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:(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?

 A: 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.
A: 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.
