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This question is somewhat related to Tilmans notorious problem in this post. Let $(M,\cdot)$ be a monoid with unit $1$ and set $$(M,\cdot)^{\times} := \lbrace x \in M \mid \exists y \in M : xy=yx=1 \rbrace.$$ Let $k$ be a field (say $\mathbb Z/ 2 \mathbb Z$) and let $k[M]$ be the monoid ring of $M$ with coefficients in $k$. Consider now $$GL(k[M]) := (k[M],\cdot)^{\times}.$$

Question:(answered by Torsten Ekedahl) Can it happen that $(M,\cdot)^{\times} = \lbrace 1 \rbrace$ but $\lbrace 1 \rbrace \subsetneq GL(k[M])$?

EDIT: Torsten Ekedahl has nicely answered the above question. However, since I was really missing a condition, I will take the opportunity to change it slightly.

Question: If $(M,\cdot)^{\times} = \lbrace 1 \rbrace$, can $k[M]$ contain an invertible element $z \in GL(k[M])$, such that the coefficient of $z$ at $1$ is zero?

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2 Answers 2

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Let $R$ be a finite dimensional algebra over $\mathbb Z/2$. Then $\{1\}\neq R^\times$ unless $R=(\mathbb Z/2)^n$. Indeed, if $N$ is the radical of $R$, then $1+N\subseteq R^\times$ so we may assume $R$ is semi-simple. Then $R=\prod_iR_i$ where the $R_i$ are simple algebras and $R^\times=\prod_iR_i^\times$ so we may assume that $R$ is simple and hence a matrix algebra over some extension field of $\mathbb Z/2$. The only such algebra with only the trivial unit is $\mathbb Z/2$.

Now, pick any finite monoid $M$ with $M^\times=\{1\}$ and apply the above to $R=\mathbb Z/2[M]$. This gives that $R^\times=\{1\}$ precisely when $M$ is a commutative monoid where every element is idempotent. As an explicit example where this is not the case we may let $M$ be the identity matrix plus all non-invertible matrices of fixed size $>1$ over some finite field.

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  • $\begingroup$ I see. This also implies that every unit in such a monoid ring contains 1 in its support. That is were I went wrong. I thought existence of non-trivial left-invertible elements would be a consequence of the existence of units in the monoid ring, but that is wrong. I hope you allow me to change the question slightly. $\endgroup$
    – Andreas Thom
    Commented Oct 23, 2010 at 12:17
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If your monoid $M$ is finite the non-invertible elements form an ideal of the monoid and hence they span an ideal of kM. So any invertible element of the algebra must have an invertible element of the monoid in its support. So the answer to your second question is no.

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  • $\begingroup$ Yes, the same holds if left-invertible is equivalent to right-invertible; which if holds in finite monoids. The question is only interesting if these notions are not equivalent. $\endgroup$
    – Andreas Thom
    Commented Jun 24, 2011 at 6:08
  • $\begingroup$ Sorry, I had the impression from the previous answers that you were looking at finite monoids. More generally, the set of elements $L$ of the monoid $M$ that are not left invertible is a proper left ideal. Therefore, the span $kL$ is a proper left ideal of $kM$. It follows that any invertible element must contain a left invertible element in its support and dually a right invertible element. I will try to think whether this situation can come up without having an invertible element. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 25, 2011 at 14:00
  • $\begingroup$ I checked in the literature and it seems that every unit in the algebra of the monoid defined by the presentation $\langle a,b\mid ab=1\rangle$ does have $1$ in its support. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 27, 2011 at 20:55
  • $\begingroup$ Benjamin, your second comment is precisely the difficult part. Why should there be an invertible element in the support? $\endgroup$
    – Andreas Thom
    Commented Jul 3, 2011 at 11:45

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