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On page 88 of Atiyah-Macdonald's "Introduction to Commutative Algebra" there is an exercise about the Grothendieck group $K(A)$ of a noetherian ring $A$. In this context to every finite ring homomorphism $f: A \rightarrow B$ of noetherian rings there is an associated group homomorphism

$$f_{!}: K(B) \rightarrow K(A)$$

which is induced by restricting a finitely generated $B$-module via $f$ to a finitely generated $A$-module. Given two finite ring homomorphisms $A \stackrel{f}\longrightarrow B \stackrel{g} \longrightarrow C$ we get

$$(g \circ f)_{!} = f_{!} \circ g_{!}$$

What I am wondering about is: why do they put the "shriek" (i.e. the symbol "$!$") into the subscript when it behaves contravariantly?

On wikipedia they say that shrieks are used either to distinguish a functor from another similar functor, or in order to warn the reader that something which intuitively behaves covariantly (contravariantly) behaves instead contravariantly (covariantly).

So what of the two, if anything, applies in my case above? It's pretty clear to me that in my case the shriek has to "turn arrows around" because we succesively restrict scalars, first along $g$ then along $f$. Would one instead, a priori, expect that the Grothendieck group functor is covariant, or is there another, well-known functor, which could easily be confused with this one?

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The notation in that exercise is a little bit unfortunate; K-theory uses projective modules, but in that exercise all f.g. modules are used for the Grothendieck group, which is usually called the G-theory. So you should call it $G_0(A)$. –  Martin Brandenburg Jun 5 '12 at 10:37
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Actually $f \mapsto f_{!}$ is covariant when considered as a functor on (affine) schemes, and similarily $f \mapsto f^*$ is contravariant. –  Martin Brandenburg Jun 5 '12 at 10:40
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What Martin is saying (allow me to translate) is that in the modern approach one usually thinks of schemes rather than rings. Affine schemes (the prototype if you want) correspond contravariantly to rings. So the notation makes sense. –  Simon Markett Jun 5 '12 at 10:42
    
@Simon Markett, Martin Brandenburg: Okay, I have a superficial aqcuaintance with schemes, so I'm willing to buy this. Still I find a commutative ring the simpler object as an affine scheme, but maybe I just have to adjust my mind.;) I just thought that there was a deeper (or maybe historical) reason for the shriek. –  Nils Matthes Jun 5 '12 at 10:52
    
@NilsMatthes I for my part (and I belive many others) still think secretly about rings rather than schemes. In the end one often does things on affines and uses some glueing afterwards. Moreover morphisms of schemes always come along with a morphism of structure sheaves which is locally just a ring homomorphism. The only non affine schemes I use regularly are the projective spaces. –  Simon Markett Jun 5 '12 at 12:08
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2 Answers 2

up vote 4 down vote accepted

When you have a map of rings $A\to B$ you get an induced functor $M(A)\to M(B)$ on the categories of finitely generated modules over $A$ and $B$, respectively. It is given by $P\mapsto P\otimes_A B $. This induces a homomorphism on $K$-theory (or $G$-theory if we are precise) which behaves covariantly. This functor is considered to be the normal one. The functor you describe is indeed the other way round.

Edit: The reason why the covariant functor is the "normal" one is imo Swan's Theorem. There is an equivalence of categories of the category of real vector bundles over a compact topological space $X$ and projective modules over the ring of continuous functions $C(X,\mathbb R)$. More specifically a vector bundle $E\to X$ is sent to the module of sections $\Gamma (X,E)$ (Exercise: this is indeed projective). Hence we have an isomorphism of the topological $K$-group of $X$ and the algebraic $K$-group of $C(X,\mathbb R)$. A map of topological spaces $X\to Y$ induces a map $K(Y)\to K(X)$ (contravariantly). In turn we want the induced map $C(Y,\mathbb R)\to C(X,\mathbb R)$ to induce a map $K(C(Y,\mathbb R))\to K(C(X,\mathbb R))$. This is done by the covariant functor as decribed above.

There are many many relations between topological and algebraic $K$-theory. In general topological K-theory is easier since we have more tools at hand. Consequently many results exist in the topological realm until we can transfer them to the algebaric world. Since everything should work essentially similar it is good to have the maps in the same direction.

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So what you mean is extension of scalars, right? Perversely the (covariant) functor this induces on $G$-Theory is denoted as $f^{!}$ on the same page of Atiyah-Macdonald (although they restrict to finitely generated $\textbf{flat}$ modules). Do you know of any reason, why extension of scalars is "the normal one"? –  Nils Matthes Jun 5 '12 at 10:49
    
Thanks alot for your answer! This is extremely interesting. I accept that answer for now and try to hack my way a little deeper into the K-Theory jungle. –  Nils Matthes Jun 5 '12 at 11:43
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In general, $f_!$ is the notation, in cohomological or cycle-theoretic contexts, for pushforward with proper supports (the relative version of cohomology with compact supports).

In this particular case, one is applying it to the proper morphism Spec $B \to $ Spec $A$ (proper because the corresponding extension of rings $A \to B$ is finite).

So the answer to your question is that there is a larger geometric context in which this particular situation considered by AM can be placed, and in that larger context $f_!$ is the traditional notation.

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Thank you! That complements Simons answer nicely as it gives a "historical" motivation. May I kindly ask you to sketch a little of this "larger geometric context" and how the AM situation fits in this larger context? –  Nils Matthes Jun 5 '12 at 14:45
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