Let k be a finite field of cardinality q.
Examples 6.1. Let F be a global field of positive characteristic, with constant field k. Let K be any other field containing k.
(a) Let K denote the bold ring consisting of K equipped with the k-linear Frobe-nius endomorphismσof K, mapping λ ∈ K toσ(λ) := λq. This is a bold field, and K itself is a base field of K.
(b) Set FK := Frac(F ⊗k K). The endomorphism σ of K induces an injective endomorphism id⊗σ of F ⊗k K, which extends uniquely to an endomor-phism, again denoted asσ, of FK. Then FK := (FK, σ) is a bold field: By our assumptions, FKis a field. Clearly,σis injective and (FK)σ =F, so the remaining conditions are fulfilled.
(c) For any placePof FK, letO(P),K ⊂ FK be the valuation ring corresponding toP. In general, this is not a bold subring of FK. However, for any place p of F, the intersection O(p),K of the rings O(P),K for all places P of FK lying overpis σ-stable. Since these primes Pare finite in number, O(p),K
is a semilocal ring. Clearly, O(p),Kσ = O(p), the local ring of F at p, and F ⊗O(p) O(p),K → FK is an isomorphism. All in all, the bold ring O(p),K is a local bold ring, and it is a bold order of the bold field FK. Furthermore, K ⊂O(p),K is a base field.
(d) For a finite non-empty set{∞1, . . . ,∞s}of places of F, let A denote the ring of elements of F integral outside the∞i, and set AK := A⊗k K. This is a Dedekind ring, and equipped with id⊗σ, we obtain the global bold ring AK. It is a bold order of the bold field FK. Furthermore, K ⊂ AK is a base field.
For every maximal primepof A, the bold ringO(p),K is a bold place of AK. For the remaining examples, we need a calculation. Fix a separable closure ksep of k. For every n ≥ 1 let kn denote the subfield of ksep of degree n over k.
Sometimes, k∞ will denote ksep. For every field K ⊃ k and element x ∈ K we set σk(x) := xqand K := (K, σk).
Proposition 6.2. For m,n ≥ 1 let δ := gcd(m,n) andµ := lcm(m,n). The two following maps are ring isomorphisms:
ix : km⊗kkn −→kµ×δ , x⊗y7→
σik(x)·yδ−1
i=0 , and iy : km⊗kkn −→kµ×δ , x⊗y7→
x·σik(y)δ−1 i=0 .
We start by checking that the given homomorphism of bold rings is an isomor-phism in two special cases.
Lemma 6.3. Ifδ=1, the homomorphism
iy : km⊗kkn →kmn, x⊗y7→ xy is an isomorphism of rings.
Proof. Since both sides are finite, the given homomorphism is bijective if it is in-jective. For this, it suffices to show that any k-linearly independent set of elements x1, . . . ,xr∈kmremains kn-linearly independent in kmn. If not, choose a counterex-amplePr
i=1xiyi = 0 in kmnwith yi ∈kn and r ≥ 1 minimal. We may assume that yr = 1. It follows thatP
ixiσm(yi)= 0, so by subtractionPr−1
i=1 xi(σm(yi)−yi) =0.
By minimality of r, we deduce σm(yi) = yi for all i. So yi ∈ km ∩kn = k, a
contradiction. ∴
Lemma 6.4. Ifδ= n, the homomorphism
iy : km⊗kkn →km×n, x⊗y7→(xσi(y))n−1i=0 is an isomorphism of rings.
Proof. Again, the given map is an isomorphism if it is injective. We must show that if given y1, . . .yr ∈ kn are k-linearly independent, then the set of vectors {(σi(yj))δ−1i=0}rj=1is km-linearly independent. If not, there exist x1, . . . ,xr∈kmwith
r
X
j=1
xjσi(yj)=0 for all 0≤i< n.
We may assume that r ≥ 1 is minimal, and that xr = 1. Applying σ to these equations, and using thatσnis the identity on kn, we deduce that
r
X
j=1
σ(xj)σi(yj)=0 for all 0≤i< n.
Hence we find that Xr−1
j=1
(σ(xj)− xj)σi(yj)=0 for all 0≤i<n.
By minimality of r, we find that all xj lie in k. So the i = 0 case of the original equation shows that the yjare linearly dependent, a contradiction. ∴ Proof of Proposition 6.2. The given homomorphism iy : km⊗k kn →kµδcoincides with the following composite isomorphism:
im,n: km⊗kkn km⊗kδ(kδ⊗kkn)−−−−−−−−−→Lemma 6.4 km⊗kδkδn (km⊗kδkn)δ −−−−−−−−−→Lemma 6.3 kδµ. Therefore it is an isomorphism of rings. The proof that ix is an isomorphism of
rings is symmetrical. ∴
Proposition 6.5. For m,n ≥ 1 let δ := gcd(m,n) and µ := lcm(m,n). Choose integers a,b such that am+ bn = δ. We consider two ring endomomorphisms σx, σy of k×δµ defined as follows.
For z=(z0, . . . ,zd)∈k×δµ , set σx(z)i :=
( zi+1, 0≤ i< δ−1 σbnk (z0), i= δ−1 and
σy(z)i :=
( zi+1, 0≤ i< δ−1 σamk (z0), i= δ−1
Then ix induces an isomorphism of bold rings km⊗k kn → (k×δµ , σx), and iy induces an isomorphism of bold rings km⊗k kn→ (k×δµ , σy).
In particular, km⊗k knis a bold field.
Proof. We start by remarking that kµ×δequipped with eitherσxorσyis a bold field.
By Proposition 6.2, iy is an isomorphism of rings. It remains to check that iy is σ-equivariant. It suffices to check that iy◦(id⊗σk) = σy◦iy on elements of the form x⊗y∈km⊗kkn. We have
iy(id⊗σk(x⊗y))= (xσik+1(y))δ−1i=0 and
(σy(iy(x⊗y))i =
( xσik+1(y) 0≤i≤δ−2 σamk (xy) i=δ−1
We have equality for the first δ − 1 components. The calculation σamk (xy) = σamk (x)σδ−bnk (x)= xσδk(y) shows that the last components also coincide. The proof
that ixisσ-equivariant is symmetrical. ∴
Corollary 6.6. For z= (z0, . . . ,zd−1) ∈k∞×d setσ0(z)i := zi+1. Then iy induces an isomorphism k∞⊗kkd →(k∞×d, σ0), whereas ixinduces an isomorphism kd⊗kk∞→ (k×d∞, σ0) of bold fields.
Proof. We have k∞⊗kd =S
d|mkm⊗kkd. By Proposition 6.5, iyis an isomorphism km⊗k kd (kdm, σ0). It follows that iy gives an isomorphism
k∞⊗ kd [
d|m
(kdm, σ0)= (k×d∞, σ0).
The case of kd⊗ k∞is symmetrical. ∴
Remark 6.7. By Proposition 6.5 and Corollary 6.6 we now know that, for 1 ≤ m,n≤ ∞with either m<∞or n< ∞, the bold ring km⊗k knis a bold field.
To leave the realm of finite fields more substantially, we quote the following results of [Jac90].
Proposition 6.8. Consider two field extensions E1,E2 of k, and assume that k is algebraically closed in E1(i.e.: every element of E1rk is transcendental over k).
(a) The tensor product E1⊗k E2is a domain.
(b) If E2is a finite extension of k, then E1⊗k E2is a field.
Proof. [Jac90, Theorem 8.50] gives item (a), and [Jac90, Theorem 8.46(2)] gives
item (b). ∴
Corollary 6.9. For every field K ⊃ k and every d ≥ 1, the bold ring (kd ⊗k K,id⊗σk) is a bold field. If K contains a copy of kd, then this bold field is isomor-phic to (K×d, σ00), whereσ00(z)i :=σk(zi+1) for z=(z0, . . . ,zd−1)∈K×d.
Proof. Let kK denote the algebraic closure of k in K. By Proposition 6.5 (and its Corollary 6.6 in case kKis infinite) the bold ring (kd⊗kkK,id⊗σk) is isomorphic to (kδµ, σx), for certain 1 ≤ µ≤ ∞andδ |d. Set r := [kµ : kK] < ∞. By Proposition 6.8(b), the ring Kr :=kµ⊗kK K is a field. It follows Kris a finite field extension of K of degree r. Therefore, we have
kd⊗k K (kµ×δ, σx)⊗kK K (Krδ, σx◦σk),
whereσx ◦σk is given by first applyingσk componentwise, and thenσx. This is indeed a bold field.
If K contains k∞, thenδ =d and r = 1, so Kr×δ = K×d, as required. Moreover, one checks that in this caseσx◦σkcoincides with the endomorphismσ00given in
the statement of this corollary. ∴
Corollary 6.10. Consider two field extension F,K of k. If either F or K contains only a finite number of roots of unity, then FK := (Frac(F⊗k K),Frac(id⊗σk)) is a bold field.
Proof. Abusing notation a little, we set kF := ksep∩ F and kK := ksep ∩K, the respective algebraic closures of k in F and K. Now (kF ⊗k kK,id⊗σk) is a bold field by Proposition 6.5 and Corollary 6.6, for certain 1 ≤µ≤ ∞and 1≤ δ <∞. In particular,
F⊗k K = F⊗kF (kF ⊗k kK)⊗kK K (F⊗kF kµ⊗kK K)×δ.
To show that FKis a bold field, it is sufficient to show that Frac(F⊗kF kµ⊗kK K) is a field, which follows if we show that F⊗kF kµ⊗kK K is a domain.
We do this in the case where F has a finite number of roots of unity, i.e. that kF is finite; the other case is symmetrical. Applying Proposition 6.8(b) to k0 =kF, E1 =F and E2 =kµ shows that F⊗kF kµ is a field. So applying Proposition 6.8(a) to k0 :=kF, E1:= K and E2 :=F⊗kFkµshows that F⊗kFkµ⊗kKK is a domain. ∴
We may now introduce further bold fields.
Examples 6.11. We continue to use the notation given in Examples 6.1.
(a) The bold ringOK,pis defined as lim
←−−nO(p),K/pn, the “completion atp” of the bold ringO(p),K. Let kp = O(p)/pand choose a local parameter t∈O(p)atp.
By the Chinese Remainder theorem we have an isomorphism OK,p ∼
−−−→(kp⊗k K)[[t]].
The σ of OK,p induces a unique endomorphism of the right hand side: It acts as the identity on t, and as id⊗σk on elements of kp⊗k K. Now kp⊗k K decomposes as finite direct product of the pairwise-isomorphic fields KP = O(P),K/Pfor those placesPof FK lying abovep, and equipped with id⊗σk
it is a bold field (Corollary 6.9). We haveOK,pσ = OFp, the valuation ring of Fp. All in all,OK,pis a local bold ring. The subfield K ⊂OK,pis a base field.
(b) Set FK,p := Frac(OK,p) and let FK,p be this ring equipped with the unique extension ofσ. By the preceding, we may identify FK,pwith
(kp⊗k K)((t)) :=(kp⊗k K)[[t]][t−1].
We have FσK,p = Fp. Again using (a), we see that FK,p is a bold field, with bold orderOK,p.
(c) Let p be a place of F, and denote by Fp the completion of F at p. Then Fp,K := (Frac(Fp ⊗k K),Frac(id⊗σk)) is a bold field by Corollary 6.10.
Clearly, Fp,K ⊂ FK,p, but it is fundamental to note that this inclusion is strict except if K is finite. The main question in this context is how we can characterize this inclusion; we shall come back to this in Chapter V.
(d) SetOp,K := Fp,K∩OK,p. This a global bold ring. It is a bold order of Fp,K and has K ⊂ Op,K as base field. Clearly, Op,Kσ = OFp. We note that we have inclusionsOFp⊗kK ⊂ Op,K ⊂ OK,p, but these inclusions are strict in general.
Let us review the most important rings for the following chapters by means of a diagram, in the case where F is a global field with field of constants K, and K is a field extension of K. Let∞,pbe two different places of F, and let A be the ring of elements of F integral outside∞. Then we have inclusions
AK //O(p),K //
Op,K //
OK,p
FK //Fp,K //FK,p
where the upper row consists of “integral” rings, whereas the lower row consists of “rational” rings. The corresponding diagram of scalar rings is
A //O(p) //
Op //
Op
F //Fp //Fp