From now on the main object (the irony of this terminology will become clear) of study is ${\mathbb{F}_1}$. I will first elaborate on what Kapranov-Smirnov calls the “folklore imagery”, trying to understand the motivation behind a statement like

A vector space of ${\mathbb{F}_1}$ is just a set.

Afterwards I will draw some conclusions from these observations, motivating some of the approaches outlined in Mapping ${\mathbb{F}_1}$-land.

Demystifying the ${\mathbb{F}_1}$-lore

In Projective geometry over ${\mathbb{F}_1}$ and the Gaussian binomial coefficient a nice build-up is given. I will summarise and give my own viewpoint.

We already know that ${\mathbb{F}_1}$ doesn’t exist as a field (because we need ${0\neq 1}$ by the axioms). But what if we loosen up the axioms and say the trivial ring should be taken as ${\mathbb{F}_1}$? In that case modules over this ring are in relationship with vector spaces over ${\mathbb{F}_1}$. But there are only trivial modules over the trivial ring, hence no nontrivial vector spaces. This approach obviously doesn’t work. A conclusion I like to make after vastly generalising this approach:

Setting ${\mathbb{F}_1}$ to be something doesn’t work.

The idea of looking at vector spaces looks promising though: given a field it is the most obvious structure built upon it, so we want to make sense of it over ${\mathbb{F}_1}$. We know that the cardinality of an ${n}$-dimensional vector space ${V}$ over a finite field ${\mathbb{F}_q}$ is ${q^n}$. Applying this to ${\mathbb{F}_1}$ we get a single point, for all ${n}$. So far so good, but we also know that a basis for ${V}$ consists of ${n}$ elements. The problem with this approach is that it’s too direct: we cannot construct objects over ${\mathbb{F}_1}$, we need to get there by an analogy that avoids contradictions like this. Applying this for instance to ${\mathbb{F}_1[t]}$ we see that this doesn’t yield any satisfying definition either. The conclusion after another round of generalisation:

Look at induced objects, not constructions.

The same applies to noncommutative geometry by the way. But let’s focus on ${\mathbb{F}_1}$ for the moment.

So we need a simple object over ${\mathbb{F}_1}$ where we can avoid an explicit construction, getting facts about that object only by analogy without running into contradictions. Let’s try this for ${\mathbb{P}^n/\mathbb{F}_1}$. The construction of this over an actual field ${k}$ consists of constructing ${\mathbb{A}^{n+1}/k}$ (an ${n+1}$-dimensional vector space over ${k}$) and setting ${\mathbb{P}^n/k}$ to be the set of lines through the origin.

If we take ${k=\mathbb{F}_q}$ the number of points in ${\mathbb{A}^{n+1}/\mathbb{F}_q}$ is ${q^{n+1}}$. The number of lines through the origin is ${(q^{n+1}-1)/(q-1)}$: just take any point in ${\mathbb{A}^{n+1}/\mathbb{F}_q\setminus\left\{ 0 \right\}}$, this defines a line through the origin, but there are ${q}$ points (including the origin) on this line, therefore we divide by ${q-1}$. If we write down the polynomial function counting the number of points in ${\mathbb{P}^n/\mathbb{F}_q}$ (i.e., ${q^n+q^{n-1}+\ldots+1}$) and evaluate in ${q=1}$ we see something that leads to a nontrivial object! To make this analogy really sound you have to write down what you actually want from a finite projective space and axiomatise it. But let’s rejoice for now, and conclude

The projective geometry ${\mathbb{P}^1/\mathbb{F}_1}$ contains ${n}$ points.

What we have actually done is changing the construction of ${\mathbb{P}^n/\mathbb{F}_1}$ from an algebraic-geometric viewpoint to a combinatorial-geometric viewpoint, something that is quintessential in finite geometry (and I guess I’m the angs+’er with the most background in this kind of stuff ). If people are interested in a write-up about this, I’d be happy to provide one but I suspect my fellow seminarians are not that into finite geometry and combinatorics. The only important observation we need to make is that a line in ${\mathbb{P}^n/\mathbb{F}_1}$ contains exactly two points in this sense, as a line in ${\mathbb{P}^n/\mathbb{F}_q}$ contains ${q+1}$ points by the axioms of a combinatorial-geometric projective space.

Now taking ${\mathbb{F}_1}$-vector spaces to be sets actually makes sense: ${\mathbb{A}^{n+1}/\mathbb{F}_1}$ is an ${(n+1)}$-set, taking any point as the distinguished base point (or origin) and considering “lines” through the origin, or more appropriately ${2}$-subsets, of which we have exactly ${n}$ as we started with ${n+1}$ elements and fixed an origin. But this (inherently geometric) idea of “fixing an origin” has its downsides in what follows, where we will adjoin an origin which is more in the sense of algebra as adding a zero vector (which doesn’t exist over ${\mathbb{F}_1}$) doesn’t change the dimension or cardinality of a base for a vector space.

Implications

Now we can switch our attention to Kapranov’s and Smirnov’s unfinished paper. If Wittgenstein were an algebraic geometer interested in ${\mathbb{F}_1}$-geometry I guess he would have written a Tractatus Absoluto-Geometricus, which could have looked (in a very crude sense) like this

1. 1 Geometry over ${\mathbb{F}_1}$ can only be understood through induced objects.
2. 2 Vector spaces over ${\mathbb{F}_1}$ are plain sets.
   1. 2.1 Dimension equals cardinality.
   2. 2.2 ${\mathrm{GL}_n(\mathbb{F}_1)=\mathrm{S}_n}$.
   3. 2.3 ${\mathrm{SL}_n(\mathbb{F}_1)=\mathrm{A}_n}$.
   4. 2.4 ${\det\colon\mathrm{GL}_n(\mathbb{F}_1)\rightarrow\mathbb{F}_1^\times}$ is the sign homomorphism.
   5. 2.5 The Grassmannian ${\mathrm{Gr}(k,n)(\mathbb{F}_1)}$ is the set of ${k}$-subsets.
   6. 2.6 There is no harm in formally adjoining a zero vector in a ${\mathbb{F}_1}$-vector space turning it into a pointed set, just be careful with interpretations.
3. 3 The polynomial ring ${\mathbb{F}_1[t]}$ can only be understood through its automorphisms.
   1. 3.1 Polynomial automorphisms are a generalisation of field automorphisms by evaluation at “zero”.
   2. 3.2 We have ${\mathrm{GL}_n(\mathbb{F}_1[t])\rightarrow\mathrm{GL}_n(\mathbb{F}_1)}$.
   3. 3.3 ${\mathrm{GL}_n(\mathbb{F}_1[t])=\mathrm{B}_n}$ the braid group on ${n}$ strings, by analogy of the canonical ${\mathrm{B}_n\rightarrow\mathrm{S}_n}$.
   4. 3.4 For more information I refer you to the grandmaster himself and his blog post ${\mathbb{F}_1}$ and braid groups.
4. 4 Finite fields have finite extensions and algebraic closures and so does ${\mathbb{F}_1}$.
   1. 4.1 ${\mathbb{F}_{1^n}}$ as a vector space over ${\mathbb{F}_1}$ is the (pointed) set ${\mu_n}$ consisting of the ${n}$-th roots of unity and an adjoined zero (which will serve as the point of the pointed set).
   2. 4.2 By choosing a primitive root in ${\mu_n}$ we get a non-canonical isomorphism ${\mathrm{C}_n\cong\mu_n}$.
   3. 4.3 ${\mathbb{A}^1/\mathbb{F}_1}$ as a scheme is ${\mathrm{Spec}\,\mathbb{F}_1[t]}$.
   4. 4.4 ${\mathrm{Spec}\,\mathbb{F}_1[t]}$ describes the algebraic closure of ${\mathbb{F}_1}$.
   5. 4.5 ${\overline{\mathbb{F}_1}}$ therefore corresponds to ${\mathrm{\mu}_\infty\cup\left\{ 0 \right\}}$.
5. 5 We can generalise linear algebra to finite extensions.
   1. 5.1 The action of ${\mathbb{F}_{1^n}}$ on ${V}$ is the action of ${\mu_n}$ on ${V\setminus\left\{ 0 \right\}}$.
   2. 5.2 ${\mathbb{F}_{1^n}}$-vector spaces are nothing but ${\mu_n}$-sets.
   3. 5.2 A ${d}$-dimensional ${\mathbb{F}_1}$-space contains ${dn}$ points.
   4. 5.4 The ${\mu_n}$-action is strictly multiplicative, there is no additive structure.
   5. 5.5 The lack of additive structure agrees with the notion of vector spaces as (pointed) sets.
   6. 5.6 A ${\mathbb{F}_{1^n}}$-basis is a set containing a representative of every orbit under the ${\mu_n}$-action.
6. 6 We can interpret other finite objects over ${\mathbb{F}_1}$.
   1. 6.1 ${\mathbb{F}_q}$ is a ${\mathbb{F}_{1^n}}$-vector space if ${q\equiv 1\bmod n}$ because ${(\mathbb{F}_q^\times,\cdot)\cong(\mathrm{C}_{nd},\cdot)}$ for ${d=(q-1)/n}$, and therefore a ${\mathbb{F}_{1^n}}$-algebra.
   2. 6.2 This vector space structure induces the (necessarily unique) ${\mathbb{F}_{1^n}}$-vector space structure on ${\mathbb{F}_q^e}$ of dimension ${ed}$.
   3. 6.3 All development of techniques should coincide with this observation.
   4. 6.4 For a construction of exact sequences over ${\mathbb{F}_1}$ I refer you to Absolute linear algebra.

For the real Wittgenstein aficionado, my apologies for not hitting the right tone and ideas. This post mostly served as a way for me to get all my ${\mathbb{F}_1}$-folklore straight, so that I can explain to a complete outsider why stuff in the ${\mathbb{F}_1}$ is taken as what it is. If you feel any gaps present, please tell me so.

If you followed me this far you should be familiar enough with the ideas and twists of mind necessary to understanding some parts of Kapranov-Smirnov, or you have made up your mind and will take all ${\mathbb{F}_1}$-stuff to be dadaist nonsense. But before tackling Kapranov-Smirnov I suggest you read the series of posts developing the same theme as this one did but in greater depth:

1. The ${\mathbb{F}_1}$ folklore
2. Absolute linear algebra
3. ${\mathbb{F}_1}$ and braid groups

For the next (and last) post of this series, the main idea will be Stuff over ${\mathbb{F}_1}$ contains multiplicative structure, not additive structure.

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Today (one of) the goals of this blog will be neglected, and we’ll focus solely on its URL. No $\mathbb{F}_{1}$ or Riemann hypothesis, just some old-fashioned ring theory.

Misplaced Poetics

I’d like write up one of (Jacobson-) Herstein’s commutativity theorems in noncommutative ring theory. It’s a beautiful testament to the power of subdividing rings into nontrivial classes, and is filled to the brim with small, simple, but extremely elegant ideas. Part of the theorem’s appeal stems from the fact that once confronted with its statement, you’re bound to ask: who cares?! It seems so useless, and that somehow makes it all the more fun. Of course, there was a good reason for proving the theorem, but I’ll just leave you to ponder some applications. I’m in a “l’art pour l’art”-mood, hoping Poe doesn’t turn in his grave.

Trivial?

Without further ado, here’s the theorem I’ve been going on about:

Theorem. A ring $R$ is commutative iff for any $a,b \in R$, there exists an integer $n(a,b)>1$ such that $(ab-ba)^{n(a,b)}=ab-ba$.

Of course, your first idea, like everyone else’s, is that this can be proven by (possible) extreme amounts of formula bashing. I encourage you to try this, as I have, and get very tired (and a tad bit frustrated) after wasting the better part of a Friday morning (and wacking half a tree). If you do manage to find a proof by pure calculation, let me know! The way we’ll tackle the theorem is by a reduction argument along the following lines:

• If $R$ is a division ring: see the next paragraph
• If $R$ is a left primitive ring: reduce to the previous case
• If $R$ is a semiprimitive ring: reduce to the previous case
• If $R$ is a (general) ring: reduce to the previous case

Just lovely, isn’t it.

Thé lemma

Crucial for proving the first part of the theorem is the following lemma, which will be applied in the next paragraph.

Lemma. Let $D$ be a division ring with nonzero characteristic. If $a$ is a non-central, periodic element in $D^{*}$, then there exists an additive commutator $b$ in $D^{*}$ such that $bab^{-1}=a^{i} \neq a$, for some $i>0$.

Taking $F_{p}$ to be the prime subfield of $D$, the $F_{p}$-algebra generated by $a$ is a finite subfield of $D$, of dimension $n$ over $F_{p}$, which we’ll denote $K$. The order of this field is $p^{n}$, and $a^{p^{n}}=a$. Since $a$ isn’t central, the inner derivation $\delta$, defined by $\delta(x)=ax-xa$, is non-zero. You can easily check that it is however a $K$-linear mapping of the $K$-vector space $D$. Using the Frobenius, one sees that $\delta^{p^{n}}=\delta$. Since any element $b \in K$ satisfies $b^{p^{n}}=b$, the following equation holds true $$t^{p^{n}} -t= \prod_{b \in K}(t-b) \in K[t].$$ Substituting $\delta$ into this equation, and remembering that $\delta \neq 0$ and all mono’s are left cancellable, there has to exist a $b_{0} \in K^{*}$ such that $\delta-b_{0}$ is not a mono. This means $\delta$ has an eigenvector $d$ with eigenvalue $b_{0}$. Using the definition of $\delta$, it follows that $dad^{-1}=a-b_{0} \in K \ \backslash \{a\}$. Since $a$ and $dad^{-1}$ have the same order in the cyclic group $K^{*}$, they generate the same subgroup, and $dad^{-1}=a^{i} \neq a$. To make sure $d$ is an additive commutator, replace it by $\delta(d)$, and you’ll see that the equation still holds up.

Division rings

Supposing $R$ is a division ring which doesn’t equal its centre $F$, you can easily see that there must exist an additive commutator which isn’t central, say $a=bb’-b’b$. Taking an element $c$ in $F^{*}$, a quick calculation shows that $ca$ is an additive commutator which can’t be central. The assumptions then imply that there is a number $k$ such that $1=a^{k}=(ca)^{k}=c^{k}a^{k}$, so any element in $F^{*}$ is periodic, and $D$ has non-zero characteristic. Using the theorem in the previous paragraph on $a$, an additive commutator $y$ exists, such that the group generated by $a$ and $y$ is a finite, periodic subgroup of $D^{*}$. This means the group has to be cyclic, which contradicts the statement $yay^{-1} \neq a$, and $R$ is commutative.

The proof of the pudding …

Let’s finish the proof. Suppose $R$ is a left primitive ring. The structure theorem for left primitive rings (basically a revamped version of Jacobson & Chevalley’s density theorem), says $R$ is isomorphic to a matrix ring over a skew field $D$, or $R$ has an infinite number of subrings $R_{m}$, each one having $M_{m}(D)$ as a quotient. Supposing $m>1$, and denoting by $E_{ij}$ the matrix with a $1$ in the $ij$-place and zeroes everywhere else, the obvious identity $$E_{11} E_{12} – E_{12} E_{11}=E_{12}$$ gives us a contradiction (just raise it to a power greater than $1$). This means $R \cong D$, and we reduced it to a previous case.

If our ring is semiprimitive, the Jacobson radical, denoted $J(R)$, is zero. Given a left primitive ideal $M_{i}$, take the quotient ring $R/M_{i}$, which is obviously left primitive, and inherits the property in the theorem because it’s a surjective image of $R$. This means each one of these rings is commutative. Now $R$ can be embedded in the product of all these rings, since the Jacobson radical, which is equal to the intersection of all left primitive ideals, is zero. Thus $R$ is commutative.

For a random ring, $R/J(R)$ is semiprimitive, and thus commutative. For $a, b$ in the ring, we know there exists an $n$ such that $(ab-ba)(1-(ab-ba)^{n-1})=0$. Since $ab-ba$ sits in $J(R)$, we know $(1-(ab-ba)^{n-1})$ is invertible, and $ab-ba=0$. Voilà!

More of this fun stuff can be found in Lam’s A first course in noncommutative rings and Lectures on modules and rings.

Here are the scans of my rough prep-notes for some of the later seminar-talks. These notes still contain mistakes, most of them were corrected during the talks. So, please, read these notes with both mercy are caution!

Hurwitz formula imples ABC : The proof of Smirnov’s argument, but modified so that one doesn’t require an $\epsilon$-term. This is known to be impossible in the number-theory case, but a possible explanation might be that not all of the Smirnov-maps $q~:~\mathsf{Spec}(\mathbb{Z}) \rightarrow \mathbb{P}^1_{\mathbb{F}_1}$ are actually covers.

Frobenius lifts and representation rings : Faithfully flat descent allows us to view torsion-free $\mathbb{Z}$-rings with a family of commuting Frobenius lifts (aka $\lambda$-rings) as algebras over the field with one element $\mathbb{F}_1$. We give several examples including the two structures on $\mathbb{Z}[x]$ and Adams operations as Frobenius lifts on representation rings $R(G)$ of finite groups. We give an example that this extra structure may separate groups having the same character table. In general this is not the case, the magic Google search term is ‘Brauer pairs’.

Big Witt vectors and Burnside rings : Because the big Witt vectors functor $W(-)$ is adjoint to the tensor-functor $- \otimes_{\mathbb{F}_1} \mathbb{Z}$ we can view the geometrical object associated to $W(A)$ as the $\mathbb{F}_1$-scheme determined by the arithmetical scheme with coordinate ring $A$. We describe the construction of $\Lambda(A)$ and describe the relation between $W(\mathbb{Z})$ and the (completion of the) Burnside ring of the infinite cyclic group.

Density theorems and the Galois-site of $\mathbb{F}_1$ : We recall standard density theorems (Frobenius, Chebotarev) in number theory and use them in combination with the Kronecker-Weber theorem to prove the result due to James Borger and Bart de Smit on the etale site of $\mathsf{Spec}(\mathbb{F}_1)$.

New geometry coming from $\mathbb{F}_1$ : This is a more speculative talk trying to determine what new features come up when we view an arithmetic scheme over $\mathbb{F}_1$. It touches on the geometric meaning of dual-coalgebras, the Habiro-structure sheaf and Habiro-topology associated to $\mathbb{P}^1_{\mathbb{Z}}$ and tries to extend these notions to more general settings. These scans are unintentionally made mysterious by the fact that the bottom part is blacked out (due to the fact they got really wet and dried horribly). In case you want more info, contact me.

why noncommutative geometry?

Some motivate noncommutative geometry as follows : assume you have a space (or variety) $X$ on which a group $G$ acts wildly so that the ‘orbit-space’ $X/G$ does not exists or has bad topological properties. Let $A$ be the ring of continuous functions on $X$ (or the coordinate ring $\mathcal{O}(X)$), then every $g \in G$ acts as an automorphism $\alpha_g$ on $A$.

Traditionally one associates the orbit-space (when possible) to the commutative fixed-point algebra $A^G$. However, when this algebra is too small to give information on the $G$-orbits in $X$ one can still associate a noncommutative algebra to the situation, the crossed product algebra $A \ast G$ which as a vectorspace is merely $A \otimes \mathbb{C} G$ but with multiplication induced by $(a \otimes g) (b \otimes h) = a \alpha_g(b) \otimes g h$. Some argue that ringtheoretical invariants of $A \ast G$ give some insight into the horrible orbit-space $X/G$.

relevant to $\mathbb{F}_1$-geometry?

We’ve defined an algebra $A$ over $\mathbb{F}_1$ to be a torsion-free $\mathbb{Z}$-ring having a commuting family of endomorphisms $\psi^n~:~A \rightarrow A$ having the property that for every prime number $p$ the endomorphism $\Psi^p$ is a lift of the Frobenius map on $A/pA$. This gives an action by endomorphisms of the multiplicative monoid $\mathbb{N}_{\times}$ on $A$.

We’ve interpreted this additional structure as descent-data from $\mathbb{Z}$ to $\mathbb{F}_1$. Now, in the case of Galois-descent between two fields $k \subset K$ with $Gal(K/k)=G$, the $k$-algebra corresponding to a $K$-algebra $A$ with descent-data $G \rightarrow Aut(A)$ is, of course, the fixed-point algebra $A^G$.

Of course, in the $\mathbb{F}_1$-setting it makes no sense to look at the fixed-point ring $A^{\mathbb{N}_{\times}}$, but we can still consider the corresponding noncommutative ring

$A \ast \mathbb{N}_{\times}$

which as a $\mathbb{Z}$-module is the tensor-product $A \otimes_{\mathbb{Z}} \mathbb{Z} [\mathbb{N}_{\times}]$ where $\mathbb{Z} [\mathbb{N}_{\times}]$ is the monoid-algebra of the commutative monoid $\mathbb{N}_{\times}$. As above, the multiplication is induced by the rule (using the variables $X_n = 1 \otimes n$)

$(a X_n) (b X_m) = a \Psi^n(b) X_{mn}$

If you are a lowly ringtheorist this is already daunting enough because the fact that the crossing is made with endos rather than autos kills most of the desired properties of your noncommutative ring (for example Noetherianness). But, if your a $C^{\ast}$-algebraist then you want to complicate matters even more as you need variables $X_n^{\ast}$ corresponding to the $X_n$ satisfying suitable properties. If this is possible, we will denote the noncommutative algebra generated by $A$, the $X_n$ and the $X_n^*$ by $A \circ \mathbb{N}_{\times}$.

the giant mashup-algebra $\mathbb{Z}[\mathbb{Q}/\mathbb{Z}] \circ \mathbb{N}_{\times}$, aka BC

Lots of papers are written trying to get novel insights into the BC-algebra by looking at its adelic-, motivic-, semi-hemi-demi-, p-adic-, $\mathbb{F}_1$-gadgety or whatever-comes-next interpretation. It is the archetypical example of the above construction.

Let’s define it by generators and relations using its ‘integral’ incarnation. Generators are $e(r)$, one for each $r \in \mathbb{Q}/\mathbb{Z}$ and elements $\tilde{\mu}_n$ and $\mu_n^*$ for $n \in \mathbb{N}_+$. The relations are

$e(r) e(s) = e(r+s)~\forall r,s \in \mathbb{Q}/\mathbb{Z}$

$\tilde{\mu}_n \tilde{\mu}_m = \tilde{\mu}_{nm}~\forall n,m \in \mathbb{N}_+$

$\mu_n^* \mu_m^* = \mu^*_{nm}~\forall n,m \in \mathbb{N}_+$

$\mu_n^* \tilde{\mu}_n = n~\quad \text{and} \quad \tilde{\mu}_n \mu^*_m = \mu^*_m \tilde{\mu}_n~\quad~\text{whenever} \quad (m,n)=1$

$\mu^*_n e(r) = e(nr) \mu^*_n~\forall r \in \mathbb{Q}/\mathbb{Z}, n \in \mathbb{N}_+$

$e(r) \tilde{\mu}_n = \tilde{\mu}_n e(nr)~\forall r \in \mathbb{Q}/\mathbb{Z}, n \in \mathbb{N}_+$

$\tilde{\mu}_n e(r) \mu^*_n = \sum_{ns=r} e(s)~\forall r \in \mathbb{Q}/\mathbb{Z}, n \in \mathbb{N}_+$

The first relations imply that the $\mathbb{Z}$-ring generated by the $e(r)$ is the integral group-ring $\mathbb{Z}[\mathbb{Q}/\mathbb{Z}]$. Taking $e(r) \mapsto e^{2 \pi i r}$ we see that this ring is isomorphic to the integral group-ring $\mathbb{Z}[\pmb{\mu}_{\infty}]$ of the multiplicative group of all roots of unity.

$\mathbb{Z}[\pmb{\mu}_{\infty}]$ is a $\lambda$-ring (actually, our best shot at the algebraic closure $\overline{\mathbb{F}}_1$) with endomorphisms $\Psi^n(e^{2 \pi i r}) = e^{2 \pi i nr}$ (which correspond to the endomorphisms $e(r) \mapsto e(nr)$ in $\mathbb{Z}[\mathbb{Q}/\mathbb{Z}]$).

Hence, we see that the subring generated by the $e(r)$ and the $\mu_n^*$ is actually isomorphic to the noncommutative crossed product $\mathbb{Z}[\pmb{\mu}_{\infty}] \ast \mathbb{N}_{\times}$ constructed before. The full BC-algebra is then what we have denoted $\mathbb{Z}[\pmb{\mu}_{\infty}] \circ \mathbb{N}_{\times}$.

More information on the (classical) BC-algebra can be found in these neverendingbook-posts : as a giant mash-up of arithmetical information and its relation to the Riemann zeta-function.

In view of the Borger-de Smit result characterizing the etale site of $\mathsf{Spec}(\mathbb{F}_1)$ it is perhaps interesting to consider the multi-variate BC-algebras $\mathbb{Z}[\pmb{\mu}_{\infty}] \otimes \cdots \otimes \mathbb{Z}[\pmb{\mu}_{\infty}] \circ \mathbb{N}_{\times}$ defined in the now obvious way.

More food-for-thought : take your favorite torsion free $\mathbb{Z}$-ring $A$ and construct your own BC-lookalike algebra $W(A) \circ \mathbb{N}_{\times}$ making clever use of the Adams operations $\Psi^n$ and the ‘Verschiebung’-operations on the ring of big Witt vectors $W(A)$.

In the previous post in this series I promised to do something with fields with characteristic two, and instead I did weird things with surreal numbers and ordinals. Neither of them has characteristic two, because we used the wrong arithmetic. In this post, I will give three new definitions of addition and multiplication in On, and prove that they are actually the same. This will turn On into a field of characteristic two, which we shall call On₂. From know on, we distinguish the ordinary operations from those in On₂ by the use of square brackets. All expressions between $[$ and $]$ are meant in the sense of ordinary arithmetic.

Arithmetic in On₂

Simplicity rules

The most obvious and at the same time unusual way of defining an addition is by starting from zero and working up. We will find the simplest addition and multiplication which make On into a field.

There is no reason why $0+0$ cannot be $0$, because there are fields (any field) with an element satisfying $x+x=x$. This is the first entry in our addition-table. This implies that $0$ must be the zero element, so we must have $0+\alpha=\alpha+0=\alpha$ for all $\alpha$. The first row and column are already filled. What about $1+1$? The least possible answer is $0$, which gives us characteristic two. Next is $1+2$. This cannot be $0$, $1$ or $2$, so we must take $3$. We can go on like this, and make sure that $\alpha + \beta$ is compatible with $\alpha’ + \beta$, $\alpha + \beta’$ and $\alpha’ + \beta’$ (with $\alpha’ < \alpha$ and $\beta’ < \beta$).

We do the same for multiplication. $0.\alpha$ can be $0$, so $0$ must be the zero of the field. Because $1.1 = 1$ is possible, $1$ is the one. The first two rows and columns are filled with $0.\alpha$, $\alpha.0$, $1.\alpha$ and $\alpha.1$. It is obvious that $2.2$ cannot be $0$, $1$ or $2$. Since there are fields (e.g. $\mathbb{F}_4$) with elements that satisfy $x^2 = x + 1$, $3$ is possible. Note that the product has to be compatible with previous entries and with the whole addition-table.

Remark

This definitions are rather difficult to work with, because we must prove a theorem every time we want to fill in an entry. Besides, it is not obvious that these definitions really define a field.

Inductive definitions

Define the minimal excluded number $\DeclareMathOperator{\mex}{mex}\mex(S)$ as the least ordinal not in the set $S$. This can be used for the following inductive definitions:

• $\alpha + \beta = \mex(\alpha’ + \beta, \alpha + \beta’)$
• $\alpha\beta = \mex(\alpha’\beta + \alpha\beta’ + \alpha’\beta’)$

It is easy to verify that $\alpha + \alpha = 0$, because $\alpha + \alpha’$ cannot be zero. We can now prove that this definitions are equivalent to the former.

If $\alpha + \beta < \mex(\alpha’ + \beta, \alpha + \beta’)$, there exists an $\alpha’ < \alpha$ such that $\alpha + \beta = \alpha’ + \beta$. This implies $\alpha = \alpha’$, which is impossible. Therefore, $$\alpha + \beta \geq \mex(\alpha’ + \beta, \alpha + \beta’).$$

If $\alpha\beta < \mex(\alpha’\beta + \alpha\beta’ + \alpha’\beta’)$, there exist $\alpha’$ and $\beta’$ so that $\alpha\beta = \alpha’\beta + \alpha\beta’ + \alpha’\beta’$. This is equivalent to $$\begin{gather}\alpha\beta + \alpha’\beta + \alpha\beta’ + \alpha’\beta’ = 0 \\ (\alpha + \alpha’)(\beta + \beta’) = 0 ,\end{gather}$$ which implies $\alpha = \alpha’$ or $\beta = \beta’$. Both are impossible. It follows that $$\alpha\beta \geq \mex(\alpha’\beta + \alpha\beta’ + \alpha’\beta’).$$

If we can prove that these inductive definitions form a field, it must be the smallest possible field, as defined before. This is a standard verification.

Nim-arithmetic

The inductive definition of the sum is known as nim-addition (frequently used in the theory of the game of Nim). An easy rule to perform nim-addition is:

1. The nim-sum of a number of distinct $2$-powers is their ordinary sum.
2. The nim-sum of two equal numbers is 0.

This rule allows us to compute the nim-sum of finite and infinite ordinals. A similar rule for nim-multiplication is:

1. The nim-product of a number of distinct Fermat $2$-powers (numbers of the form $2^{2^n}$) is their ordinary product.
2. The square of a Fermat $2$-power is its sesquimultiple (multiplying by $\frac{3}{2}$ in the ordinary sense).

Unfortunately, this rule applies only to finite ordinals. A more general rule is explained at neverendingbooks.

Groups in On₂

The ordinals that are groups are precisely the $2$-powers. This can be proved with the simplest extension theorems.

Theorem 1. If $\Delta$ is not a group (under addition), then $\Delta = \alpha + \beta$, where $(\alpha, \beta)$ is any lexicographically earliest pair of numbers in $\Delta$ whose sum is not in $\Delta$.

Theorem 2. If $\Delta$ is a group, we have $[\Delta\alpha] + \beta = [\Delta\alpha + \beta]$, for all $\alpha$, and all $\beta \in \Delta$.

If $\Delta$ is a group, and $\Gamma$ is a group with $\Delta < \Gamma < [\Delta.2]$, we can write $\Gamma = [\Delta + \delta]$ with $\delta < \Delta$. This is a contradiction, because $\Gamma > \Delta + \delta$ follows from the inductive definitions, and $[\Delta + \delta]  = \Delta + \delta$ according to Theorem 2.

If $\alpha, \beta \in [\Delta.2]$, there are three possible cases.

1. $\alpha < \Delta$ and $\beta < \Delta$. Then $\alpha + \beta < \Delta < [\Delta.2]$
2. $\alpha \geq \Delta$ and $\beta < \Delta$. By Theorem 2:
   $\alpha + \beta = [\Delta + \delta] + \beta = \Delta + \delta + \beta = \Delta + \delta’ = [\Delta + \delta'] < [\Delta.2]$
3. $\alpha \geq \Delta$ and $\beta \geq \Delta$. By Theorem 2 and $\alpha + \alpha = 0$:
   $\alpha + \beta = [\Delta + \delta] + [\Delta + \delta'] = \Delta + \Delta + \delta + \delta’ = \delta + \delta’ < \Delta < [\Delta.2]$

This proves that if $\Delta$ is any group, then the next group is $[\Delta.2]$. Because $2$ is a group, it follows that the groups are the $2$-powers. This justifies the rule for the calculation of nim-sums.

Fields in On₂

Similar theorems exist for fields in On₂. Complete proofs can be found in Conway’s On Numbers and Games.

Theorem 3. If $\Delta$ is a group but not a ring, then $\Delta = \alpha\beta$, where $(\alpha, \beta)$ is any lexicographically earliest pair of numbers in $\Delta$ whose product is not in $\Delta$.

Theorem 4. If $\Delta$ is a ring but not a field, then $\Delta = \alpha^{-1}$, where $\alpha$ is the earliest non-zero number in $\Delta$ which has no inverse in $\Delta$.

Theorem 5. If $\Delta$ is a field but not algebraically closed, then $\Delta$ is a root of the lexicographically earliest polynomial having no root in $\Delta$.

Finite ordinals

We will prove by induction that the finite ordinals that are fields are precisely the Fermat $2$-powers. We suppose that the following statements are true for $n$, and prove them for $n + 1$:

1. $[2^{2^n}]$ is a field
2. $[2^{2^{n-1}}]^2 = [\frac{3}{2}2^{2^{n-1}}]$
3. $x^2 + x$ takes precisely the values $0, 1, \dotsc, [2^{2^n-1}-1]$ as $x$ varies in $[2^{2^n}]$

The lexicographically earliest irreducible polynomial over $[2^{2^n}]$ is $x^2 + x = [2^{2^n-1}]$, because $x^2 = \alpha$ always has a root in finite field of characteristic $2$, and $x^2 + x = \alpha$ has a root for earlier $\alpha$ according to statement 3. We know by Theorem 5 that $[2^{2^n}]$ is a root of $x^2 + x = [2^{2^n-1}]$, hence $$\textstyle [2^{2^n}]^2 = [2^{2^n}] + [2^{2^n-1}] = [2^{2^n} + 2^{2^n-1}] = [\frac{3}{2}2^{2^n}].$$ We obtain the field $[2^{2^{n+1}}]$ as a vector space over $[2^{2^n}]$ with typical element $X = [2^{2^n}]x + y$. We examine the polynomial $$\begin{align}X^2 + X &= ([2^{2^n}]x + y)^2 + [2^{2^n}]x + y \\ &= [2^{2^n}]^2 x^2 + y^2 + [2^{2^n}]x + y \\ &= [2^{2^n}](x^2 + x) + ([2^{2^n-1}]x^2 + y^2 + y).\end{align}$$ By induction,  $x^2 + x$ can take any value in $[2^{2^n-1}]$. Note that $x^2 + x$ remains unchanged when we replace $x$ by $x + 1$. The same is true for $y^2 + y$. It follows that $[2^{2^n-1}]x^2 + y^2 + y$ can be made to take any value in $[2^{2^n}]$ without affecting the value of $x^2 + x$. This implies that the values of $X^2 + X$ can be written as $[2^{2^n}]\alpha + \beta$, where $\alpha < [2^{2^n-1}]$ and $\beta < [2^{2^n}]$, which are precisely the values less than $[2^{2^{n+1}-1}]$.

This and Theorem 6 justify the rule for the calculation of nim-products.

Infinite ordinals

Consider the sequence $$[\omega^{\omega^k}], [\omega^{\omega^k p_k}], [\omega^{\omega^k p_k^2}], \dotsc, [\omega^{\omega^k p_k^n}], \dotsc$$ where $p_k$ is the $(k+1)$’st prime. Then the following statements are true for each $k > 0$:

1. Each term in the sequence is a field
2. The field $[\omega^{\omega^k p_k^n}]$ is the union of all finite fields $\mathbb{F}_{2^{p_0^{n_0} p_1^{n_1} \dotsm p_k^{n_k}}}$ with $n_i < \omega$ for $0 \leq i \leq k – 1$ and $n_k \leq n$
3. Each term is the $p_k$’th power of its successor, and $[\omega^{\omega^k}]$ is the $p_k$’th root of $\alpha_{p_k}$, which is the least number in $[\omega^{\omega^k}]$ with no $p_k$’th root in $[\omega^{\omega^k}]$.

We will prove this by induction on $k$.

$\boldsymbol{n = 0}$

$[\omega^{\omega^{k+1}}]$ is the union of all fields $[\omega^{\omega^k p_k^n}]$. It is obvious that this defines a field, and there are no fields in between. This proves statement 1, and statement 2 follows immediately. Because of Theorem 5, $[\omega^{\omega^{k+1}}]$ is the root of the lexicographically earliest polynomial having no root in $[\omega^{\omega^{k+1}}]$. If $f(x)$ is a polynomial of degree $d < p_{k+1}$, all coefficients are contained in a finite field $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_k^{n_k}}}$. Therefore, the root of $f(x)$ is an element of the field $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_k^{n_k} d}} = \mathbb{F}_{2^{p_0^{m_0} \dotsm p_k^{m_k}}}$, which is a subfield of $[\omega^{\omega^{k+1}}]$. It follows that the earliest irreducible polynomial is $x^{p_{k+1}} = \alpha_{p_{k+1}}$ with $\alpha_{p_{k+1}}$ as defined in statement 3.

$\boldsymbol{n > 0}$

Assume $\Gamma = [\omega^{\omega^{k+1} p_{k+1}^{n-1}}]$ is a field, and $\Delta$ is the lexicographically earliest algebraic extension. The field $\Gamma$ is not closed for polynomials of degree $p_{k+1}$, because $\mathbb{F}_{2^{p_{k+1}^n}}$ is a field extension of $\mathbb{F}_{2^{p_{k+1}^{n-1}}} \subset \Gamma$ of degree $p_{k+1}$, and $\mathbb{F}_{2^{p_{k+1}^n}}$ is not contained in $\Gamma$. This means that $[\Delta:\Gamma]$ is at most $p_{k+1}$. Therefore, every element $\alpha \in \Delta$ is the root of a polynomial $f(x)$ of degree $d \leq p_{k+1}$. By induction, all coefficients of $f(x)$ are contained in a field $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_{k+1}^{n_{k+1}}}}$ with $n_{k+1} \leq n – 1$. It follows that the root of $f(x)$ is contained in $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_{k+1}^{n_{k+1}}d}} = \mathbb{F}_{2^{p_0^{m_0} \dotsm p_{k+1}^{m_{k+1}}}}$ with $m_{k+1} \leq n$. If $d < p_{k+1}$, this is a subfield of $\Gamma$ and $f(x)$ is not irreducible. We can conclude that $[\Delta:\Gamma] = p_{k+1}$, and $\Delta = [\omega^{\omega^{k+1} p_{k+1}^{n}}]$. This proves statements 1 and 2.

If $f(x)$ is a polynomial $x^{p_{k+1}} = \alpha$ with $\alpha \in [\omega^{\omega^{k+1} p_{k+1}^{n-1}}]$, we know by induction that $\alpha$ is contained in $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_{k+1}^{n_{k+1}}}}$ with $n_{k+1} \leq n – 1$. The root of $f(x)$ is thus contained in $\mathbb{F}_{2^{p_0^{m_0} \dotsm p_{k+1}^{m_{k+1}}}}$ with $m_{k+1} \leq n$, which is a subfield of $[\omega^{\omega^{k+1} p_{k+1}^{n}}]$. It follows that the earliest irreducible polynomial is $x^{p_{k+1}} = [\omega^{\omega^{k+1} p_{k+1}^{n-1}}]$. By Theorem 5, $[\omega^{\omega^{k+1} p_{k+1}^n}]$ is a root of this polynomial. This proves statement 3.

From this, we can conclude that $[\omega^{\omega^\omega}]$ is the algebraic closure of $2$.

Remark

The computation of $\alpha_p$ is not a trivial task. Conway did stop at $\alpha_7$. Hendrik Lenstra described an effective method in his paper On the algebraic closure of two, and computed $\alpha_p$ for $p \leq 43$. Lieven Le Bruyn extended the list.

Last time, I proved that the Hasse principle is valid for Kronecker-Weber Theorem. So in order to prove Kronecker-Weber, the local version still has to be proved. As noted before, you only need to prove it for Galois extensions $K/\mathbb{Q}_p$ with Galois group $\mathbb{Z}/q^m\mathbb{Z}$, where $q$ is a prime number that can be equal to $p$. There will be 3 cases to prove:

• The extension is unramified.
• The extension is ramified, but the ramification degree $e$ is not divisible by $p$ (also known as tamely ramified).
• The extension is ramified with ramification degree $e$ divisible by $p$ (wildly ramified).

In all these cases, we will have need for Hensel’s Lemma:

Hensel’s Lemma Let $L$ be a local field, with ring of integers $\mathcal{O}_L$ and maximal ideal $\mathfrak{b}$. Let $l$ be it’s residue field. Suppose $f \in \mathcal{O}_L [x]$, monic with restriction $\bar{f}$ in $l[x]$. Suppose$\bar{f}$ has a root $\alpha$ in $l$, such that $\bar{f}’(\alpha) \neq 0$. Then there exists a root $\beta \in \mathcal{O}_L$ of $f$, with $\beta \equiv \alpha \bmod \mathfrak{b}$.

Let us start the unramified case, which will be the easiest. We will use the notation from the previous post.

$K/\mathbb{Q}_p$ is unramified

There is a much stronger result for this, which I will prove here. Taking $K = \mathbb{Q}_p$ will give what we want.

Theorem Suppose $L/K$ is an unramified, finite Galois extension, where $L$ and $K$ are finite extensions of $\mathbb{Q}_p$ (not necessarily Galois over $\mathbb{Q}_p$). Then there exists an $n \in \mathbb{N},p \nmid n$ so that $L = K(\zeta_n)$. Furthermore, $Gal(L/K)$ will be cyclic.

Proof Take $L/K$ like in the theorem, with prime $\mathfrak{p}$ in both extensions, since $e=1$. Let $l$ and $k$ be the residue fields of respectively $L$ and $K$. Since $e = 1$, $Gal(L/K) = Gal(l/k)$. The right hand side is a Galois group of an extension in finite fields, hence it is cyclic. Since $l/k$ is a Galois extension, there exists a primitive element $\alpha$ so that $l = k[\alpha]$. Consequently, there exists an $n$, with $\gcd(n,p)=1$, such that $X^n-1$ has a root in $l$, namely $\alpha$. Using Hensel’s Lemma, we find a $\beta \in \mathcal{O}_L$ as a root of $X^n-1$ and $\beta \equiv \alpha \bmod \mathfrak{p}$. Since $\beta$ is a root of $1$ and, due to being equivalent to $\alpha$, has order at least $n$, we have $$\left[K(\beta):K\right] \geq \left[k[\alpha]:k\right] = \left[l:k\right] = \left[L:K\right].$$ But $\beta \in L$, so we get $L = K(\beta)$ and $\beta = \zeta_n$.

This was the easy one, the next two cases are harder.

$K/\mathbb{Q}_p$ is tamely ramified.

This case follows from 3 lemmas. I will give the proofs depending on how difficult or technical it is.

Lemma Let $p\neq 2$ be a prime. Then $\mathbb{Q}_p$ contains the $p-1$-th roots of unity, but none other.

Proof Let $\mu_p$ be the set of roots of unity in $\mathbb{Q}_p$. Thanks to Hensel’s Lemma, we have a surjective morphism $\phi:\mu_p \rightarrow \mathbb{F}_p^*$, so we need to prove that it is injective. So suppose we have a root of unity of $(1+tp)^n$. Using Newton’s binomial formula, we get $$\sum_{i = 1}^n \binom{n}{i} (tp)^{i}=0.$$ Now, there are 2 cases: $t = 0$ and we are done, or $$ \sum_{i = 1}^n \binom{n}{i} (tp)^{i-1}=0,$$ but then, since $p|0$, $p|\binom{n}{1}=n.$ Since $\mathbb{Q}_p$ contains the $n$-th root of $1$ and $p|n$, $\mu_p$ must contain the $p$-th roots of unity. So we are reduced to $(1+tp)^p = 1$. But then we have $$\sum_{i = 1}^p \binom{p}{i} (tp)^{i}=0.$$ If $t \neq 0$, we have $$\sum_{i = 1}^p \binom{p}{i} (tp)^{i-1}=0.$$But every term is divisible by $p^2$ except for the first one, which is only divisible by $p$, so we have a contradiction. This concludes the proof.

In case of $p=2$, $\mathbb{Q}_2$ only has $-1,1$ as roots of unity. This is proved by noticing that $(\mathbb{Z}/8\mathbb{Z})^* = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

Lemma Let $L/K/\mathbb{Q}_p$ be a tower of finite extensions, with $L/K$ Galois. Let $\mathfrak{p}_K$ be the unique maximal ideal of $K$ and suppose that that $L/K$ is totally ramified of degree $e$. Then there exists a $\pi \in K$ a generator of $\mathfrak{p}_K$ and a root $\alpha$ of the polynomial $f(X) = X^e – \pi$, such that $L = K(\alpha)$.

Proof Use of the non-archimedean absolute value, choosing a uniformizing parameter and using the fact that elements of the Galois group preserve absolute values, this is a long but not very difficult proof and will be omitted.

Lemma $\mathbb{Q}_p((-p)^{\frac{1}{p-1}}) = \mathbb{Q}_p(\zeta_p)$

Proof It can be proved that the maximal ideal of $\mathbb{Q}_p(\zeta_p)$ is given by $(1-\zeta_p)$ and that $(p) = (1-\zeta_p)^{p-1}$. Consequently, $(1-\zeta_p)^p = (p(1-\zeta_p))$. Take the polynomial $$g(X) = \frac{(X+1)^p-1}{X}=\sum_{i=1}^{p} \binom{p}{i}X^{i-1},$$ which is the minimal polynomial of $\zeta_p-1$, so we get $$0 = g(\zeta_p) \equiv (\zeta_p-1)^{p-1}+p \bmod (\zeta_p-1)^p.$$ This can be modified to $$u = \frac{(\zeta_p-1)^{p-1}}{-p} \equiv 1 \bmod (\zeta_p-1).$$ Now look at the polynomial $f(X) = X^{p-1}- u$. Then $f(1) \equiv 0 \bmod (\zeta_p – 1)$ and $(\zeta_p – 1)\nmid f’(1)$. Hensel’s Lemma (again) implies that there exists an $\alpha \in \mathbb{Q}_p(\zeta_p)$ as a root of $f$. But $(-p)^{\frac{1}{p-1}} = \frac{\zeta_p-1}{\alpha}$, so $(-p)^{\frac{1}{p-1}} \in \mathbb{Q}_p(\zeta_p)$. Now, the minimal polynomial of $(-p)^{\frac{1}{1-p}}$ is $X^{p-1}+p$, since it is irreducible due to Eisenstein’s criterion, so $(-p)^{\frac{1}{1-p}}$ and $\zeta_p$ have the same degree over $\mathbb{Q}_p$. This gives $\mathbb{Q}_p((-p)^{\frac{1}{1-p}}) = \mathbb{Q}_p(\zeta_{p})$.

Now, for the proof of the tamely ramified case. Let $L/\mathbb{Q}_p$ be a tamely ramified abelian extension of ramification degree $e, p\nmid e$, with $K/\mathbb{Q}_p$ the maximal unramified subextension. We already know that $K \subset \mathbb{Q}_p(\zeta_n)$ for some $n$. We also have $L/K$ is totally ramified with degree $e$. From one of the above lemmas, we know that $L = K(\pi^{\frac{1}{e}})$ with $\pi$ a generator of the unique maximal ideal $\mathfrak{p}_K$ of $\mathcal{O}_K$. $K/\mathbb{Q}_p$ is unramified, so $p$ is also a generator of $\mathfrak{p}_K$. This gives $\pi = -up$ for some unit $u$ of $\mathcal{O}_K$. $u$ is a unit, $p \nmid e$, so the discriminant of $f(X) = X^e-u$ is not divisible by $p$. This implies that $K(u^{\frac{1}{e}})/K$ is unramified. $K \subset \mathbb{Q}_p(\zeta_n)$, so $$K(u^{\frac{1}{e}}) \subset K(\zeta_M) \subset \mathbb{Q}_p(\zeta_{Mn}),$$ with $M$ an integer. Let $T$ be the compositum of $\mathbb{Q}_p(\zeta_{Mn})$ and $L$. $T$ is an abelian extension of $\mathbb{Q}_p$ and $u^{\frac{1}{e}},\pi^{\frac{1}{e}} \in T$, which gives $(-p)^{\frac{1}{e}} \in T$. Since $\mathbb{Q}_p((-p)^{\frac{1}{e}}) \subset T$ and $T/\mathbb{Q}_p$ is abelian, it is an Galois extension over $\mathbb{Q}_p$. Since all the roots of $X^e +p$ are $\zeta_e^k (-p)^{\frac{1}{e}}, k=0 \ldots e-1$, we know that $\zeta_e \in \mathbb{Q}_p((-p)^{\frac{1}{e}})$. $\mathbb{Q}_p((-p)^{\frac{1}{e}})$ is totally ramified and since every subfield of a totally ramified subfield is totally ramified in case of local fields, $\mathbb{Q}_p(\zeta_e)$ is also totally ramified. But $p \nmid e$, so it can’t ramify, except when it’s the trivial extension, so $\zeta_e \in \mathbb{Q}_p$ and $e | p-1$. We have a tower of extensions $$\mathbb{Q}_p((-p)^{\frac{1}{e}}) \subset \mathbb{Q}_p((-p)^{\frac{1}{p-1}}) \subset \mathbb{Q}_p(\zeta_p).$$ Consequently, $$L = K(\pi^{\frac{1}{e}}) \subset K(u^{\frac{1}{e}},(-p)^{\frac{1}{e}}) \subset \mathbb{Q}_p(\zeta_{Mnp}),$$ which concludes this case.

$K/\mathbb{Q}_p$ is widely ramified

This is the hardest case, so I won’t give the entire proof, but I will describe the way the proof works. We know that we just have to work with extensions with Galois group $\mathbb{Z}/q^{m}\mathbb{Z}$, with $q$ a prime. Since we work in the widely ramified case, this breaks down to $q = p$. There are 2 cases to consider: $p = q = 2$ or $p = q \neq 2$ (2 is always a bad prime). Let us first proof the last case.

Take an abelian widely ramified extension $L/\mathbb{Q}_p$ of degree $p^m$, with cyclic Galois group. We first construct 2 Galois extensions of $\mathbb{Q}_p$ of degree $p^m$, one totally ramified, the other unramified:

• Consider the extension $\mathbb{Q}_p(\zeta_{p^{m+1}})$ with Galois group $\mathbb{Z}/p^{m+1}\mathbb{Z}$ and let $K_r$ be the fixed field by the subgroup of order $p-1$. This will be a totally ramified extension of degree $p^m$ with cyclic Galois group.
• Take an irreducible polynomial in $\mathbb{F}_p[X]$ of degree $p^m$, lift it to an irreducible polynomial in $\mathbb{Z}_p[X]$ and construct the corresponding Galois extension. This will give an unramified extension $K_u$ with Galois group $\mathbb{Z}/p^m\mathbb{Z}$.

As we know from the unramified case, $K_u \subset \mathbb{Q}_p(\zeta_n)$ for some $n$. Since one extension is unramified and the other is totally ramified, we necessarily have $K_u \cap K_v = \mathbb{Q}_p$ and consequently, $Gal(K_uK_v/\mathbb{Q}_p) = (\mathbb{Z}/p^m\mathbb{Z}^2)$. Suppose that $L \not\subset \mathbb{Q}_p(\zeta_n,\zeta_{p^{m+1}})$, then we should have $$Gal(L(\zeta_n,\zeta_{p^{m+1}})) = (\mathbb{Z}/p^m\mathbb{Z})^2 \times\mathbb{Z}/p^k\mathbb{Z},$$ with $k > 0$. This implies that there is a Galois extension of $\mathbb{Q}_p$ with Galois group $(\mathbb{Z}/p\mathbb{Z})^3$. The next part is to prove that there doesn’t exist such an extension, but this is very hard and requires knowledge of Kummer Theory..

The case $p=q=2$ is similar. One concludes that there should exist a Galois extension of $\mathbb{Q}_2$ with Galois group $(\mathbb{Z}/2\mathbb{Z})^4$ or $(\mathbb{Z}/4\mathbb{Z})^3$ and once again it is proved that this is impossible.

Finite fields might seem to be easy to understand, but they are definitely not. Some examples of our limited knowledge are:

• We represent the field $\mathbb{F}_{p^n}$ by $\mathbb{F}_p[x]/(f(x))$ whith $f(x)$ a monic irreducible polynomial of degree $n$. The polynomial $f$ is not unique, and there is no best option.
• We can’t extend $\mathbb{F}_{p^n}$ to $\mathbb{F}_{p^{n+1}}$ with compatible addition and multiplication.
• Simple problems like placing knights at a table are too difficult.
• We know what the algebraic closure of a finite field is, but how can we do calculations in such an exotic thing?

Conway polynomials are a good example of the danger of solving these problems.

Therefore a better description of finite fields is necessary. For fields with characteristic two there is a good candidate using ordinal numbers with nim-addition and nim-multiplication. In this series I will work out this example. Let’s start by defining numbers.

Numbers

In ONAG, Conway defines (surreal) numbers as follows:

Definition. If $L$, $R$ are any two sets of numbers, and no member of $L$ is $\geq$ any member of $R$, then there is a number $\{L \vert R\}$. All numbers are constructed in this way.

The last sentence is somewhat informal. It means there is no sequence of numbers $x_i = \{L_i \vert R_i\}$ with $x_{i+1} \in L_i \cup R_i$ for all $i \in \mathbb{N}$.

Notation. If $x = \{L \vert R\}$ we write $x^L$ and $x^R$ for the typical member of $L$ resp. $R$. For $x$ itself we then write $\{x^L \vert x^R\}$.

Note that all the definitions are inductive, and don’t need a basis because they start with the empty set. Each element of the empty set has all desired properties because the empty set has no members. This allows us to make a lot of proofs very short. If we can proof that for a property $P$, $P(x)$ is implied by $P(x^L)$ and $P(x^R)$, then $P$ must be true for all numbers. Otherwise, there would be a number $x$ for which $P(x)$ is false. Hence, there is a $x^L$ or $x^R$ for which $P(x^\bullet)$ is false. We can go on like this, but since that would create an infinite descending sequence of numbers, $P(x)$ has to be true.

The relations on numbers are defined as:

• $x \leq y$ iff $x < y^R$ and $x^L < y$
• $x \geq y$ iff $x > y^L$ and $x^R > y$
• $x = y$ iff $x \leq y$ and $x \geq y$

This is a total order. A remarkable property is that $x^L < x < x^R$ for all numbers $x$. We now define an addition and multiplication. We know that $x + y$ must lie between both $x^L + y$ and $x + y^L$ (on the left) and $x^R + y$ and $x + y^R$ (on the right). From $x – x^L > 0$ and $y – y^L > 0$ we can deduce $(x – x^L)(y – y^L) > 0$, so that we must have $xy > x^Ly + xy^L – x^Ly^L$. This motivates following definition:

• $x + y = \{x^L + y, x + y^L \vert x^R + y, x + y^R \}$
• $xy = \{ x^Ly + xy^L – x^Ly^L, x^Ry + xy^R – x^Ry^R \vert x^Ly + xy^R – x^Ly^R, x^Ry + xy^L – x^Ry^L \}$

With this addition and multiplication, the Class No of all surreal numbers forms a Field, but not a field (because No is not a set).

Examples of numbers

According to the definition, every number is constructed as two sets of earlier constructed numbers. The only way to get it off the ground is by using empty sets. We call $\{ \vert \}$ the number $0$, born on the zeroth day.

Starting from $0$, the next generation of numbers can be constructed. We obtain $1 = \{ 0 \vert \}$ and $-1 = \{ \vert 0 \}$. It is easy to verify that $\{ 0 \vert 0 \}$ is not a number. We have $-1 < 0 < 1$. This was the first day.

The number $\{ 0 \vert 1 \}$ lies somewhere between $0$ and $1$. We call this number $\frac{1}{2}$. What about $\{ -1, 0 \vert 1 \}$? We can verify the inequalities $\{ -1, 0 \vert 1 \} \leq \{ 0 \vert 1 \}$ and $\{ -1, 0 \vert 1 \} \geq \{ 0 \vert 1 \}$, so we have two expressions for the same number. They are equal, but not identical.

We can go on like this with $\frac{1}{4} = \{ 0 \vert \frac{1}{2} \}$ and $\frac{3}{4} = \{ \frac{1}{2} \vert 1 \}$. The whole number $n$ is born on day $n$ as $\{ n-1 \vert 0 \}$. All dyadic rationals and whole numbers are born on finite days. On day $\omega$, $L$ and $R$ can be infinite sets. Examples are:

• $\frac{1}{3} = \{ \frac{1}{4}, \frac{1}{4} + \frac{1}{16}, \frac{1}{4} + \frac{1}{16} + \frac{1}{64}, \ldots \vert \frac{1}{2}, \frac{1}{2} – \frac{1}{8}, \ldots \}$
• $\omega = \{ 0, 1, 2, 3, \ldots \vert \}$
• $\frac{1}{\omega} = \{ 0 \vert 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots \}$

On day $\omega + 1$ it’s starting to get a little strange. The number $x = \{ 0, 1, 2, 3, \ldots \vert \omega \}$ satisfies $n < x < \omega$ for all finite integers $n$, and $x = \omega – 1$.

Subsets

The surreal numbers are the largest possible totally ordered field. The rationals, the reals, the ordinals and all other ordered fields are subsets of No. An interesting subclass of No is On, the class of all ordinal numbers.

Definition. $\alpha$ is an ordinal number if $\alpha$ has an expression of the form $\alpha = \{L\vert \}$.

No is a proper Class (not a set), but for every ordinal $x$ the subclass $\{ a : a < x \}$ is a set. Because $\alpha = \{ \beta : \beta < \alpha \vert \}$, we can treat $\alpha$ as the set of all lesser ordinals.

October 21st : Dear Diary,

today’s seminar was fun, though a bit unconventional. My goal was to explain faithfully flat descent, but at the last moment i had this urge to let students discover the main idea themselves (in the easiest of examples) by means of a thought experiment :

“I am an alien (laughter…), and a very stubborn alien at that. There’s just one field, the complex numbers $\mathbb{C}$, and all rings are $\mathbb{C}$-algebras. I’ve heard strange rumours that you humans believe in a geometry ‘hidden under $\mathbb{C}$’, something called real manifolds. What then is an algebra over this obviously virtual ‘real’ field?”

Their first hurdle was to convey the concept of complex conjugation as the alien(me) was unwilling to decompose a complex number $c$ into two ‘ghost components’ $a+bi$. Still i had to concede that i knew about addition and multiplication, i had a $1$ and a square root of $-1$, which for some reason they preferred to call $i$.

‘Oh, but then you know about $\mathbb{Z}[i]$! You just add a number of times $1$’s with $i$’s.’

‘Why are you humans so obsessed with counting? We do not count! We can’t! We have neither fingers nor toes!’

Admittedly a fairly drastic intervention, but i had to keep them on the path leading to Galois descent… After a while we agreed on a map (they called it conjugation) sending sums to sums, products to products and taking a root of unity to its inverse.

Next, they asked me to be a bit flexible and allow for ‘generalized’ fields such as the one consisting of all elements fixed under conjugation! Clearly, the alien refused : ‘We’re not going on that slippery road called generalization, we’ve seen the havock caused by it in human-mathematics.’

It took them a while to realize they would never be able to sell me an $\mathbb{R}$-algebra $A$, but perhaps they could try to sell me the complex algebra $B= A \otimes_{\mathbb{R}} \mathbb{C}$?

Alien : ‘But, how do i recognize one of your algebras among mine? Is there a test to detect them?’

Humans : ‘Yes, they have a map (which we know to be the map $a \otimes c \mapsto a \otimes \overline{c}$, but you cannot see it) sending sums to sums, products to products that extends the conjugation on $\mathbb{C}$.’

Alien : ‘But if i take a basis for any of my algebras and apply conjugation to all its coordinates, then surely all my algebras have this property, not?’

Humans : ‘No, such maps are good for sums, but not always for products. For example, take $\mathbb{C}[x]/(x^2-c)$ for $c$ a complex-number not fixed under conjugation.’

Alien : ‘Point taken. But then, your algebras are just a subclass of my algebras, right?’

Humans : ‘No! An algebra can have several of such additional maps. For example, take $B = \mathbb{C} \times \mathbb{C}$ then there is one sending $(a,b)$ to $(\overline{a},\overline{b})$ and another sending it to $(\overline{b},\overline{a})$. (because we know there are two distinct real algebras $\mathbb{R} \times \mathbb{R}$ and $\mathbb{C}$ of dimension two, tensoring both to $\mathbb{C} \times \mathbb{C}$.)’

By now, the alien and humans agreed on a dictionary : what to humans is the $\mathbb{R}$-algebra $A$ is to the alien the complex algebra $B=A \otimes \mathbb{C}$ together with a map $\gamma_B : B \rightarrow B$ sending sums to sums, products to products and extending conjugation on $\mathbb{C}$ (this extra structure, the map $\gamma_B$, is called the ‘descent data’).

A human-observed $\mathbb{R}$-algebra morphism $\phi : A \rightarrow A’$ is to the alien the $\mathbb{C}$-algebra morphism $\Phi = \phi \otimes id_{\mathbb{C}} : B \rightarrow B’$ which commutes with the extra structures, that is, $\Phi \circ \gamma_B = \gamma_{B’} \circ \Phi$.

Phrased differently (the alien didn’t want to hear any of this) : there is an equivalence of categories between the category $\mathbb{R}-\mathsf{algebras}$ of commutative $\mathbb{R}$-algebras and the category $\gamma-\mathsf{algebras}$ consisting of complex commutative algebras $B$ together with a ringmorphism $\gamma_B$ extending complex conjugation and with morphisms $\mathbb{C}$-algebra morphisms compatible with the $\gamma$-structure.

Further, what to humans is the base-extension (or tensor) functor

$- \otimes_{\mathbb{R}} \mathbb{C}~:~\mathbb{R}-\mathsf{algebras} \rightarrow \mathbb{C}-\mathsf{algebras}$

is (modulo the above equivalence) to the alien merely the forgetful functor

$\mathsf{Forget}~:~\gamma-\mathsf{algebras} \rightarrow \mathbb{C}-\mathsf{algebras}$

stripping off the descent-data.

After the break (yes, it took us that long to get here) we used this idea to properly define obviously non-existing rings living ‘under $\mathbb{Z}$’, or if you like silly terminology, algebras over the field with one element $\mathbb{F}_1$.

Alien : ‘Ha-ha-ha, a field with one element? Surely you’re joking Mr. Human’

Note to self : Dare to waste time like this in a seminar.

Remember we ended the last post with a question: is it possible to associate with a prime $p$ not dividing $\Delta(f)$, an element $\sigma_{p} \in G$ such that the decomposition type of $f \ \mathsf{mod} \ p$ is the same as the cycle type of $\sigma_{p}$. I told you this could be done up to conjugacy, which I’ll now explain. The element $\sigma_{p}$ will be called the Frobenius substitution of $p$.

Galois theory for finite fields

The Frobenius, as an element of $Aut(\overline{\mathbb{F}}_{p})$, permutes the zeros of any polynomial with coefficients in $\mathbb{F}_{p}$. For such a polynomial $g$, the cycle pattern of the Frobenius, as a permution of the zeros of $g$, is the same as the decomposition type of $g$ over $\mathbb{F}_{p}$. To see this, it’s sufficient to suppose $g$ irreducible, of degree $n$ say. Then you know there is a root $\alpha \in \mathbb{F}_{p^{n}}$, and that the other roots are $\alpha^{p}, \alpha^{p^{2}},…,\alpha^{p^{n-1}}$. The Frobenius obviously acts on $g$ as the permutation $(12…n)$, and since $g$ is irreducible, its decomposition type is also $n$. Returning to the situation at hand, the statement we have just proven closely resembles the one in the introduction. To lift the Frobenius to a Frobenius substitution, which will be an element of $Aut(\mathbb{Q}(\alpha_{1},…,\alpha_{d}))$, we’ll need to connect the fields they’re acting on. Notice that this means we want a way to reduce elements of $\mathbb{Q}(\alpha_{1},…,\alpha_{d})$ modulo $p$. This is exactly what the notion of a place will do for us.

All over the place

A place of $\mathbb{Q}(\alpha_{1},…,\alpha_{d})$ over $p$ is a map $\psi:\mathbb{Q}(\alpha_{1},…,\alpha_{d}) \to \overline{\mathbb{F}}_{p}
\cup \{ \infty \}$, with the following properties:

1. $\psi^{-1} \overline{\mathbb{F}}_{p}$ is a subring of $\mathbb{Q}(\alpha_{1},…,\alpha_{d})$, and $\psi:\psi^{-1} \overline{\mathbb{F}}_{p} \to \overline{\mathbb{F}}_{p}$ is a ring morphism
2. $\psi(x)=\infty$ iff $\psi(x^{-1})=0$, for an $x \in \mathbb{Q}(\alpha_{1},…,\alpha_{d})-\{0\}$

This is a concept frequently used in algebraic number theory, albeit in a different context. To motivate the definition a little, think about this as an extension of the classic reduction mod $p$. Since $p \equiv 0 \ \mathsf{mod} \ p$, it’s only natural to introduce $\infty$, since we would like $1/p \equiv 1/0 \equiv \infty$. Taking the polynomial $f$, and one of its roots $\alpha_{1}$, $\psi(f(\alpha_{1}))=0$. Since we would like a ring morphism, this would imply $$\sum_{i=0}^{d} \ \overline{a_{i}} \ \psi(\alpha_{1})^{i}=0,$$ with $f=\sum_{i=0}^{d} \ a_{i}X^{i}$, and $a_{i} \in \mathbb{Z}$. This shows why the codomain has to contain $\overline{\mathbb{F}}_{p}$.

Frobenius substitution

The places we have introduced have three important properties:

1. a place of $\mathbb{Q}(\alpha_{1},…,\alpha_{d})$ over $p$ exists, for every prime $p$
2. if $\psi$ and $\psi’$ are two places over $p$, then there is a $\tau \in G$ such that $\psi’=\psi \circ \tau$
3. if $p \nmid \Delta(f)$, then $\tau$ is unique

Taking a $p \nmid \Delta(f)$, and $\psi$ a place of $K$ over $p$, the map $\psi’=Frob \circ \psi$ is again a place, since $Frob$ is an $\overline{\mathbb{F}}_{p}$-automorphism. If we apply property $2$ to $\psi$ and $\psi’$, one finds a unique element $Frob_{\psi}$ in $G$ such that $\psi \circ Frob_{\psi} = Frob \circ \psi$. This element is going to be our Frobenius substitution. The end of the last paragraph showed us that the $\psi(\alpha_{i})$ are the zeros of $f \ \mathsf{mod} \ p$ in $\overline{\mathbb{F}}_{p}$, and the characterizing property of $Frob_{\psi}$ tells us that it permutes $\alpha_{1},…,\alpha_{d}$ in exactly the same way as $Frob$ permutes $\psi(\alpha_{1}),…,\psi(\alpha_{d})$. This means that the cycle pattern of $Frob_{\psi}$ is equal to the decomposition type of $f \ \mathsf{mod} \ p$.

There’s no reason why $Frob_{\psi}$ shouldn’t depend on the choice of place $\psi$. Taking $\eta$ to be another place over $p$, we know by property $2$ that $\eta=\psi \circ \tau$, and thus $$\psi \circ \tau \circ Frob_{\psi \circ \tau}=Frob \circ \psi \circ \tau=\psi \circ Frob_{\psi} \circ \tau.$$ Taking $Frob_{\psi \circ \tau}$ equal to $\tau^{-1} \circ Frob_{\psi} \circ \tau$ does the trick, and by property $3$, it’s unique. This means that $Frob_{\psi}$ ranges over a conjugacy class as $\psi$ varies over all the places over $p$. We’ll denote a typical element of this conjugacy class by $\sigma_{p}$.

Chebotarëv density

All of this allows for the formulation of the announced theorem

Theorem. Let $f$ be a monic polynomial with integer coefficients and nonzero discriminant $\Delta(f)$. Let $C$ be a conjugacy class of the Galois group of $f$. Denote by $S$ the following set, $$S=\{ p \ \vert \ p \ prime, \ p \nmid \Delta(f), \ \sigma_{p} \in C \}.$$ Then $S$ has a density, which is equal to $\vert C \vert / \vert G \vert$.

Let’s see how exactly this generalizes our previous two theorems. First off, Dirichlet. The only thing we have to do in this case, is pick $f$ to be equal to $X^{m}-1$. Since the Galois group is abelian, the conjugacy classes are the elements, so choose $a \in G \cong (\mathbb{Z}/m\mathbb{Z})^{*}$, and thus $a$ coprime with $m$. Now what is the unique element $\sigma_{p}$? For $\psi$ a place over $p$, we know that $\psi(\sigma_{p}(x))=\psi(x)^{p}$, for all $x \in \mathbb{Q}(\zeta)$. Picking $x$ to be $\zeta$, and knowing that $\sigma_{p}(\zeta)=\zeta^{a}$ (this is how you identify $G$ with $(\mathbb{Z}/m\mathbb{Z})^{*}$), we get that $\psi(\zeta)^{a}=\psi(\zeta)^{p}$. Since $\psi(\zeta)$ is a primitive $m$-th root of unity, one has $a \equiv p$ mod $m$, and Dirichlet pops out.

The way it generalizes Frobenius is slightly trickier, since it needs a reformulation of Frobenius employing divisions of $G$. A division is sort of a conjugacy up to a power. Given an element $g \in G$ of order $m$, the division of $g$ consists of all elements that are conjugate to any $g^{n}$, as long as $n$ and $m$ are coprime. Obviously, a conjugacy class is part of a division. The converse isn’t necessarily true; two element can belong to the same division without being conjugated. A simple abelian example is given by the elements $3$ mod $10$ and $7$ mod $10$, which aren’t equal in $\mathbb{Z}/10\mathbb{Z}$, and thus not conjugated, but $7^{3}=3$, so they’re in the same division. The alternative statement of Frobenius density then says exactly what Chebotarëv density says, with ‘conjugacy class’ replaced by ‘division’. Often, Chebotarëv can distinguish between primes that get stuck together with Frobenius.

Application

I just can’t let you go without a small, beautiful, and rather mysterious application of Chebotarëv. The statement is this:

Corollary. If $f \in \mathbb{Z}[X]$ is an irreducible polynomial that has a $0$ modulo almost all primes, then $f$ is linear.

Proof. Assume that $deg(f)>1$. Since the Galois group of $f$ acts transitively on the roots $\Omega$, and for almost all primes $p$, $f \ \mathsf{mod} \ p$ has a zero in $\mathbb{F}_{p}$, almost all Frobenius substitutions fix a root of $f$. Taking $S$ to be the set of all $g \in G$ that fix at least one element of $\Omega$, we see that $$S=\bigcup_{\omega \in \Omega} Stab(\omega)=\bigcup_{g \in G} g \ Stab(x) \ g^{-1},$$ with $x$ a fixed element of $\Omega$. The first equality is trivial; the second one as well, if you remember to use transitivity. Since no finite group is the union of conjugates of a proper subgroup, $G$ has to contain elements that don’t fix a root of $f$. These elements can be the Frobenius subsitution of only finitely many primes, which is impossible by Chebotarëv, which says there should always should be a density, and we know a finite set has density $0$. $\square$

Of course, one would also like a completely algebraic proof of the theorem, not requiring a big cannon like Chebotarëv (the proof of which uses lots of L-functions and complex analysis). The surprising thing is that the only known algebraic proofs make use of none other than the classification of finite simple groups!

Let me just provide you with the glue that relates $\mathbb{F}_{1}$-geometry to recreational number theory (my state of mind upon writing the first post seems not to have been overly optimistic):

Recreational number theory is that branch of number theory which is too difficult for serious study.

By now I’m sure you can guess who it was that uttered this phrase.

Last time on the seminar, we used the Kronecker-Weber Theorem to discuss number theory in $\mathbb{F}_1$. Due to my almost nonexistent knowledge of number theory, I thought it would be a great exercise to work this theorem out. It is anything but trivial and it requires a lot of lemmas and fundamental theorems, which I will write about this time.

The theorem and notations

Theorem (Global Kronecker-Weber Theorem) If $K/\mathbb{Q}$ is a finite abelian Galois extension, then there exists an $n$ such that $K \subseteq \mathbb{Q}(\zeta_n)$, with $\zeta_n$ a primitive root of $1$.

There is also a local version of this theorem, given by replacing $\mathbb{Q}$ by $\mathbb{Q}_p$ for some prime $p$. This theorem has been proved in various ways, the “easiest” one by using class field theory. I will give another proof, based on the book “Introduction to cyclotomic fields” by Lawrence C. Washington. It will use the Hasse principle, which we will prove in the first step of the proof.

Theorem If the local Kronecker-Weber Theorem is true for every prime $p$, then so is the global Kronecker-Weber Theorem.

For the remainder of this series, $p$ is always a prime and $\mathbb{Q}_p$ are the $p$-adic numbers with ring of integers $\mathbb{Z}_p$. For an arbitrary number or local field $K$, one can define $K_\mathfrak{p}$ as the completion of $K$ with respect to a prime ideal $\mathfrak{p}$. For 2 extensions $L_1$ and $L_2$ of a number or local field $K$, we denote the compositum as $L_1L_2$.

Preliminary work

Next proposition is probably one of the most important theorems in number theory and is worth a post itself, but one has to choose a subject and stick to it.

Proposition Let $L/K$ be a finite Galois extension, where $K$ may be a number field or a local field. Let $\mathfrak{p}$ be a prime of $K$. Then we have the following equality $$\mathfrak{p}=\mathfrak{b}_1^e \mathfrak{b}_2^e\ldots \mathfrak{b}_g^e,$$ with $\mathfrak{b}_i$ primes in $L$ lying above $K$. $e$ is called the ramification index. Let $f$ be the degree of the extension $\mathcal{O}_L \bmod \mathfrak{b}_i / \mathcal{O}_K \bmod \mathfrak{p}$. Then we have $$ \left[L:K\right] = efg.$$ $\mathfrak{p}$ is said to be totally ramified in $L$ if $e =n $ and to be unramified if $e=1$. Moreover, for all $1 \leq i<j \leq g$, there exists a $\sigma \in Gal(L/K)$ such that $\sigma(\mathfrak{b}_i) = \mathfrak{b}_j$.

In local fields, it is easy to see that if $K \subset L \subset M$ with ramification indices $e_{LK},e_{ML},e_{MK}$ and with extension degrees of the residue fields $f_{LK},f_{ML},f_{MK}$, then we have $e_{LK}e_{ML} = e_{MK}$ and $f_{LK}f_{ML}=f_{MK}$. In general, one sees that if $K \subset L \subset M$ and a prime $\mathfrak{p}$ of $K$ ramifies in $L$, then it ramifies in $M$.

In case of cyclotomic fields, next theorem says something about ramification.

Theorem A prime $\mathfrak{p}$ of $K$ ramifies in a cyclotomic extension $K(\zeta_n)$ if and only if $p|n$.

One may hope that there are extensions of $\mathbb{Q}$ that aren’t ramified at any prime. Unfortunately, there aren’t any: if $K/\mathbb{Q}$ is unramified at all primes, $K = \mathbb{Q}$. This is easily proved using a theorem of Minkowski (for a proof, see here) and the fact that if a prime divides the discriminant of a number field $K$, it always ramifies (the converse is also true).

A useful theorem is the following, where $\mathbb{Q}$ can be replaced by $\mathbb{Q}_p$ in both this theorem and its corollary:

Theorem Let $K$ and $L$ be finite Galois extensions of $\mathbb{Q}$. Then $Gal(KL/\mathbb{Q}) \cong H$, with $H$ a subgroup of $Gal(K/\mathbb{Q}) \times Gal(L/\mathbb{Q})$ consisting of all elements $(\phi,\psi)$ such that $\phi|_{K \cap L} = \psi|_{K \cap L}$.

Corollary Let $L/\mathbb{Q}$ be an abelian Galois extension, with $$Gal(L/\mathbb{Q}) = \prod_{i=1}^m G_i.$$ Then $$L = \prod_{i=1}^m L^{G_i}.$$

Since all abelian groups are products of cyclic groups of prime order, we can reduce Kronecker-Weber to Galois extensions with Galois group a cyclic group of prime order, since previous corollary gives the general case.

Definition Let $L/K$ be a Galois extension, with $\mathfrak{p}$ a prime of $K$ and $\mathfrak{b}$ a prime lying above $\mathfrak{p}$. Define $$D_\mathfrak{b}=\left\{\sigma \in Gal(L/K)|\sigma(\mathfrak{b}) = \mathfrak{b}\right\}$$ as the decomposition group of $\mathfrak{b}$ with inertia group $$I_\mathfrak{b} = \left\{\sigma \in D_\mathfrak{b}|\sigma(x) \equiv x \bmod \mathfrak{b} \text{ for all } \alpha \in \mathcal{O}_L\right\}.$$

In general, 2 decomposition groups $D_{\mathfrak{b}_1}$ and $D_{\mathfrak{b}_2}$ with $\mathfrak{b}_1$ and $\mathfrak{b}_2$ primes lying above $\mathfrak{p}$, will be conjugated. Moreover, we have: $$ \sigma D_{\mathfrak{b}_1} \sigma^{-1} = D_{\mathfrak{b}_2} \Rightarrow \sigma I_{\mathfrak{b}_1} \sigma^{-1} = I_{\mathfrak{b}_2}.$$
Knowing this, it’s easy to see that if you work in abelian extensions, there is only one decomposition group and one inertia group. In this case, it’s easier to write $I_{\mathfrak{p}}$ instead of $I_{\mathfrak{b}}$, since the inertia group doesn’t depend on $\mathfrak{b}$ anymore.
If you work in a local field $K$ which is an extension of $\mathbb{Q}_p$, you can discard the index $\mathfrak{b}$ and write $I_K$ instead.

The order of $D_\mathfrak{b}$ is $ef$, since the number of elements in the orbit of $\mathfrak{b}$ is $g$.  The order of $I_\mathfrak{b}$ will be $e$ and since $$\{id_L\}\subseteq I_\mathfrak{b} \subseteq D_\mathfrak{b} \subseteq Gal(L/K),$$ we have a tower of extensions $$L\supseteq L^{I_\mathfrak{b}} \supseteq L^{D_\mathfrak{b}} \supseteq K,$$ on which we can use the index calculation $$n = \left[L:K\right] = \left[L:L^{I_\mathfrak{b}}\right]\left[L^{I_\mathfrak{b}}:L^{D_\mathfrak{b}}\right]\left[L^{D_\mathfrak{b}}:K\right]=efg.$$
Furthermore, $\mathfrak{p}$ will be unramified in $L^{I_\mathfrak{b}}$. In case of a unique inertia group, $L^{I_\mathfrak{b}}$ is the largest field between $L$ and $K$ that has this property. Also, $I_\mathfrak{b}$ is always a normal subgroup of $D_\mathfrak{b}$ (not only in case of abelian extensions).

Next four theorems will prove vital for proving the ‘local implies global’ Kronecker-Weber Theorem.

Theorem Let $L/K$ be a Galois extension of number fields, with $\mathfrak{p}$ a prime of $K$ with a prime $\mathfrak{b}$ lying over. Then we have $$Gal(L_\mathfrak{b}/K_\mathfrak{p}) \cong D_\mathfrak{b}$$ and the inertia groups of $\mathfrak{b}$ are isomorphic in both extensions.

Theorem Let $L/K/\mathbb{Q}_p$ be a tower of finite Galois extensions. Then there exists a surjective homomorphism $f:I_L \rightarrow I_K$.

Theorem The inertia group of $\mathbb{Q}_p(\zeta_n)/\mathbb{Q}_p$ is $(\mathbb{Z}/p^e\mathbb{Z})^*$, where $e$ is determined by $p^e|n, p^{e+1} \nmid n.$

Theorem The Galois group of an abelian extension $K/\mathbb{Q}$ is generated by the inertia groups $I_{p}$’s, with $p$ running over all primes of $\mathbb{Q}$.

Local implies global

All these theorem imply the first step in the proof of Kronecker-Weber.

Theorem If the local Kronecker-Weber Theorem is true for every prime $p$, then so is the global Kronecker-Weber Theorem.

Proof Let $K/\mathbb{Q}$ be an abelian extension and $p$ a rational prime that ramifies, with $\mathfrak{b}$ a prime lying above $p$. Take completions $K_\mathfrak{b}$ and $\mathbb{Q}_p$. We know that $Gal(K_\mathfrak{b}/\mathbb{Q}_p) = D_\mathfrak{b}$, so it is necessarily abelian, since it is a subset of $Gal(K/\mathbb{Q})$. Assuming local Kronecker-Weber, we know that $K_\mathfrak{b} \subset \mathbb{Q}_p(\zeta_{n_p})$ for some $n_p$. Now, let $p^{e_p}$ be the exact power of $p$ that divides $n_p$ and let $$n = \prod_{p \text{ ramifies in } K} p^{e_p}.$$ We will now prove that $K \subset \mathbb{Q}(\zeta_n)$.
Take $L = K(\zeta_n)$. The main objective will be to prove that $\left[L:\mathbb{Q}\right] \leq \phi(n)$, since $\left[\mathbb{Q}(\zeta_n):\mathbb{Q}\right] = \phi(n)$ and $\zeta_n \in L$ implies that $\mathbb{Q}(\zeta_n) = L$ and so $K \subset \mathbb{Q}(\zeta_n)$.
We know that a prime $p$ is unramified in $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ if $p$ doesn’t divide $n$, so the primes that ramify in $K$ and $\mathbb{Q}(\zeta_n)$ are the same. So $L$ and $K$ have the same ramifying primes. Take a prime $p$ of $\mathbb{Q}$ that ramifies in $L$, with a prime $\mathfrak{b}$ lying over it. By one of the above theorems, we can calculate $I_{\mathfrak{b}}$ locally. Since we are working in abelian extensions and consequently there is only one inertia group, we denote $I_p$ instead of $I_{\mathfrak{b}}$. Taking completions and using that $K_\mathfrak{b} \subset \mathbb{Q}_p(\zeta_{n_p})$, we get $$K_\mathfrak{b}(\zeta_n) \subset \mathbb{Q}_p(\zeta_{n_p},\zeta_n) = \mathbb{Q}_p(\zeta_m),$$ with $m = lcm(n_p,n)$. Since the highest power of $p$ that divides both $m$ and $n$ is $e_p$, we have that the inertia groups of both $\mathbb{Q}_p(\zeta_m)$ and $\mathbb{Q}_p(\zeta_n)$ are $(\mathbb{Z}/p^{e_p}\mathbb{Z})^*$. Since $\mathbb{Q}_p(\zeta_n) \subset K_\mathfrak{b}(\zeta_n) \subset \mathbb{Q}_p(\zeta_m)$, we have that $I_p = (\mathbb{Z}/p^{e_p}\mathbb{Z})^*$ and so $|I_p| = \phi(p^{e_p})$. Since all the $I_p$ generate $Gal(L/\mathbb{Q})$ and we work in abelian extensions, we get $$|Gal(L/\mathbb{Q})|\leq \prod_{p \text{ ramifies}} |I_p|  = \phi(n).$$ So $\left[L:\mathbb{Q}\right] \leq \phi(n)$ and we are done !

Next time, I’ll post (some of) the proof for the local Kronecker-Weber Theorem.