# 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.

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If $M$ is a ring scheme, do we know a priori that the invertible elements form a scheme? (By which I mean, is the functor $R \mapsto M(R)^*$ on $\mathrm{CRing}$ a scheme?) – Akhil Mathew Mar 1 '11 at 0:45
@Akhil - I believe so. It is a general result that the invertible elements in a monoid object in a category can be described as a finite limit. Sch being a finitely complete category, we are done. – David Roberts Mar 1 '11 at 0:49
Please make your messages self-contained, not referencing the title as an integral part of them. The title is like the title of a book in the spine. What would you think of a book that says "Please read the spine to understand the next comment"? – Arturo Magidin Mar 1 '11 at 4:30

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 ... .)

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Brilliant! So for the present we get the result is true for smooth monoid schemes of finite type over a field. Can we weaken to Noetherian schemes? I don't have a particular category of schemes in mind, but the more general the better. – David Roberts Mar 1 '11 at 6:22
In the absence of a better answer, I'm accepting this, but I hold out for better answers. – David Roberts Jun 14 '11 at 7:01