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INTRODUCTION TO ARITHMETIC GEOMETRY
(NOTES FROM 18.782, FALL 2009)
BJORN POONEN
- What is arithmetic geometry? Contents
- Absolute values on fields
- The p-adic absolute value on Q
- Ostrowski’s classification of absolute values on Q
- Cauchy sequences and completion
- Inverse limits
- Defining Zp as an inverse limit
- Properties of Zp
- The field of p-adic numbers
- p-adic expansions
- Solutions to polynomial equations
- Hensel’s lemma
- Structure of Q× p
- Squares in Q× p
- 14.1. The case of odd p
- 14.2. The case p =
- p-adic analytic functions
- Algebraic closure
- Finite fields
- Inverse limits in general
- Profinite groups
- 19.1. Order
- 19.2. Topology on a profinite group
- 19.3. Subgroups
- Review of field theory
- Infinite Galois theory
- 21.1. Examples of Galois groups
- Affine varieties
- 22.1. Affine space
- 22.2. Affine varieties
- 22.3. Irreducible varieties
- 22.4. Dimension
- 22.5. Smooth varieties
- Projective varieties
- 23.1. Motivation
- 23.2. Projective space
- 23.3. Projective varieties
- 23.4. Projective varieties as a union of affine varieties
- Morphisms and rational maps
- Quadratic forms
- 25.1. Equivalence of quadratic forms
- 25.2. Numbers represented by quadratic forms
- Local-global principle for quadratic forms
- 26.1. Proof of the Hasse-Minkowski theorem for quadratic forms in 2 or 3 variables
- Rational points on conics
- Sums of three squares
- Valuations on the function field of a curve
- 29.1. Closed points
- Review
- Curves and function fields
- Divisors
- 32.1. Degree of a divisor
- 32.2. Base extension
- 32.3. Principal divisors
- 32.4. Linear equivalence and the Picard group
- Genus
- 33.1. Newton polygons of two-variable polynomials
- Riemann-Roch theorem
- Weierstrass equations
- Elliptic curves
- Group law
- 37.1. Chord-tangent description
- 37.2. Torsion points
- Mordell’s theorem
- The weak Mordell-Weil theorem
- Height of a rational number
(Abs3) ‖x + y‖ ≤ ‖x‖ + ‖y‖ (“triangle inequality”)
Examples:
- R with the usual | |
- C with the usual | | (or any subfield of this)
- any field k with
‖x‖ :=
1 , if x 6 = 0 0 , if x = 0. This is called the trivial absolute value.
Definition 2.2. An absolute value ‖ ‖ satisfying
(Abs3′) ‖x + y‖ ≤ max(‖x‖, ‖y‖) (“nonarchimedean triangle inequality”)
is said to be nonarchimedean. Otherwise it is said to be archimedean.
(Abs3′) is more restrictive than (Abs3), since max(‖x‖, ‖y‖) ≤ ‖x‖ + ‖y‖. (Abs3′) is strange from the point of view of classical analysis: it says that if you add many copies of a “small” number, you will never get a “large” number, no matter how many copies you use. This is what gives p-adic analysis its strange flavor. Of the absolute values considered so far, only the trivial absolute value is nonarchimedean. But we will construct others soon. In fact, most absolute values are nonarchimedean!
- The p-adic absolute value on Q The fundamental theorem of arithmetic (for integers) implies that every nonzero rational number x can be factored as
x = u
p
pnp^ = u 2 n^2 3 n^3 5 n^5 · · ·
where u ∈ { 1 , − 1 }, and np ∈ Z for each prime p, and np = 0 for almost all p (so that all but finitely many factors in the product are 1, making it a finite product).
Definition 3.1. Fix a prime p. The p-adic valuation is the function
vp : Q×^ → Z x 7 → vp(x) := np,
that gives the exponent of p in the factorization of a nonzero rational number x. If x = 0, then by convention, vp(0) := +∞. Sometimes the function is called ordp instead of vp.
Another way of saying the definition: If x is a nonzero rational number, it can be written in the form pn^ r s
, where r and s are integers not divisible by p, and n ∈ Z, and then vp(x) := n.
Example 3.2. We have v 2 (5/24) = −3, since 5/24 = 2−3 5 3 = 2−^33 −^151.
Properties:
(Val1) vp(x) = +∞ if and only if x = 0 (Val2) vp(xy) = vp(x) + vp(y) (Val3) vp(x + y) ≥ min(vp(x), vp(y))
These hold even when x or y is 0, as long as one uses reasonable conventions for +∞, namely:
- (+∞) + a = +∞
- +∞ ≥ a
- min(+∞, a) = a
for any a, including a = +∞. Property (Val2) says that if we disregard the input 0, then vp is a homomorphism from the multiplicative group Q×^ to the additive group Z.
Proof of (Val3). The cases where x = 0 or y = 0 or x + y = 0 are easy, so assume that x, y, and x + y are all nonzero. Write
x = pn^ rs (and) y = pm^ uv
with r, s, u, v not divisible by p, so vp(x) = n and vp(y) = m. Without loss of generality, assume that n ≤ m. Then
x + y = pn
r s
m−nu v
= pn^ sv N.
Here sv is not divisible by p, but N might be so N might contribute some extra factors of p. Thus all we can say is that
vp(x + y) ≥ n = min(n, m) = min(vp(x), vp(y)).
Definition 3.3. Fix a prime p. The p-adic absolute value of a rational number x is defined by |x|p := p−vp(x).
If x = 0 (i.e., vp(x) = +∞), then we interpret this as | 0 |p := 0.
Properties (Val1), (Val2), (Val3) for vp are equivalent to properties (Abs1), (Abs2), (Abs3′) for | |p. In particular, | |p really is an absolute value on Q.
- Ostrowski’s classification of absolute values on Q On Q we now have absolute values | | 2 , | | 3 , | | 5 ,... , and the usual absolute value | |, which is also denoted | |∞, for reasons having to do with an analogy with function fields that we will not discuss now. Ostrowski’s theorem says that these are essentially all of them.
so
‖n‖ ≥ ‖bs+1‖ − ‖bs+1^ − n‖ = b(s+1)α^ − ‖bs+1^ − n‖ (since ‖b‖ = bα) ≥ b(s+1)α^ − (bs+1^ − n)α^ (by the previous paragraph) ≥ b(s+1)α^ − (bs+1^ − bs)α^ (since bs^ ≤ n < bs+1)
= b(s+1)α
[
b
)α]
= (bn)α
[
1 − (^1) b
)α]
= cnα,
where c is a positive real number independent of n. This inequality, ‖n‖ ≥ cnα^ holds for all positive integers n, so as before, we may substitute n = nN^ , take N th^ roots, and take the limit as N → ∞ to deduce ‖n‖ ≥ nα. Combining the previous two paragraphs yields ‖n‖ = nα^ for any positive integer n. If m is another positive integer, then
‖n‖ · ‖m/n‖ = ‖m‖ ‖m/n‖ = ‖m‖/‖n‖ = mα/nα^ = (m/n)α.
Thus ‖q‖ = qα^ for every positive rational number. Finally, if q is a positive rational number, then ‖ − q‖ = ‖ − 1 ‖ · ‖q‖ = qα^ = | − q|α
so ‖x‖ = |x|α^ holds for all x ∈ Q (including 0). Case 2: ‖b‖ = 1 for all positive integers b. Then as in the previous paragraph, the axioms of absolute values imply that ‖x‖ = 1 for all x ∈ Q×, contradicting the assumption that ‖ ‖ is a nontrivial absolute value. Case 3: ‖n‖ ≤ 1 for all positive integers n, and there exists a positive integer b such that ‖b‖ < 1. Assume that b is the smallest such integer. If it were possible to write b = rs for some smaller positive integers r and s, then ‖r| = 1 and ‖s‖ = 1 by definition of b, but then ‖b‖ = ‖r‖ · ‖s‖ = 1, a contradiction; thus b is a prime p. We prove (by contradiction) that p is the only prime satisfying ‖p‖ < 1. Suppose that q were another such prime. For any positive integer N , the integers pN^ and qN^ are relatively prime, so there exist integers u, v such that
upN^ + vqN^ = 1,
and then
1 = ‖ 1 ‖ = ‖upN^ + vqN^ ‖ ≤ ‖u‖ · ‖p‖N^ + ‖v| · ‖q‖N ≤ ‖p‖N^ + ‖q‖N^.
This is a contradiction if N is large enough. So ‖q‖ = 1 for every prime q 6 = p. Since 0 < ‖p‖ < 1 and 0 < |p|p < 1, there exists a positive real number α such that ‖p‖ = |p|αp. Now, for any nonzero rational number
x = ±
primes q including p
qnq
property (Abs2) (and ‖ − 1 ‖ = 1) imply
‖x‖ =
primes q including p
‖q‖nq^ = ‖p‖np
since all the other factors are 1. Since ‖p‖ = |p|αp , this becomes
‖x‖ = |p|np pα= |x|αp.
- Cauchy sequences and completion Let k be a field equipped with an absolute value ‖ ‖.
Definition 5.1. A sequence (ai) in k converges if there exists ∈ k such that for every > 0, the terms ai are eventually within of: i.e., for every > 0, there exists a positive integer N such that for all i ≥ N , the distance bound ‖ai − ‖ < holds. In this case, is called the limit of the sequence.
Equivalently (ai) converges to if and only if ‖ai −‖ → 0 as i → ∞. The limit is unique if it exists: if (ai) converges to both and′, then
‖′^ −‖ ≤ ‖ai − ‖ + ‖ai −‖ → 0 + 0 = 0,
so ‖′^ −‖ = 0, so ′^ =.
Definition 5.2. A sequence (ai) in k is a Cauchy sequence if for every > 0, the terms are eventually within of each other; i.e., for every > 0, there exists a positive integer N such that for all i, j ≥ N , the distance bound ‖ai − aj ‖ < holds.
Proposition 5.3. If a sequence converges, it is a Cauchy sequence.
Proof. Use the triangle inequality.
Unfortunately, the converse can fail.
Example 5.7. The completion of Q with respect to the usual absolute value | | is the field R of real numbers.
Proposition 5.8. Let k be a subfield of a complete field L. Then
(1) The inclusion k ↪→ L extends to an embedding ˆk ↪→ L. (2) If every element of L is a limit of a sequence in k, then the embedding ˆk ↪→ L is an isomorphism.
Proof. (1) Given an element a ∈ ˆk, represented as the limit of (ai) with ai ∈ k, map a to the limit of (ai) in L. This defines a ring homomorphism ˆk → L, which is automatically injective since these are fields. (2) Suppose that every element of L is a limit of a sequence in k. Given ∈ L, choose a sequence (ai) in k converging to. Then (ai) is Cauchy, so it also converges to an element a ∈ ˆk. This a maps to `, by definition of the embedding. So the embedding is surjective as well as injective; hence it is an isomorphism.
- Inverse limits
Definition 6.1. An inverse system of sets is an infinite sequence of sets (An) with maps between them as follows:
· · · → An+1 →^ fn An → · · · →f^1 A 1 →^ f^0 A 0.
Definition 6.2. The inverse limit A = lim ←− An of an inverse system of sets (An), (fn) as above is the set A whose elements are the infinite sequences (an) with an ∈ An for each n ≥ 0 satisfying the compatibility condition fn(an+1) = an for each n ≥ 0. It comes with a projection map n : A → An that takes the nth^ term in the sequence.
Remark 6.3. If the An are groups and the fn are group homomorphisms, then the inverse limit A has the structure of a group: multiply sequences term-by-term. If the An are rings and the fn are ring homomorphisms, then the inverse limit A has the structure of a ring.
- Defining Zp as an inverse limit Fix a prime p. Let An be the ring Z/pnZ. Let fn be the ring homomorphism sending ¯b := b + pn+1Z to ¯b := b + pnZ. The ring of p-adic integers is Zp := lim ←− An. For example, if p = 3, then a sequence like
0 mod 1, 2 mod 3, 5 mod 9, 23 mod 27, · · ·
defines an element of Z 3.
- Properties of Zp Recall that a sequence of group homomorphism is exact if at the group in each position, the kernel of the outgoing arrow equals the image of the incoming arrow. For example,
0 → A →f B →g C → 0
is called a short exact sequence if f is injective, g is surjective, and g induces an isomorphism from B/A (or more precisely, B/f (A)) to C.
Proposition 8.1. For each m ≥ 0 ,
0 → Zpp
m → Zp →m Z/pmZ → 0
is exact. (Here the first map is the multiplication-by-pm^ map, sending (an)n≥ 0 to (pman)n≥ 0 ., and m maps (an)n≥ 0 to am.)
Proof. First let us check that multiplication-by-p on Zp is injective. Suppose that a = (an) ∈ Zp is in the kernel. Then pa = 0, so pan = 0 in Z/pnZ for all n. In particular, pan+1 = 0 in Z/pn+1Z. That means that an+1 = pnyn+1 for some yn+1 ∈ Z/pn+1Z. But then an = fn(an+1) = pnfn(yn+1) = 0 in Z/pnZ. This holds for all n, so a = 0. Exactness on the left: Since multiplication-by-p is injective, composing this with itself m times shows that multiplication-by-pm^ is injective. Exactness on the right: Given an element β ∈ Z/pmZ, choose an integer b that represents β. Then the constant sequence b represents an element of Zp mapping to β. Exactness in the middle: If a ∈ Zp, then m(pma) = pm(a) = 0 in Z/pmZ. Thus the image of the incoming arrow (multiplication-by-pm) is contained in the kernel of the outgoing arrow (m). Conversely, suppose that x = (xn) is in the kernel of m. So xm = 0. Then for all n ≥ m, we have xn ∈ p pmnZZ. So there is a unique yn−m mapping to xn via the isomorphism
Z pn−mZ −→^ pm pmZ pnZ.
These yn−m are compatible (because the xn are), so as n ranges through integers ≥ m, they form an element y ∈ Zp such that pmy = x. So x is in the image of multiplication-by-pm.
Proposition 8.2.
(1) An element of Zp is a unit if and only if it is not divisible by p. In other words, the group of p-adic units Z× p equals Zp − pZp. (2) Every nonzero a ∈ Zp can be uniquely expressed as pnu with n ∈ Z≥ 0 and u ∈ Z× p.
Proof.
of the subsequence Sn. Finally, a Cauchy sequence with a convergent subsequence converges. (2) Let a ∈ Qp. By multiplying by a suitable power of p, we reduce to the case where a ∈ Zp. Write a = (an) with an ∈ Z/pnZ. Choose an integer bn ∈ Z representing an. Then vp(a − bn) ≥ n, so |a − bn| ≤ p−n, so the sequence (bn) converges to a in Qp.
Combining Propositions 5.8 and 9.2 shows that Qp is the completion of Q with respect to | |p.
- p-adic expansions
Definition 10.1. Say that a series
n=1 an^ of^ p-adic numbers^ converges^ if and only if the sequence of partial sums converges with respect to | |p.
Theorem 10.2.
(1) Each a ∈ Zp has a unique expansion a = b 0 +b 1 p+b 2 p^2 +· · · with bn ∈ { 0 , 1 ,... , p− 1 } for all n. (2) Each a ∈ Qp has a unique expansion a =
n∈Z bnpn^ in which^ bn^ ∈ {^0 ,^1 ,... , p^ −^1 } and bn = 0 for all sufficiently negative n. (3) For either expansion, vp(a) is the least integer n such that bn 6 = 0. (If no such n exists, then a = 0 and vp(a) = +∞.)
Proof.
(1) Existence: Write a = (an) with an ∈ Z/pnZ. Choose sn ∈ { 0 , 1 ,... , pn^ − 1 } repre- senting an. Write sn = b 0 + b 1 p + · · · + bn− 1 pn−^1 with bi ∈ { 0 , 1 ,... , p − 1 }. The compatibility condition on the an implies that the bi so defined are independent of n; i.e., the base-p expansion of sn+1 extends the base-p expansion of sn by one term bnpn. Then sn → a in Qp, so
b 0 + b 1 p + b 2 p^2 + · · · = a.
Uniqueness: If b′ n ∈ { 0 , 1 ,... , p − 1 } also satisfy
b′ 0 + b′ 1 p + b′ 2 p^2 + · · · = a,
then we get
b 0 + b 1 p + · · · + bn− 1 pn−^1 ≡ b′ 0 + b′ 1 p + · · · + b′ n− 1 pn−^1 (mod pn),
but both sides are integers in { 0 , 1 ,... , pn^ − 1 }, so they are equal, and this forces bi = b′ i for all i.
(2) Existence: If a ∈ Qp, then there exists m ∈ Z such that pma ∈ Zp. Write
pma = b 0 + b 1 p + b 2 p^2 + · · ·
with bi ∈ { 0 , 1 ,... , p − 1 } and divide by pm. Uniqueness: Follows from uniqueness for Zp. (3) If a = b 0 + b 1 p + · · · ∈ Zp with bi ∈ { 0 , 1 ,... , p − 1 }, and b 0 6 = 0, then a has nonzero image in Z/pZ, so a is a unit, and vp(a) = 0. The general case follows from this one by multiplying by pn^ for an arbitrary n ∈ Z.
- Solutions to polynomial equations
Lemma 11.1 (“Compactness argument”). Let · · · → S 2 → S 1 → S 0 be an inverse system of finite nonempty sets. Then lim ←− Si is nonempty.
Proof. Let Ti, 0 be the image of Si → · · · → S 0. Then
· · · ⊆ T 2 , 0 ⊆ T 1 , 0 ⊆ T 0 , 0 ,
but these are finite nonempty sets, so Ti, 0 must be constant for sufficiently large i. Let E 0 be this “eventual image”. Define Ti, 1 and E 1 in the same way, and define E 2 , and so on. Then the Ei form an inverse system in which the maps Ei+1 → Ei are surjective. Choose e 0 ∈ E 0 , choose a preimage e 1 ∈ E 1 of e 0 , choose a preimage e 2 ∈ E 2 of e 1 , and so on: this defines an element of lim ←− Si.
Proposition 11.2. Let f ∈ Zp[x] be a polynomial. Then the following are equivalent:
(1) The equation f (x) = 0 has a solution in Zp. (2) The equation f (x) = 0 has a solution in Z/pnZ for every n ≥ 0.
Proof. Let Sn be the set of solutions in Z/pnZ. Then lim ←− Sn ⊆ lim ←− Z/pnZ = Zp is the set of solutions in Zp. We have lim ←− Sn 6 = ∅ if and only if all the Sn are nonempty, by Lemma 11.1.
- Hensel’s lemma Hensel’s lemma says that approximate zeros of polynomials can be improved to exact zeros.
Theorem 12.1 (Hensel’s lemma). Let f ∈ Zp[x]. Suppose that f (a) ≡ 0 (mod p), and f ′(a) 6 ≡ 0 (mod p). (That is, a is a simple root of (f mod p).) Then there exists a unique b ∈ Zp with b ≡ a (mod p) such that f (b) = 0.
Lemma 13.1. The quotients in the filtration are:
(1) Z× p /U 1 ' F× p , and (2) Un/Un+1 ' Z/pZ for all n ≥ 1.
Proof. The first of these has already been proved. For the second, observe that
Un → Z/pZ 1 + pnz 7 → (z mod p)
is surjective and has kernel Un+1.
Corollary 13.2. The order of U 1 /Un is pn−^1.
Proposition 13.3. Let μp− 1 be the set of solutions to xp−^1 = 1 in Z× p. Then μp− 1 is a group (under multiplication) mapping isomorphically to F× p , and Z× p = U 1 × μp− 1.
Proof. The set μp− 1 is the kernel of the (p − 1)th^ power map from Z× p to itself, so it is a group. Given a ∈ { 1 , 2 ,... , p − 1 }, Hensel’s lemma shows that μp− 1 contains a unique p-adic integer congruent to a modulo p. And there are no elements of μp− 1 congruent to 0 mod p. So reduction modulo p induces an isomorphism μp− 1 → F× p. We have U 1 ∩ μp− 1 = { 1 } (by Hensel’s lemma, there is only one solution to xp−^1 − 1 = 0 congruent to 1 modulo p). Also, U 1 · μp− 1 = Z× p , since any a ∈ Z× p can be divided by an element of μp− 1 congruent to a modulo p to land in U 1. Thus the direct product U 1 × μp− 1 is equal to Z× p.
Lemma 13.4. Let p be a prime. If p 6 = 2, let n ≥ 1 ; if p = 2, let n ≥ 2. If x ∈ Un − Un+1, then xp^ = Un+1 − Un+2.
Proof. We have x = 1 + kpn^ for some k not divisible by p. Then
xp^ = 1 +
p 1
kpn^ +
p 2
k^2 p^2 n^ + · · · + kppnp
≡ 1 + kpn+1^ (mod pn+2).
so xp^ ∈ Un+1 − Un+2.
Proposition 13.5. If p 6 = 2, then U 1 ' Zp. If p = 2, then U 1 = {± 1 } × U 2 and U 2 ' Z 2.
Proof. Suppose then p 6 = 2. Let α = 1 + p ∈ U 1 − U 2. By the previous lemma applied repeatedly, αpi ∈ Ui+1 − Ui+2. Let αn be the image of α in U 1 /Un. Then α npn −^26 = 1 but α npn −^1 = 1, so αn has exact order pn−^1. On the other hand, the group it belongs to, U 1 /Un, also has order pn−^1. So U 1 /Un is cyclic, generated by αn. We have an isomorphism of inverse
systems · · · //Z/pnZ //
Z/pn−^1 Z //
· · · //U 1 /Un+1 //U 1 /Un //· · ·
Taking inverse limits shows that Zp ' U 1. For p = 2, the same argument with α = 1 + 4 works to prove that Z 2 ' U 2. Now {± 1 } and U 2 have trivial intersection, and they generate U 1 (since U 2 has index 2 in U 1 ), so the direct product {± 1 } × U 2 equals U 1.
Theorem 13.6.
(1) The group Z× p is isomorphic to Z/(p − 1)Z × Zp if p 6 = 2, and to Z/ 2 Z × Z 2 if p = 2. (2) The group Q× p is isomorphic to Z × Z/(p − 1)Z × Zp if p 6 = 2, and to Z × Z/ 2 Z × Z 2 if p = 2.
Proof.
(1) Combine Propositions 13.3 and 13.5. (2) The map Z × Z× p → Q× p (n, u) 7 → pnu is an isomorphism of groups. Now substitute the known structure of Z× p into this.
- Squares in Q× p
14.1. The case of odd p.
Theorem 14.1.
(1) An element pnu ∈ Q× p (with n ∈ Z and u ∈ Z× p ) is a square if and only if n is even and u mod p is a square in F× p. (2) We have Q× p /Q× p 2 ' (Z/ 2 Z)^2. (3) For any c ∈ Z× p with c mod p /∈ F× p 2 , the images of p and c generate Q× p /Q× p 2.
Proof.
(1) We have Q× p = pZ^ × F× p × Zp, and 2Zp = Zp, so Q× p 2 = p^2 Z^ × F× p 2 × Zp. Thus an element pnu is a square if and only if n is even and u mod p ∈ F× p 2. (2) Using the same decomposition, Q× p /Q× p 2 = (Z/ 2 Z) × (F× p /F× p 2 ) × { 0 } ' (Z/ 2 Z)^2 since F× p is cyclic of even order.
power series in either order gives z implies that the corresponding p-adic analytic functions are inverses of each other when both converge.
- Algebraic closure Given a field k, let k[x]≥ 1 be the set of polynomials in k[x] of degree at least 1.
Definition 16.1. A field k is algebraically closed if and only if every f ∈ k[x]≥ 1 has a zero in k.
Definition 16.2. An algebraic closure of a field k is a algebraic field extension k of k that is algebraically closed.
Example 16.3. The field C of complex numbers is an algebraic closure of R. But C is not an algebraic closure of Q because some elements of C (like e and π) are not algebraic over Q.
Theorem 16.4. Every field k has an algebraic closure, and any two algebraic closures of k are isomorphic over k (but the isomorphism is not necessarily unique).
Step 1: Given f ∈ k[x]≥ 1 , there exists a field extension E ⊇ k in which f has a zero.
Proof. Choose an irreducible factor g of f. Define E := k[x]/(g(x)). Then E is a field extension of k, and the image of x in E is a zero of f.
Step 2: Given f 1 ,... , fn ∈ k[x]≥ 1 , there exists a field extension E ⊇ k in which each fi has a zero.
Proof. Step 1 and induction.
Step 3: There exists a field extension k′^ ⊇ k containing a zero of every f ∈ k[x]≥ 1.
Proof. Define a commutative ring
A := k(f[{ (XXf^ :^ f^ ∈^ k[x]≥^1 }] f ) :^ f^ ∈^ k[x]≥ 1 )
Suppose that A is the zero ring. Then 1 is in the ideal generated by the f (Xf ). So we have an equation 1 = g 1 f 1 (Xf 1 ) + · · · + gnfn(Xfn ).
for some polynomials gi. By Step 2, there exists a field extension F ⊇ k containing a zero αi of each fi. Evaluating the previous equation at Xfi = αi yields 1 = 0 in F , contradicting the fact that F is a field. Thus A is not the zero ring. So A has a maximal ideal m. Let k′^ := A/m. Then k′^ is a field extension of k, and the image of Xfi in k′^ is a zero of Fi.
Step 4: There exists an algebraically closed field E ⊇ k.
Proof. Iterate Step 3 to obtain a chain of fields
k ⊆ k′^ ⊆ k′′^ ⊆ · · · ⊆ k(n)^ ⊆ · · ·.
Let E be their union. Any polynomial in E[x]≥ 1 has coefficients in some fixed k(n), and hence has a zero in k(n+1), so it has a zero in E. Thus E is algebraically closed.
Step 5: There exists an algebraic closure k of k.
Proof. Let E be as in Step 4. Let k be the set of α ∈ E that are algebraic over k. Since algebraic elements are closed under addition, multiplication, etc., the set k is a subfield of E. And of course, k is algebraic over k. If f ∈ k[x]≥ 1 , let β be a zero of f in E; then β is algebraic over the field k(coefficients of f ), which is algebraic over k, so β is algebraic over k, so β ∈ k. Thus k is algebraically closed.
Step 6: If E is an algebraic extension of k, and L is an algebraically closed field then any embedding k ↪→ L extends to an embedding E ↪→ L.
Proof. If E is generated by one element α, then E ' k[x]/(f (x)) for some f ∈ k[x]≥ 1. Choose a zero α′^ ∈ L of f , and define E ↪→ L by mapping α to α′. If E is generated by finitely many elements, extend the embedding in stages, adjoining one element at a time. In general, use transfinite induction (Zorn’s lemma). Step 7: Any two algebraic closures of k are isomorphic over k.
Proof. Let E and L be two algebraic closures of k. Step 6 extends k ↪→ L to E ↪→ L. If E 6 = L, then the minimal polynomial of an element of L − E would be a polynomial in E[x]≥ 1 , contradicting the assumption that E is algebraically closed.
- Finite fields Let Fp be Z/pZ viewed as a field.
Theorem 17.1. For each prime p, choose an algebraic closure Fp of Fp.
(1) Given a prime power q = pn, there exists a unique subfield of Fp of order q, namely Fq := {x ∈ Fp : xq^ = x}. (2) Every finite field is isomorphic to exactly one Fq. (3) Fpm ⊆ Fpn if and only if m|n. (4) Gal(Fqn^ /Fq) ' Z/nZ, and it is generated by Frobq : Fqn → Fqn x 7 → xq.
Proof.
