Unknown
If you face any problem to loading this page, click here for PDF file of this chapter. The definition of global field varies in the literature, but all definitions include our primary
source of examples, number fields. The other fields that are of interest in algebraic number
theory are the local fields, which are complete with respect to a discrete valuation. This
terminology will be explained as we go along.
9.1 Absolute Values and Discrete Valuations
9.1.1 Definitions and Comments
An absolute value on a field k is a mapping x → |x| from k to the real numbers, such that
for every x, y ∈ k,
1. |x| ≥ 0, with equality if and only if x =0;
2. |xy| = |x| |y|;
3. |x + y| ≤ |x| + |y|.
The absolute value is nonarchimedean if the third condition is replaced by a stronger
version:
3 . |x + y| ≤ max(|x|, |y|).
As expected, archimedean means not nonarchimedean.
The familiar absolute values on the reals and the complex numbers are archimedean.
However, our interest will be in nonarchimedean absolute values. Here is where most of
them come from.
A discrete valuation on k is a surjective map v : k → Z∪{∞}, such that for every x, y ∈ k,
(a) v(x) = ∞ if and only if x =0;
(b) v(xy) = v(x) + v(y);
(c) v(x + y) ≥ min(v(x), v(y)).
A discrete valuation induces a nonarchimedean absolute value via |x| = cv(x), where c
is a constant with 0 < c < 1.
9.1.2 Example
Let A be a Dedekind domain with fraction field K, and let P be a nonzero prime ideal of
A. Then [see the proof of (4.1.6)] the localized ring AP is a discrete valuation ring (DVR)
with unique maximal ideal (equivalently, unique nonzero prime ideal) PAP . Choose a
generator π of this ideal; this is possible because a DVR is, in particular, a PID. Now if
x ∈ K∗, the set of nonzero elements of K, then by factoring the principal fractional ideal
(x)AP , we find that x = uπn, where n ∈ Z and u is a unit in AP . We define vP (x) = n,
with vP (0) = ∞. We can check that vP is a discrete valuation, called the P-adic valuation
on K. Surjectivity and conditions (a) and (b) follow directly from the definition. To verify
(c), let x = uπm, y = vπn with m ≥ n. Then x + y = (v−1uπm−n + 1)vπn, and since
the term in parentheses belongs to AP , the exponent in its prime factorization will be
nonnegative. Therefore vP (x + y) ≥ n =min( vP (x), vP (y)).
Now consider the special case A = Z, K = Q, P = (p). If x is rational and x = pra/b
where neither a nor b is divisible by p, then we get the p-adic valuation on the rationals,
given by vp(pra/b) = r.
Here are some of the basic properties of nonarchimedean absolute values. It is often
convenient to exclude the trivial absolute value, given by |x| =1 for x
=0, and |0| =0.
Note also that for any absolute value, |1| = | − 1| = 1, | − x| = |x|, and |x−1| = 1/|x| for
x
=0. (Observe that 1 × 1 = (−1) × (−1) = x × x−1 =1.)
9.1.3 Proposition
Let | | be a nonarchimedean absolute value on the field K. Let A be the corresponding
valuation ring, defined as {x ∈ K : |x| ≤ 1}, and P the valuation ideal {x ∈ K : |x| < 1}.
Then A is a local ring with unique maximal ideal P and fraction field K. If u ∈ K, then
u is a unit of A if and only if |u| =1. If the trivial absolute value is excluded, then A is
not a field.
Proof.
1. A is a ring, because it is closed under addition, subtraction and multiplication, and
contains the identity.
2. K is the fraction field of A, because if z is a nonzero element of K, then either z or its
inverse belongs to A.
3. A is a local ring with unique maximal ideal P. It follows from the definition that P
is a proper ideal. If Q is any proper ideal of A, then Q ⊆ P, because A \ P ⊆ A \ Q.
(If x ∈ A \ P, then |x| =1, hence |x−1| =1, so x−1 ∈ A. Thus x ∈ Q implies that
xx−1 = 1 ∈ Q, a contradiction.)
4. If u ∈ K, then u is a unit of A iff |u| =1. For if u and v belong to A and uv =1, then
|u| |v| =1. But both |u| and |v| are at most 1, hence they must equal 1. Conversely, if
|u| =1, then |u−1| =1. But then both u and its inverse belong to A, so u is a unit of A.
5. If | | is nontrivial, then A is not a field. For if x
=0 and |x|
=1, then either |x| < 1
and |x−1| > 1, or |x| > 1 and |x−1| < 1. Either way, we have an element of A whose
inverse lies outside of A. ♣
9.1.4 Proposition
If the nonarchimedean and nontrivial absolute value | | on K is induced by the discrete
valuation v, then the valuation ring A is a DVR.
Proof. In view of (9.1.3), we need only show that A is a PID. Choose an element π ∈ A
such that v(π) =1. If x is a nonzero element of A and v(x) = n ∈ Z, then v(xπ−n) = 0,
so xπ−n has absolute value 1 and is therefore a unit u by (9.1.3). Thus x = uπn. Nowif I
is any proper ideal of A, then I will contain an element uπn with |n| as small as possible,
say |n| = n0. Either πn0 or π−n0 will be a generator of I (but not both since I is proper).
We conclude that every ideal of A is principal. ♣
The proof of (9.1.4) shows that A has exactly one nonzero prime ideal, namely (π).
9.1.5 Proposition
If | | is a nonarchimedean absolute value , then |x|
= |y| implies |x + y| =max( |x|, |y|).
Hence by induction, if |x1| > |xi| for all i = 2, . . . , n, then |x1 + · · · + xn| = |x1|.
Proof. We may assume without loss of generality that |x| > |y|. Then
|x| = |x + y − y| ≤ max(|x + y|, |y|) = |x + y|,
otherwise max(|x+y|, |y|) = |y| < |x|, a contradiction. Since |x+y| ≤ max(|x|, |y|) = |x|,
the result follows. ♣
9.1.6 Corollary
With respect to the metric induced by a nonarchimedean absolute value, all triangles are
isosceles.
Proof. Let the vertices of the triangle be x, y and z. Then |x − y| = |(x − z) + (z − y)|.
If |x − z| = |z − y|, then two side lengths are equal. If |x − z|
= |z − y|, then by (9.1.5),
|x − y| =max( |x − z|, |z − y|), and again two side lengths are equal. ♣
9.1.7 Proposition
The absolute value | | is nonarchimedean if and only if |n| ≤ 1 for every integer n =
1 ± ·· ·±1, equivalently if and only if the set {|n| : n ∈ Z} is bounded.
Proof. If the absolute value is nonarchimedean, then |n| ≤ 1 by repeated application of
condition 3 of (9.1.1). Conversely, if every integer has absolute value at most 1, then it
suffices to show that |x + 1| ≤ max(|x|, 1) for every x. (Apply this result to x/y, y
=0.)
By the binomial theorem,
|x + 1|n =
n
r=0
n
r
xr ≤
n
r=0
n
r
|x|r.
By hypothesis, the integer
n
r
has absolute value at most 1. If |x| > 1, then |x|r ≤ |x|n
for all r = 0, 1, . . . , n. If |x| ≤ 1, then |x|r ≤ 1. Consequently,
|x + 1|n ≤ (n + 1) max(|x|n, 1).
For Section 9.1
1. Show that every absolute value on a finite field is trivial.
2. Show that a field that has an archimedean absolute value must have characteristic 0.
3. Two nontrivial absolute values | |1 and | |2 on the same field are said to be equivalent
if for every x, |x|1 < 1 if and only if |x|2 < 1. [Equally well, |x|1 > 1 if and only if
|x|2 > 1; just replace x by 1/x if x
=0.] This says that the absolute values induce the
same topology (because they have the same sequences that converge to 0). Show that two
nontrivial absolute values are equivalent if and only if for some real number a, we have
|x|a1
= |x|2 for all x.
9.2 Absolute Values on the Rationals
In (9.1.2), we discussed the p-adic absolute value on the rationals (induced by the p-adic
valuation, with p prime), and we are familiar with the usual absolute value. In this section,
we will prove that up to equivalence (see Problem 3 of Section 9.1), there are no other
nontrivial absolute values on Q.
9.2.1 Preliminary Calculations
Fix an absolute value | | on Q. If m and n are positive integers greater than 1, expand
m to the base n. Then m = a0 + a1n + · · · + arnr, 0 ≤ ai ≤ n − 1, ar
=0.
(1) r ≤ log m/log n.
This follows because nr ≤ m.
(2) For every positive integer l we have |l| ≤ l, hence in the above base n expansion,
|ai| ≤ ai < n.
This can be done by induction: |1| = 1, |1 + 1| ≤ |1| + |1|, and so on.
There are 1 + r terms in the expansion of m, each bounded by n[max(1, |n|)]r. [We
must allow for the possibility that |n| < 1, so that |n|i decreases as i increases. In this
case, we will not be able to claim that |a0| ≤ n(|n|r).] With the aid of (1), we have
(3) |m| ≤ (1 + log m/log n)n[max(1, |n|)]log m/log n.
Replace m by mt and take the tth root of both sides. The result is
(4) |m| ≤ (1 + t log m/log n)1/tn1/t[max(1, |n|)]log m/log n.
Let t→∞ to obtain our key formula:
(5) |m| ≤ [max(1, |n|)]log m/log n.
9.2.2 The Archimedean Case
Suppose that |n| > 1 for every n > 1. Then by (5), |m| ≤ |n|log m/log n, and therefore
log |m| ≤ (log m/log n) log |n|. Interchanging m and n gives the reverse inequality, so
log |m| =(log m/log n) log |n|. It follows that log |n|/ log n is a constant a, so |n| = na.
Since 1 < |n| ≤ n [see (2)], we have 0 < a ≤ 1. Thus our absolute value is equivalent to
the usual one.
9.2.3 The Nonarchimedean Case
Suppose that for some n > 1 we have |n| ≤ 1. By (5), |m| ≤ 1 for all m > 1, so
|n| ≤ 1 for all n ≥ 1, and the absolute value is nonarchimedean by (9.1.7). Excluding
the trivial absolute value, we have |n| < 1 for some n > 1. (If every nonzero integer
has absolute value 1, then every nonzero rational number has absolute value 1.) Let
P = {n ∈ Z : |n| < 1}. Then P is a prime ideal (p). (Note that if ab has absolute value
less than 1, so does either a or b.) Let c = |p|, so 0 < c < 1.
Now let r be the exact power of p dividing n, so that pr divides n but pr+1 does not.
Then n/pr /∈
P, so |n|/cr =1, |n| = cr. Note that n/pr+1 also fails to belong to P, but
this causes no difficulty because n/pr+1 is not an integer.
To summarize, our absolute value agrees, up to equivalence, with the p-adic absolute
value on the positive integers, hence on all rational numbers. (In going from a discrete
valuation to an absolute value, we are free to choose any constant in (0,1). A different
constant will yield an equivalent absolute value.)
Problems For Section 9.2
If vp is the p-adic valuation on Q, let p be the associated absolute value with the
particular choice c = 1/p. Thus pr p = p−r. Denote the usual absolute value by ∞.
1. Establish the product formula: If a is a nonzero rational number, then
p
a p = 1
where p ranges over all primes, including the “infinite prime” p = ∞.
9.3 Artin-Whaples Approximation Theorem
The Chinese remainder theorem states that if I1, . . . In are ideals in a ring R that are
relatively prime in pairs, and ai ∈ Ii, i = 1, . . . ,n, then there exists a ∈ R such that
a ≡ ai mod Ii for all i. We are going to prove a result about mutually equivalent absolute
values that is in a sense analogous. The condition a ≡ ai mod Ii will be replaced by
the statement that a is close to ai with respect to the ith absolute value. First, some
computations.
9.3.1 Lemma
Let | | be an arbitrary absolute value. Then
(1) |a| < 1 ⇒ an → 0;
(2) |a| < 1 ⇒ an/(1 + an) → 0;
(3) |a| > 1 ⇒ an/(1 + an) → 1.
Proof. The first statement follows from |an| = |a|n. To prove (2), use the triangle
inequality and the observation that 1 + an = 1− (−an) to get
1 − |a|n ≤ |1 + an| ≤ 1 + |a|n,
so by (1), |1+an| → 1. Since |α/β| = |α|/|β|, another application of (1) gives the desired
result. To prove (3), write
1 − an
1 + an =
1
1 + an = a−n
1 + a−n
→ 0 by (2). ♣
Here is the key step in the development.
9.3.2 Proposition
Let | |1, . . . ,| |n be nontrivial, mutually inequivalent absolute values on the same field.
Then there is an element a such that |a|1 > 1 and |a|i < 1 for i = 2, . . . ,n.
Proof. First consider the case n =2. Since | |1 and | |2 are inequivalent, there are
elements b and c such that |b|1 < 1, |b|2 ≥ 1, |c|1 ≥ 1, |c|2 < 1. If a = c/b, then |a|1 > 1
and |a|2 < 1.
Now if the result holds for n−1, we can choose an element b such that |b|1 > 1, |b|2 <
1, . . . , |b|n−1 < 1. By the n =2 case, we can choose c such that |c|1 > 1 and |c|n < 1.
Case 1. Suppose |b|n ≤ 1. Take ar = cbr, r ≥ 1. Then |ar|1 > 1, |ar|n < 1, and |ar|i → 0
as r→∞ for i = 2, . . . ,n − 1. Thus we can take a = ar for sufficiently large r.
Case 2. Suppose |b|n > 1. Take ar = cbr/(1 + br). By (3) of (9.3.1), |ar|1 → |c|1 > 1 and
|ar|n → |c|n < 1 as r→∞. If 2 ≤ i ≤ n − 1, then |b|i < 1, so by (2) of (9.3.1), |ar|i → 0
as r→∞. Again we can take a = ar for sufficiently large r. ♣
9.3.3 Approximation Theorem
Let | |1, . . . ,| |n be nontrivial mutually inequivalent absolute values on the field k.
Given arbitrary elements x1, . . . , xn ∈ k and any positive real number !, there is an
element x ∈ k such that |x − xi|i < ! for all i = 1, . . . , n.
Proof. By (9.3.2), ∀i ∃yi ∈ k such that |yi|i > 1 and |yi|j < 1 for j
= i. Take zi =
yr
i /(1+yr
i ). Given δ > 0, it follows from (2) and (3) of (9.3.1) that for r sufficiently large,
|zi − 1|i < δ and |zj | < δ, j
= i.
Our candidate is
x = x1z1 + · · · xnzn.
To show that x works, note that x − xi =
j =i xjzj + xi(zi − 1). Thus
|x − xi|i ≤ δ
j =i
|xj |i + δ|xi|i = δ
n
j=1
|xj |i .
Choose δ so that the right side is less than !, and the result follows. ♣
Problems For Section 9.3
1. Let | |1, . . . ,| |n be nontrivial mutually inequivalent absolute values on the field k.
Fix r with 0 ≤ r ≤ n. Show that there is an element a ∈ k such that |a|1 > 1, . . . , |a|r > 1
and |a|r+1, . . . , |a|n < 1.
2. There is a gap in the first paragraph of the proof of (9.3.2), which can be repaired by
showing that the implication |a|1 < 1 ⇒ |a|2 < 1 is sufficient for equivalence. Prove this.
9.4 Completions
You have probably seen the construction of the real numbers from the rationals, and
the general process of completing a metric space using equivalence classes of Cauchy
sequences. If the metric is induced by an absolute value on a field, then we have some
additional structure that we can exploit to simplify the development. If we complete the
rationals with respect to the p-adic rather than the usual absolute value, we get the p-adic
numbers, the most popular example of a local field.
9.4.1 Definitions and Comments
Let K be a field with an absolute value | |, and let C be the set of Cauchy sequences
with elements in K. Then C is a ring under componentwise addition and multiplication.
Let N be the set of null sequences (sequences converging to 0). Then N is an ideal of C
(because every Cauchy sequence is bounded). In fact N is a maximal ideal, because every
Cauchy sequence not in N is eventually bounded away from 0, hence is a unit in C. The
completion of K with respect to the given absolute value is the field ˆK = C/N. We can
embed K in ˆK via c → {c, c, . . . } + N.
We now extend the absolute value on K to ˆK . If (cn)+N ∈ ˆK , then (|cn|) is a Cauchy
sequence of real numbers, because by the triangle inequality, |cn| − |cm| has (ordinary)
absolute value at most |cn −cm| → 0 as n,m→∞. Thus |cn| converges to a limit, which
we take as the absolute value of (cn) + N. Since the original absolute value satisfies the
defining conditions in (9.1.1), so does the extension.
To simplify the notation, we will denote the element (cn) + N of ˆK by (cn). If
cn = c ∈ K for all n, we will write the element as c.
9.4.2 Theorem
K is dense in ˆK and ˆK is complete.
Proof. Let α = (cn) ∈ ˆK, with αn = cn. Then
|α − αn| =lim
m→∞
|cm − cn| → 0 as n→∞,
proving that K is dense in ˆK . To prove completeness of ˆK, let (αn) be a Cauchy sequence
in ˆK . Since K is dense, for every positive integer n there exists cn ∈ K such that
|αn−cn| < 1/n. But then (cn) is a Cauchy sequence in ˆK, hence in K, and we are assured
that α = (cn) is a legal element of ˆK . Moreover, |αn − α| → 0, proving completeness. ♣
9.4.3 Uniqueness of the Completion
Suppose K is isomorphic to a dense subfield of the complete field L, where the absolute
value on L extends that of (the isomorphic copy of) K. If x ∈ ˆK, then there is a sequence
xn ∈ K such that xn → x. But the sequence (xn) is also Cauchy in L, hence converges
to an element y ∈ L. If we define f(x) = y, then f is a well-defined homomorphism of
fields, necessarily injective. If y ∈ L, then y is the limit of a Cauchy sequence in K, which
converges to some x ∈ ˆK . Consequently, f(x) = y. Thus f is an isomorphism of ˆK and
L, and f preserves the absolute value.
9.4.4 Power Series Representation
We define a local field K as follows. There is an absolute value on K induced by a discrete
valuation v, and with respect to this absolute value, K is complete. For short, we say
that K is complete with respect to the discrete valuation v. Let A be the valuation ring
(a DVR), and P the valuation ideal; see (9.1.3) and (9.1.4) for terminology. If α ∈ K,
then by (9.1.4) we can write α = uπr with r ∈ Z, u a unit in A and π an element of
A such that v(π) =1. Often, π is called a prime element or a uniformizer. Note that
A = {α ∈ K : v(α) ≥ 0} and P = {α ∈ K : v(α) ≥ 1} = Aπ.
Let S be a fixed set of representatives of the cosets of A/P. We will show that each
α ∈ K has a Laurent series expansion
α = a−mπ
−m + · · · + a−1π
−1 + a0 + a1π + a2π2 + · · · , ai ∈ S,
and if ar is the first nonzero coefficient (r may be negative), then v(α) = r.
The idea is to expand the unit u in a power series involving only nonnegative powers of
π. For some a0 ∈ S we have u−a0 ∈ P. But then v(u−a0) ≥ 1, hence v((u−a0)/π) ≥ 0,
so (u − a0)/π ∈ A. Then for some a1 ∈ S we have [(u − a0)/π] − a1 ∈ P, in other words,
u − a0 − a1π
π
∈ P.
Repeating the above argument, we get
u − a0 − a1π
π2
∈ A.
Continue inductively to obtain the desired series expansion. Note that by definition of S,
the coefficients ai are unique. Thus an expansion of α that begins with a term of degree
r in π corresponds to a representation α = uπr and a valuation v(α) = r. Also, since
|π| < 1, high positive powers of π are small with respect to the given absolute value. The
partial sums sn of the series form a coherent sequence, that is, sn ≡ sn−1 mod (π)n.
9.4.5 Proposition
Let
an be any series of elements in a local field. Then the series converges if and only
if an → 0.
Proof. If the series converges, then an → 0 by the standard calculus argument, so assume
that an → 0. Since the absolute value is nonarchimedean, n ≤ m implies that
|
m
i=n
ai| ≤ max(an, . . . ,am) → 0 as n→∞. ♣
9.4.6 Definitions and Comments
The completion of the rationals with respect to the p-adic valuation is called the field of
p-adic numbers, denoted by Qp. The valuation ring A = {α : v(α) ≥ 0} is called the ring
of p-adic integers, denoted by Zp. The series representation of a p-adic integer contains
only nonnegative powers of π = p. If in addition, there is no constant term, we get the
valuation ideal P = {α : v(α) ≥ 1}. The set S of coset representatives may be chosen to
be {0, 1, . . . ,p − 1}. (Note that if a
= b and a ≡ b mod p, then a − b ∈ P, so a and b
cannot both belong to S. Also, a rational number can always be replaced by an integer
with the same valuation.) Arithmetic is carried out via polynomial multiplication, except
that there is a “carry”. For example, if p =7, then 3 + 6 =9 =2 +p. For some practice,
see the exercises.
We adopt the convention that in going from the p-adic valuation to the associated
absolute value |x| = cv(x), 0 < c < 1, we take c = 1/p. Thus |pr| = p−r.
Problems For Section 9.4
1. Show that a rational number a/b (in lowest terms) is a p-adic integer if and only if p
does not divide b.
2. With p =3, express the product of (2 + p + p2) and (2 + p2) as a p-adic integer.
3. Express the p-adic integer -1 as an infinite series.
4. Show that the sequence an = n! of p-adic integers converges to 0.
5. Does the sequence an = n of p-adic integers converge?
6. Show that the p-adic power series for log(1+x), namely
∞
n=1(−1)n+1xn/n, converges
in Qp for |x| < 1 and diverges elsewhere. This allows a definition of a p-adic logarithm:
logp(x) =log [1 + (x − 1)].
In Problems 7-9, we consider the p-adic exponential function.
7. Recall from elementary number theory that the highest power of p dividing n! is ∞
i=1
n/pi . (As an example, let n =15 and p =2. Calculate the number of multiples
of 2, 4,and 8 in the integers 1-15.) Use this result to show that the p-adic valuation of n!
is at most n/(p − 1).
8. Show that the p-adic valuation of (pm)! is (pm − 1)/(p − 1).
9. Show that the exponential series
∞
n=0 xn/n! converges for |x| < p−1/(p−1) and diverges
elsewhere.
9.5 Hensel’s Lemma
9.5.1 The Setup
Let K be a local field with valuation ring A and valuation ideal P. By (9.1.3) and (9.1.4),
A is a local ring, in fact a DVR, with maximal ideal P. The field k = A/P is called the
residue field of A or of K. If a ∈ A, then the coset a+P ∈ k will be denoted by a. If f is
a polynomial in A[X], then reduction of the coefficients of f mod P yields a polynomial
f in k[X]. Thus
f(X) =
d
i=0
aiXi ∈ A[X], f(X) =
d
i=0
aiXi ∈ k[X].
Hensel’s lemma is about lifting a factorization of f from k[X] to A[X]. Here is the precise
statement.
9.5.2 Hensel’s Lemma
Assume that f is a monic polynomial of degree d in A[X], and that the corresponding
polynomial F = f factors as the product of relatively prime monic polynomials G and H
in k[X]. Then there are monic polynomials g and h in A[X] such that g = G, h = H and
f = gh.
Proof. Let r be the degree of G, so that degH = d − r. We will inductively construct
gn, hn ∈ A[X], n = 1, 2, . . . , such that deg gn = r, deg hn = d − r, gn = G, hn = H, and
f(X) − gn(X)hn(X) ∈ Pn[X].
Thus the coefficients of f − gnhn belong to Pn.
The basis step: Let n =1. Choose monic g1, h1 ∈ A[X] such that g1 = G and h1 = H.
Then deg g1 = r and deg h1 = d − r. Since f = g1h1, we have f − g1h1 ∈ P[X].
T he inductive step: Assume that gn and hn have been constructed. Let f(X)−gn(X)hn(X) =
d
i=0 ciXi with the ci ∈ Pn. Since G = gn and H = hn are relatively prime, for each
i = 0, . . . , d there are polynomials vi and wi in k[X] such that
Xi = vi(X)gn(X) + wi(X)hn(X).
Since gn has degree r, the degree of vi is at most d − r, and similarly the degree of wi is
at most r. Moreover,
Xi − vi(X)gn(X) − wi(X)hn(X) ∈ P[X]. (1)
We define
gn+1(X) = gn(X) +
d
i=0
ciwi(X), hn+1(X) = hn(X) +
d
i=0
civi(X).
Since the ci belong to Pn ⊆ P, it follows that gn+1 = gn = G and hn+1 = hn = H. Since
the degree of gn+1 is at most r, it must be exactly r, and similarly the degree of hn+1 is
d − r. To check the remaining condition,
f − gn+1hn+1 = f − (gn +
i
ciwi)(hn +
i
civi)
= (f − gnhn −
i
ciXi) +
i
ci(Xi − gnvi − hnwi) −
i,j
cicjwivj .
By the induction hypothesis, the first grouped term on the right is zero, and, with the
aid of Equation (1) above, the second grouped term belongs to PnP[X] = Pn+1[X]. The
final term belongs to P2n[X] ⊆ Pn+1[X], completing the induction.
Finishing the proof. By definition of gn+1, we have gn+1 − gn ∈ Pn[X], so for any
fixed i, the sequence of coefficients of Xi in gn(X) is Cauchy and therefore converges.
To simplify the notation we write gn(X) → g(X), and similarly hn(X) → h(X), with
g(X), h(X) ∈ A[X]. By construction, f − gnhn ∈ Pn[X], and we may let n→∞ to get
f = gh. Since gn = G and hn = H for all n, we must have g = G and h = H. Since
f,G and H are monic, the highest degree terms of g and h are of the form (1+a)Xr and
(1 + a)−1Xd−r respectively, with a ∈ P. (Note that 1 + a must reduce to 1 mod P.) By
replacing g and h by (1 + a)−1g and (1 + a)h, respectively, we can make g and h monic
without disturbing the other conditions. The proof is complete. ♣
9.5.3 Corollary
With notation as in (9.5.1), let f be a monic polynomial in A[X] such that f has a simple
root η ∈ k. Then f has a simple root a ∈ A such that a = η.
Proof. We may write f(X) = (X − η)H(X) where X − η and H(X) are relatively prime
in k[X]. By Hensel’s lemma, we may lift the factorization to f(X) = (X − a)h(X) with
h ∈ A[X], a ∈ A and a = η. If a is a multiple root of f, then η is a multiple root of f,
which is a contradiction. ♣
Problems For Section 9.5
1. Show that for any prime p, there are p − 1 distinct (p − 1)th roots of unity in Zp.
2. Let p be an odd prime not dividing the integer m. We wish to determine whether m
is a square in Zp. Describe an effective procedure for doing this.
3. In Problem 2, suppose that we not only want to decide if m is a square in Zp, but
to find the series representation of
√
m explicitly. Indicate how to do this, and illustrate
with an example.
If you face any problem to loading this page, click here for PDF file of this chapter.
You Might Also Like
Posted in: Linear Algebra
