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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 $\lbrace 1 \rbrace \subsetneq GL(k[M])$?

Clearly

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

Question: If $\lbrace (M,\cdot)^{\times} = \lbrace 1 \rbrace rbrace$, can $k[M]$ contain an invertible element $z \subsetneq in GL(k[M])$, then $M$ must contain right- and left-invertible elements; but such that is not enough to ensure the existence coefficient of non-trivial invertible elements.$z$ at $1$ is zero?

2 added 12 characters in body; edited tags

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

Clearly, if $\lbrace 1 \rbrace \subsetneq GL(k[M])$, then $M$ must contain right- and left-invertible elements; but that is not enough to ensure the existence of non-trivial invertible elements.

1

# Invertible elements in monoid rings of unital monoids without non-trivial invertible elements

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

Clearly, if $\lbrace 1 \rbrace \subsetneq GL(k[M])$, then $M$ must contain right- and left-invertible elements; but that is not enough to ensure the existence of invertible elements.