Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
@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
1  
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
add comment

1 Answer 1

up vote 5 down vote accepted
+100

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 ... .)

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.