example of a ring with IBN where the answer to (1) is negative.
Motivation:
Arturo's answer to my previous question provides a reference to an article of Cohn (which I could only get a quarter of). However, he proves that there is no noetherian, artinian, or commutative (which we have already seen by different means) where (1) is negative.
The example of a ring which does not satisfy (1) that he gives does not have IBN.
But the cases in which he proves it affirmative are all classes of rings that have IBN! Can it be, then, that the answer to (1) is positive for every ring with IBN, or is there a ring with IBN where the answer to (1) is negative? (It would have to be a non noetherian, non artinian, noncommutative ring!)
 A: On Bruno's request I'm making my comment into an answer.
Invariant basis number and many related properties are studied in section 1.§1 of Lam's Lectures on modules and rings, they are summarized in this diagram on page 17 in which none of the implications can be reversed:


The example showing that invariant basis number does not imply the rank condition is as follows (it appears on page 11 after proposition 1.24:
Let $R$ be the $\mathbb{Q}$-algebra with generators $a,b,c,d$ and relations
$$
ac=1, \quad bd=1, \quad ad = 0 = bc
$$
so that we have
$$
\begin{bmatrix} a \\ d \end{bmatrix} \begin{bmatrix} c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.
$$
This shows that $\begin{bmatrix} a \\ d\end{bmatrix}: R \to R^2$ is an epimorphism so that the rank condition fails.
The proof that $R$ does have invariant basis number is given in P.M. Cohn, Some remarks on the invariant basis property, Topology 5 (1966), 215–228, MR197511.
Cohn denotes this ring by $U_{1,2}$ in §5. The proof that it has invariant basis number involves deriving a normal form for elements of $R$ and showing that the trace of $1 \in R$ (i.e., the image of $1$ in the abelian group $R/[R,R]$) has infinite order. It is a bit too laborious for me to reproduce it here. The argument is given before Lemma 5.1. By the Corollary in §3 (or Lam's Exercise 5 on page 17) this implies that $R$ has invariant basis number.

See also the comments to this related MO thread.
