Is the inclusion of the group of units in an algebraic monoid an open immersion? Given an algebraic monoid $M$ (by which I mean a monoid in the category of schemes), one has  the group $M^\times$ of invertible elements (which by general abstract nonsense is also a scheme, see my comment below), which comes with a map of algebraic monoids $M^\times \to M$. Is this map an open immersion?
I have only the following evidence to work from: one knows that the general linear group over $\mathbb{C}$ is open in the monoid of matrices (with the usual topology), and more generally the general linear group $GL_n$ is Zariksi open in all matrices $M_n$. One can then consider subvarieties of $M_n$ which are also submonoids, and the invertible elements thereof should again be Zariski open.
Unless I've got things terribly wrong, the inclusion $M^\times \hookrightarrow M$ is a monomorphism of schemes (it's described as a certain equaliser), and so I suppose we just need to show it's smooth. But my algebraic geometry is non-existent, so help, please!
Edit: I've added a bounty, in case anyone can sharpen Matt E's result, in other words, work out if we need to restrict to smooth schemes or not.
 A: I think that (assuming say that $M$ is finite type over a field $k$) the map
$M^{\times} \to M$ should be an open immersion.
Here is an attempt at a proof, hopefully not too bogus:
First of all, let's begin by carefully defining $M^{\times}$.
We have have the multiplication map $\mu:M \times M \to M,$ the identity
element $e: \mathrm{Spec} k \to M$, and the tranposition $\tau: M\times M
\to M \times M$ (switching factors).  (Here I am assuming I have a monoid in the 
category of $k$-schemes, for some field $k$.)
We may form the product $\nu:=\mu\times (\mu\circ \tau): M\times M \to M\times M,$
which in terms of $S$-valued points (for any $k$-scheme $S$) maps
the pair $(x,y)$ to the pair $(xy, yx).$
If we form the fibre product of $\nu$ and $e\times e: \mathrm{Spec} k \to M\times M$, we obtain a closed subscheme
$X \subset M\times M$, whose $S$-points consist of 
pairs $(x,y)$ such that $x y = yx = e.$  
Now in fact we can make $X$ a group scheme, by defining (on $S$-valued points) the multiplication $(x,y)\cdot (u,v) =
(xu,vy),$ the identity element $(e,e)$, and the inverse operation $(x,y)^{-1} = (y,x)$. Intuitively, we have that $y = x^{-1}$, and this is just the product
on invertible elements.
Thus we should write $X = M^{\times}$; this is the formation of the group
of units in a categorical manner.
Now projection onto the first factor gives a map $X \to M$,
which is a monomorphism (look at $S$-valued points, and observe by the usual proof for monoids in sets that $x$ determines $y$).    
Everything so far is purely categorical.
Now we have to do some algebraic geometry. I'll assume that $M$ is finite type
over $k$, and argue with Zariski tangent spaces.
Consider the Zariski tangent space to $M$ at $e$.  This is equal to
morphisms $\mathrm{Spec} k[\epsilon]/(\epsilon^2) \to M$ lying above
the map $e: \mathrm{Spec}k \to M$, and so 
is naturally a monoid in the category of $k$-vector spaces,
with zero acting as the identity.  A standard argument then shows
that the monoid structure on the Zariski tangent space structure coincides
with the underlying addition (coming from the a priori vector space
structure).
Now this vector space is a group under addition, and so we can lift this map
$\mathrm{Spec} k[\epsilon]/(\epsilon^2)$ from a map to $M$ to a map to $M^{\times}$.  
In conclusion: the map $M^{\times} \to M$ induces an isomorphism of Zariski tangent spaces the identity.  Since $M^{\times}$ is homogeneous, it must do so
at every (say $\overline{k}$-valued) point of its domain.  (Alternatively, we could just repeat the above argument as these points.)
So $M^{\times} \to M$ is a monomorphism that induces an isomorphism on Zariski tangent spaces at every $\overline{k}$-valued point of its domain.
This should imply that $M^{\times} \to M$ is an etale monomorphism, and hence is an open immersion.  (Hopefully this last step is correct.  If $M$ is smooth it is okay; if $M$ is not smooth then it also seems okay to me at the moment, but perhaps I am blundering ... .)
