geometry

Lambda-rings for formula-phobics

Feb 5th

In 1956, Alexander Grothendieck (middle) introduced $\lambda$ -rings in an algebraic-geometric context to be commutative rings A equipped with a bunch of operations $\lambda^i$ (for all numbers $i \in \mathbb{N}_+$ ) satisfying a list of rather obscure identities. From the easier ones, such as

$\lambda^0(x)=1, \lambda^1(x)=x, \lambda^n(x+y) = \sum_i \lambda^i(x) \lambda^{n-i}(y)$

to those expressing $\lambda^n(x.y)$ and $\lambda^m(\lambda^n(x))$ via specific universal polynomials. An attempt to capture the essence of $\lambda$ -rings without formulas?

Lenstra’s elegant construction of the 1-power series rings $~(\Lambda(A),\boxplus,\boxtimes)$ requires only one identity to remember

$~(1-at)^{-1} \boxtimes (1-bt)^{-1} = (1-abt)^{-1}$ .

Still, one can use it to show the existence of ringmorphisms $\gamma_n~:~\Lambda(A) \rightarrow A$ , for all numbers $n \in \mathbb{N}_+$ . Consider the formal ‘logarithmic derivative’

$\gamma = \frac{t u(t)'}{u(t)} = \sum_{i=1}^\infty \gamma_i(u(t))t^i~:~\Lambda(A) \rightarrow \Lambda(A)$

where u(t)' is the usual formal derivative of a power series. As this derivative satisfies the chain rule, we have

$\gamma(u(t) \boxplus v(t)) = \frac{t (u(t)v(t))'}{u(t)v(t)} = \frac{t(u(t)'v(t)+u(t)v(t)'}{u(t)v(t))} = \frac{tu(t)'}{u(t)} + \frac{tv(t)'}{v(t)} = \gamma(u(t)) + \gamma(v(t))$

and so all the maps $\gamma_n~:~\Lambda(A) \rightarrow A$ are additive. To show that they are also multiplicative, it suffices by functoriality to verify this on the special 1-series $~(1-at)^{-1}$ for all $a \in A$ . But,

$\gamma((1-at)^{-1}) = \frac{t \frac{a}{(1-at)^2}}{(1-at)} = \frac{at}{(1-at)} = at + a^2t^2 + a^3t^3+\hdots$

That is, $\gamma_n((1-at)^{-1}) = a^n$ and Lenstra’s identity implies that $\gamma_n$ is indeed multiplicative! A first attempt :

hassle-free definition 1 : a commutative ring is a $\lambda$ -ring if and only if there is a ringmorphism $s_A~:~A \rightarrow \Lambda(A)$ splitting $\gamma_1$ , that is, such that $\gamma_1 \circ s_A = id_A$ .

In particular, a $\lambda$ -ring comes equipped with a multiplicative set of ring-endomorphisms $s_n = \gamma_n \circ s_A~:~A \rightarrow A$ satisfying $s_m \circ s_m = s_{mn}$ . One can then define a $\lambda$ -ringmorphism to be a ringmorphism commuting with these endo-morphisms.

The motivation being that $\lambda$ -rings are known to form a subcategory of commutative rings for which the 1-power series functor is the right adjoint to the functor forgetting the $\lambda$ -structure. In particular, if is a $\lambda$ -ring, we have a ringmorphism $A \rightarrow \Lambda(A)$ corresponding to the identity morphism.

But then, what is the connection to the usual one involving all the operations $\lambda^i$ ? Well, one ought to recover those from $s_A(a) = (1-\lambda^1(a)t+\lambda^2(a)t^2-\lambda^3(a)t^3+...)^{-1}$ .

For s_A to be a ringmorphism will require identities among the $\lambda^i$ . I hope an expert will correct me on this one, but I’d guess we won’t yet obtain all identities required. By the very definition of an adjoint we must have that s_A is a morphism of $\lambda$ -rings, and, this would require defining a $\lambda$ -ring structure on $\Lambda(A)$ , that is a ringmorphism $s_{AH}~:~\Lambda(A) \rightarrow \Lambda(\Lambda(A))$ , the so called Artin-Hasse exponential, to which I’d like to return later.

For now, we can define a multiplicative set of ring-endomorphisms $f_n~:~\Lambda(A) \rightarrow \Lambda(A)$ from requiring that $f_n((1-at)^{-1}) = (1-a^nt)^{-1}$ for all $a \in A$ . Another try?

hassle-free definition 2 : is a $\lambda$ -ring if and only if there is splitting s_A to $\gamma_1$ satisfying the compatibility relations $f_n \circ s_A = s_A \circ s_n$ .

But even then, checking that a map $s_A~:~A \rightarrow \Lambda(A)$ is a ringmorphism is as hard as verifying the lists of identities among the $\lambda^i$ . Fortunately, we get such a ringmorphism for free in the important case when A is of ‘characteristic zero’, that is, has no additive torsion. Then, a ringmorphism $A \rightarrow \Lambda(A)$ exists whenever we have a multiplicative set of ring endomorphisms $F_n~:~A \rightarrow A$ for all $n \in \mathbb{N}_+$ such that for every prime number the morphism F_p is a lift of the Frobenius, that is, $F_p(a) \in a^p + pA$ .

Perhaps this captures the essence of $\lambda$ -rings best (without the risk of getting an headache) : in characteristic zero, they are the (commutative) rings having a multiplicative set of endomorphisms, generated by lifts of the Frobenius maps.

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fun, Grothendieck, lambda rings, Lenstra, numbers

big Witt vectors for everyone (1/2)

Feb 2nd

Posted by lievenlb in geometry

2 comments

Next time you visit your math-library, please have a look whether these books are still on the shelves : Michiel Hazewinkel‘s Formal groups and applications, William Fulton’s and Serge Lange’s Riemann-Roch algebra and Donald Knutson’s lambda-rings and the representation theory of the symmetric group.

I wouldn’t be surprised if one or more of these books are borrowed out, probably all of them to the same person. I’m afraid I’m that person in Antwerp…

Lately, there’s been a renewed interest in $\lambda$ -rings and the endo-functor W assigning to a commutative algebra its ring of big Witt vectors, following Borger’s new proposal for a geometry over the absolute point.

However, as Hendrik Lenstra writes in his 2002 course-notes on the subject Construction of the ring of Witt vectors : “The literature on the functor W is in a somewhat unsatisfactory state: nobody seems to have any interest in Witt vectors beyond applying them for a purpose, and they are often treated in appendices to papers devoting to something else; also, the construction usually depends on a set of implicit or unintelligible formulae. Apparently, anybody who wishes to understand Witt vectors needs to construct them personally. That is what is now happening to myself.”

Before doing a series on Borger’s paper, we’d better run through Lenstra’s elegant construction in a couple of posts. Let A be a commutative ring and consider the multiplicative group of all ‘one-power series’ over it $\Lambda(A)=1+t A[[t]]$ . Our aim is to define a commutative ring structure on $\Lambda(A)$ taking as its ADDITION the MULTIPLICATION of power series.

That is, if $u(t),v(t) \in \Lambda(A)$ , then we define our addition $u(t) \boxplus v(t) = u(t) \times v(t)$ . This may be slightly confusing as the ZERO-element in $\Lambda(A),\boxplus$ will then turn be the constant power series 1…

We are now going to define a multiplication $\boxtimes$ on $\Lambda(A)$ which is distributively with respect to $\boxplus$ and turns $\Lambda(A)$ into a commutative ring with ONE-element the series $~(1-t)^{-1}=1+t+t^2+t^3+\hdots$ .

We will do this inductively, so consider $\Lambda_n(A)$ the (classes of) one-power series truncated at term n, that is, the kernel of the natural augmentation map between the multiplicative group-units $~A[t]/(t^{n+1})^* \rightarrow A^*$ . Again, taking multiplication in $A[t]/(t^{n+1})$ as a new addition rule $\boxplus$ , we see that $~(\Lambda_n(A),\boxplus)$ is an Abelian group, whence a $\Z$ -module.

For all elements $a \in A$ we have a scaling operator $\phi_a$ (sending $t \rightarrow at$ ) which is an A-ring endomorphism of $A[t]/(t^{n+1})$ , in particular multiplicative wrt. $\times$ . But then, $\phi_a$ is an additive endomorphism of $~(\Lambda_n(A),\boxplus)$ , so is an element of the endomorphism-RING $End_{\Z}(\Lambda_n(A))$ . Because composition (being the multiplication in this endomorphism ring) of scaling operators is clearly commutative ( $\phi_a \circ \phi_b = \phi_{ab}$ ) we can define a commutative RING being the subring of $End_{\Z}(\Lambda_n(A))$ generated by the operators $\phi_a$ .

The action turns $~(\Lambda_n(A),\boxplus)$ into an E-module and we define an E-module morphism $E \rightarrow \Lambda_n(A)$ by $\phi_a \mapsto \phi_a((1-t)^{-1}) = (1-at)^{-a}$ .

All of this looks pretty harmless, but the upshot is that we have now equipped the image of this E-module morphism, say L_n(A) (which is the additive subgroup of $~(\Lambda_n(A),\boxplus)$ generated by the elements $~(1-at)^{-1}$ ) with a commutative multiplication $\boxtimes$ induced by the rule $~(1-at)^{-1} \boxtimes (1-bt)^{-1} = (1-abt)^{-1}$ .

Explicitly, L_n(A) is the set of one-truncated polynomials u(t) with coefficients in such that one can find elements $a_1,\hdots,a_k \in A$ such that $u(t) \equiv (1-a_1t)^{-1} \times \hdots \times (1-a_k)^{-1}~mod~t^{n+1}$ . We multiply u(t) with another such truncated one-polynomial v(t) (taking elements $b_1,b_2,\hdots,b_l \in A$ ) via

$u(t) \boxtimes v(t) = ((1-a_1t)^{-1} \boxplus \hdots \boxplus (1-a_k)^{-1}) \boxtimes ((1-b_1t)^{-1} \boxplus \hdots \boxplus (1-b_l)^{-1})$

and using distributivity and the multiplication rule this gives the element $\prod_{i,j} (1-a_ib_jt)^{-1}~mod~t^{n+1} \in L_n(A)$ . Being a ring-qutient of we have that $~(L_n(A),\boxplus,\boxtimes)$ is a commutative ring, and, from the construction it is clear that L_n behaves functorially.

For rings such that $L_n(A)=\Lambda_n(A)$ we are done, but in general L_n(A) may be strictly smaller. The idea is to use functoriality and do the relevant calculations in a larger ring $A \subset B$ where we can multiply the two truncated one-polynomials and observe that the resulting truncated polynomial still has all its coefficients in .

Here’s how we would do this over $\Z$ : take two irreducible one-polynomials u(t) and v(t) of degrees r resp. s smaller or equal to n. Then over the complex numbers we have $u(t)=(1-\alpha_1t) \hdots (1-\alpha_rt)$ and $v(t)=(1-\beta_1) \hdots (1-\beta_st)$ . Then, over the field $K=\mathbb{Q}(\alpha_1,\hdots,\alpha_r,\beta_1,\hdots,\beta_s)$ we have that $u(t),v(t) \in L_n(K)$ and hence we can compute their product $u(t) \boxtimes v(t)$ as before to be $\prod_{i,j}(1-\alpha_i\beta_jt)^{-1}~mod~t^{n+1}$ . But then, all coefficients of this truncated K-polynomial are invariant under all permutations of the roots $\alpha_i$ and the roots $\beta_j$ and so is invariant under all elements of the Galois group. But then, these coefficients are algebraic numbers in $\mathbb{Q}$ whence integers. That is, $u(t) \boxtimes v(t) \in \Lambda_n(\Z)$ . It should already be clear from this that the rings $\Lambda_n(\Z)$ contain a lot of arithmetic information!

For a general commutative ring we will copy this argument by considering a free overring $A^{(\infty)}$ (with 1 as one of the base elements) by formally adjoining roots. At level 1, consider M_0 to be the set of all non-constant one-polynomials over and consider the ring

$A^{(1)} = \bigotimes_{f \in M_0} A[X]/(f) = A[X_f, f \in M_0]/(f(X_f) , f \in M_0)$

The idea being that every one-polynomial $f \in M_0$ now has one root, namely $\alpha_f = \overline{X_f}$ in $A^{(1)}$ . Further, $A^{(1)}$ is a free A-module with basis elements all $\alpha_f^i$ with $0 \leq i < deg(f)$ .

Good! We now have at least one root, but we can continue this process. At level 2, M_1 will be the set of all non-constant one-polynomials over $A^{(1)}$ and we use them to construct the free overring $A^{(2)}$ (which now has the property that every $f \in M_0$ has at least two roots in $A^{(2)}$ ). And, again, we repeat this process and obtain in succession the rings $A^{(3)},A^{(4)},\hdots$ . Finally, we define $A^{(\infty)} = \underset{\rightarrow}{lim}~A^{(i)}$ having the property that every one-polynomial over A splits entirely in linear factors over $A^{(\infty)}$ .

But then, for all $u(t),v(t) \in \Lambda_n(A)$ we can compute $u(t) \boxtimes v(t) \in \Lambda_n(A^{(\infty)})$ . Remains to show that the resulting truncated one-polynomial has all its entries in A. The ring $A^{(\infty)} \otimes_A A^{(\infty)}$ contains two copies of $A^{(\infty)}$ namely $A^{(\infty)} \otimes 1$ and $1 \otimes A^{(\infty)}$ and the intersection of these two rings in exactly (here we use the freeness property and the additional fact that 1 is one of the base elements). But then, by functoriality of L_n , the element $u(t) \boxtimes v(t) \in L_n(A^{(\infty)} \otimes_A A^{(\infty)})$ lies in the intersection $\Lambda_n(A^{(\infty)} \otimes 1) \cap \Lambda_n(1 \otimes A^{(\infty)})=\Lambda_n(A)$ . Done!

Hence, we have endo-functors $\Lambda_n$ in the category of all commutative rings, for every number n. Reviewing the construction of L_n one observes that there are natural transformations $L_{n+1} \rightarrow L_n$ and therefore also natural transformations $\Lambda_{n+1} \rightarrow \Lambda_n$ . Taking the inverse limits $\Lambda(A) = \underset{\leftarrow}{lim} \Lambda_n(A)$ we therefore have the ‘one-power series’ endo-functor $\Lambda~:~\wis{comm} \rightarrow \wis{comm}$ which is ‘almost’ the functor W of big Witt vectors. Next time we’ll take you through the identification using ‘ghost variables’ and how the functor $\Lambda$ can be used to define the category of $\lambda$ -rings.

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fun, Galois, geometry, google, Grothendieck, groups, lambda rings, Lenstra, numbers, Riemann, Witt

The odd knights of the round table

Jan 28th

Posted by lievenlb in games

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Here’s a tiny problem illustrating our limited knowledge of finite fields : “Imagine an infinite queue of Knights $\{ K_1,K_2,K_3,\hdots \}$ , waiting to be seated at the unit-circular table. The master of ceremony (that is, you) must give Knights K_a and K_b a place at an odd root of unity, say $\omega_a$ and $\omega_b$ , such that the seat at the odd root of unity $\omega_a \times \omega_b$ must be given to the Knight $K_{a \otimes b}$ , where $a \otimes b$ is the Nim-multiplication of and . Which place would you offer to Knight $K_{16}$ , or Knight K_n , or, if you’re into ordinals, Knight $K_{\omega}$ ?”

What does this have to do with finite fields? Well, consider the simplest of all finite field $\mathbb{F}_2 = \{ 0,1 \}$ and consider its algebraic closure $\overline{\mathbb{F}_2}$ . Last year, we’ve run a series starting here, identifying the field $\overline{\mathbb{F}_2}$ , following John H. Conway in ONAG, with the set of all ordinals smaller than $\omega^{\omega^{\omega}}$ , given the Nim addition and multiplication. I know that ordinal numbers may be intimidating at first, so let’s just restrict to ordinary natural numbers for now. The Nim-addition of two numbers $n \oplus m$ can be calculated by writing the numbers n and m in binary form and add them without carrying. For example, $9 \oplus 1 = 1001+1 = 1000 = 8$ . Nim-multiplication is slightly more complicated and is best expressed using the so-called Fermat-powers $F_n = 2^{2^n}$ . We then demand that $F_n \otimes m = F_n \times m$ whenever m < F_n and $F_n \otimes F_n = \frac{3}{2}F_n$ . Distributivity wrt. $\oplus$ can then be used to calculate arbitrary Nim-products. For example, $8 \otimes 3 = (4 \otimes 2) \otimes (2 \oplus 1) = (4 \otimes 3) \oplus (4 \otimes 2) = 12 \oplus 8 = 4$ . Conway’s remarkable result asserts that the ordinal numbers, equipped with Nim addition and multiplication, form an algebraically closed field of characteristic two. The closure $\overline{\mathbb{F}_2}$ is identified with the subfield of all ordinals smaller than $\omega^{\omega^{\omega}}$ . For those of you who don’t feel like going transfinite, the subfield $~(\mathbb{N},\oplus,\otimes)$ is identified with the quadratic closure of $\mathbb{F}_2$ .

The connection between $\overline{\mathbb{F}_2}$ and the odd roots of unity has been advocated by Alain Connes in his talk before a general public at the IHES : “L’ange de la géométrie, le diable de l’algèbre et le corps à un élément” (the angel of geometry, the devil of algebra and the field with one element). He describes its content briefly in this YouTube-video

At first it was unclear to me which ‘coupling-problem’ Alain meant, but this has been clarified in his paper together with Caterina Consani Characteristic one, entropy and the absolute point. The non-zero elements of $\overline{\mathbb{F}_2}$ can be identified with the set of all odd roots of unity. For, if x is such a unit, it belongs to a finite subfield of the form $\mathbb{F}_{2^n}$ for some n, and, as the group of units of any finite field is cyclic, x is an element of order 2^n-1 . Hence, $\mathbb{F}_{2^n}- \{ 0 \}$ can be identified with the set of 2^n-1 -roots of unity, with $e^{2 \pi i/n}$ corresponding to a generator of the unit-group. So, all elements of $\overline{\mathbb{F}_2}$ correspond to an odd root of unity. The observation that we get indeed all odd roots of unity may take you a couple of seconds¹.

Assuming we succeed in fixing a one-to-one correspondence between the non-zero elements of $\overline{\mathbb{F}_2}$ and the odd roots of unity $\mu_{odd}$ respecting multiplication, how can we recover the addition on $\overline{\mathbb{F}_2}$ ? Well, here’s Alain’s coupling function, he ties up an element x of the algebraic closure to the element s(x)=x+1 (and as we are in characteristic two, this is an involution, so also the element tied up to x+1 is s(x+1)=(x+1)+1=x. The clue being that multiplication together with the coupling map s allows us to compute any sum of two elements as $x+y=x \times s(\frac{y}{x}) = x \times (\frac{y}{x}+1)$ . For example, all information about the finite field $\mathbb{F}_{2^4}$ is encoded in this identification with the 15-th roots of unity, together with the pairing s depicted as

Okay, we now have two identifications of the algebraic closure $\overline{\mathbb{F}_2}$ : the smaller ordinals equipped with Nim addition and Nim multiplication and the odd roots of unity with complex-multiplication and the Connes-coupling s. The question we started from asks for a general recipe to identify these two approaches.

To those of you who are convinced that finite fields (LOL, even characteristic two!) are objects far too trivial to bother thinking about : as far as I know, NOBODY knows how to do this explicitly, even restricting the ordinals to merely the natural numbers!

Please feel challenged! To get you started, I’ll show you how to place the first 15 Knights and give you a procedure (though far from explicit) to continue. Here’s the Nim-picture compatible with that above

To verify this, and to illustrate the general strategy, I’d better hand you the Nim-tables of the first 16 numbers. Here they are

It is known that the finite subfields of $~(\mathbb{N},\oplus,\otimes)$ are precisely the sets of numbers smaller than the Fermat-powers F_n . So, the first one is all numbers smaller than F_1=4 (check!). The smallest generator of the multiplicative group (of order 3) is 2, so we take this to correspond to the unit-root $e^{2 \pi i/3}$ . The next subfield are all numbers smaller than F_2 = 16 and its multiplicative group has order 15. Now, choose the smallest integer k which generates this group, compatible with the condition that $k^{\otimes 5}=2$ . Verify that this number is 4 and that this forces the identification and coupling given above.

The next finite subfield would consist of all natural numbers smaller than F_3=256 . Hence, in this field we are looking for the smallest number k generating the multiplicative group of order 255 satisfying the extra condition that $k^{\otimes 17}=4$ which would fix an identification at that level. Then, the next level would be all numbers smaller than F_4=65536 and again we would like to find the smallest number generating the multiplicative group and such that the appropriate power is equal to the aforementioned k, etc. etc.

Can you give explicit (even inductive) formulae to achieve this? I guess even the problem of placing Knight 16 will give you a couple of hours to think about… (to be continued).

If m is odd, then (2,m)=1 and so 2 is a unit in the finite cyclic group $~(\mathbb{Z}/m\mathbb{Z})^*$ whence , so the m-roots of unity lie within those of order [↩]

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arxiv, Connes, Conway, fun, Galois, games, geometry, ONAG, ordinals, simples

Grothendieck’s functor of points

Sep 29th

Posted by lievenlb in geometry

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Brave New Geometries

A comment-thread well worth following while on vacation was Algebraic Geometry without Prime Ideals at the Secret Blogging Seminar. Peter Woit became lyric about it :

My nomination for the all-time highest quality discussion ever held in a blog comment section goes to the comments on this posting at Secret Blogging Seminar, where several of the best (relatively)-young algebraic geometers in the business discuss the foundations of the subject and how it should be taught.

I follow far too few comment-sections to make such a definite statement, but found the contributions by James Borger and David Ben-Zvi of exceptional high quality. They made a case for using Grothendieck’s ‘functor of points’ approach in teaching algebraic geometry instead of the ‘usual’ approach via prime spectra and their structure sheaves.

The text below was written on december 15th of last year, but never posted. As far as I recall it was meant to be part two of the ‘Brave New Geometries’-series starting with the Mumford’s treasure map post. Anyway, it may perhaps serve someone unfamiliar with Grothendieck’s functorial approach to make the first few timid steps in that directions.

Allyn Jackson’s beautiful account of Grothendieck’s life “Comme Appele du Neant, part II” (the first part of the paper can be found here) contains this gem :

“One striking characteristic of Grothendieck’s mode of thinking is that it seemed to rely so little on examples. This can be seen in the legend of the so-called “Grothendieck prime”.

In a mathematical conversation, someone suggested to Grothendieck that they should consider a particular prime number. “You mean an actual number?” Grothendieck asked. The other person replied, yes, an actual prime number. Grothendieck suggested, “All right, take 57.”

But Grothendieck must have known that 57 is not prime, right? Absolutely not, said David Mumford of Brown University. “He doesn’t think concretely.””

We have seen before how Mumford’s doodles allow us to depict all ‘points’ of the affine scheme $\wis{spec}(\Z[x])$ , that is, all prime ideals of the integral polynomial ring $\Z[x]$ . Perhaps not too surprising, in view of the above story, Alexander Grothendieck pushed the view that one should consider all ideals, rather than just the primes. He achieved this by associating the ‘functor of points’ to an affine scheme.

Consider an arbitrary affine integral scheme with coordinate ring $\Z[X] = \Z[t_1,\hdots,t_n]/(f_1,\hdots,f_k)$ , then any ringmorphism $\phi~:~\Z[t_1,\hdots,t_n]/(f_1,\hdots,f_k) \rightarrow R$ is determined by an n-tuple of elements $~(r_1,\hdots,r_n) = (\phi(t_1),\hdots,\phi(t_n))$ from which must satisfy the polynomial relations $f_i(r_1,\hdots,r_n)=0$ . Thus, Grothendieck argued, one can consider $~(r_1,\hdots,r_n)$ an an ‘-point’ of and all such tuples form a set h_X(R) called the set of -points of . But then we have a functor

$h_X~:~\wis{commutative rings} \rightarrow \wis{sets} \qquad R \mapsto h_X(R)=Rings(\Z[t_1,\hdots,t_n]/(f_1,\hdots,f_k),R)$

So, what is this mysterious functor in the special case of interest to us, that is when $X = \wis{spec}(\Z[x])$ ? Well, in that case there are no relations to be satisfied so any ringmorphism $\Z[x] \rightarrow R$ is fully determined by the image of which can be any element $r \in R$ . That is, $Ring(\Z[x],R) = R$ and therefore Grothendieck’s functor of points $h_{\wis{spec}(\Z[x]}$ is nothing but the forgetful functor.

But, surely the forgetful functor cannot give us interesting extra information on Mumford’s drawing? Well, have a look at the slightly extended drawing below :

What are these ‘smudgy’ lines and ‘spiky’ points? Well, before we come to those let us consider the easier case of identifying the -points in case is a domain. Then, for any $r \in R$ , the inverse image of the zero prime ideal of under the ringmap $\phi_r~:~\Z[x] \rightarrow R$ must be a prime ideal of $\Z[x]$ , that is, something visible in Mumford’s drawing. Let’s consider a few easy cases :

For starters, what are the $\Z$ -points of $\wis{spec}(\Z[x])$ ? Any natural number $n \in \Z$ determines the surjective ringmorphism $\phi_n~:~\Z[x] \rightarrow \Z$ identifying $\Z$ with the quotient $\Z[x]/(x-n)$ , identifying the ‘arithmetic line’ $\wis{spec}(\Z) = \{ (2),(3),(5),\hdots,(p),\hdots, (0) \}$ with the horizontal line in $\wis{spec}(\Z[x])$ corresponding to the principal ideal ~(x-n) (such as the indicated line ~(x) ).

When $\mathbb{Q}$ are the rational numbers, then $\lambda = \tfrac{m}{n}$ with m,n coprime integers, in which case we have $\phi_{\lambda}^{-1}(0) = (nx-m)$ , hence we get again an horizontal line in $\wis{spec}(\Z[x])$ . For $\overline{\mathbb{Q}}$ , the algebraic closure of $\mathbb{Q}$ we have for any $\lambda$ that $\phi_{\lambda}^{-1}(0) = (f(x))$ where f(x) is a minimal integral polynomial for which $\lambda$ is a root. But what happens when $K = \mathbb{C}$ and $\lambda$ is a trancendental number? Well, in that case the ringmorphism $\phi_{\lambda}~:~\Z[x] \rightarrow \mathbb{C}$ is injective and therefore $\phi_{\lambda}^{-1}(0) = (0)$ so we get the whole arithmetic plane!

In the case of a finite field $\mathbb{F}_{p^n}$ we have seen that there are ‘fat’ points in the arithmetic plane, corresponding to maximal ideals ~(p,f(x)) (with f(x) a polynomial of degree which remains irreducible over $\mathbb{F}_p$ ), having $\mathbb{F}_{p^n}$ as their residue field. But these are not the only $\mathbb{F}_{p^n}$ -points. For, take any element $\lambda \in \mathbb{F}_{p^n}$ , then the map $\phi_{\lambda}$ takes $\Z[x]$ to the subfield of $\mathbb{F}_{p^n}$ generated by $\lambda$ . That is, the $\mathbb{F}_{p^n}$ -points of $\wis{spec}(\Z[x])$ consists of all fat points with residue field $\mathbb{F}_{p^n}$ , together with slightly slimmer points having as their residue field $\mathbb{F}_{p^m}$ where is a divisor of . In all, there are precisely p^n (that is, the number of elements of $\mathbb{F}_{p^n}$ ) such points, as could be expected.

Things become quickly more interesting when we consider -points for rings containing nilpotent elements.

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introducing : the n-geometry cafe

Jul 17th

Posted by lievenlb in general

3 comments

It all started with this comment on the noncommutative geometry blog by “gabriel” :

Even though my understanding of noncommutative geometry is limited, there are some aspects that I am able to follow. I was wondering, since there are so few blogs here, why don’t you guys forge an alliance with neverending books, you blog about noncommutative geometry anyways. That way you have another(n-category cafe) blogspot and gives well informed views(well depending on how well defined a conversational-style blog can be).

The technology to set up a ‘conversational-style blog’, where anyone can either leave twitter-like messages or more substantial posts, is available thanks to the incredible people from Automattic.

For starters, they have the sensational p2 wordpress theme : “blogging at the speed of thought”

A group blog theme for short update messages, inspired by Twitter. Featuring: Hassle-free posting from the front page. Perfect for group blogging, or as a liveblog theme. Dynamic page updates. Threaded comment display on the front page. In-line editing for posts and comments. Live tag suggestion based on previously used tags. A show/hide feature for comments, to keep things tidy. Real-time notifications when a new comment or update is posted. Super-handy keyboard shortcuts.

Next, any lively online community is open for intense debate : “supercharge your community”

Fire up the debate with commenter profiles, reputation scores, and OpenID. With IntenseDebate you’ll tap into a whole new network of sites with avid bloggers and commenters. And that’s just the tip of the iceberg!

And finally, as we want to talk math, both in posts and comments, they provide us with the WP-LaTeX plugin.

All these ingredients make up the n-geometry cafe¹ to be found at noncommutative.org (explaining the ‘n’).

Anyone can walk into a Cafe and have his/her say, that’s why you’ll get automatic author-privileges if you register.

Fill in your nick and email (please take your IntenseDebate setting and consider signing up with Gravator.com to get a nice image next to your contributions), invent your own password, show that you’re human by answering the reCapcha question and you’ll get a verification email within minutes². This will take you to your admin-page, allowing you to start blogging. For more info, check out the FAQ-pages.

I’m well aware of the obvious dangers of non-moderated sites, but also a strong believer in any Cafe’s self-regulating powers…

If you are interested in noncommutative geometry, and feel like sharing, please try it out.

with apologies to the original cafe but I simply couldn’t resist… [↩]
if you don’t get an email within the hour, please notify me [↩]

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geometry

Lambda-rings for formula-phobics

big Witt vectors for everyone (1/2)

The odd knights of the round table

Grothendieck’s functor of points

Brave New Geometries

introducing : the n-geometry cafe

My latest tweets

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geometry

Lambda-rings for formula-phobics

big Witt vectors for everyone (1/2)

The odd knights of the round table

Grothendieck’s functor of points

Brave New Geometries

introducing : the n-geometry cafe

My latest tweets

User Login

Blogroll

Series of posts

Who's Online

About Me