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Home , Quotient ring

Integral Domains Arising as Rings of Z[[x]]

By

James M. McDonough

A DISSERTATION SUBMITTED IN PARTIAL FULFILLMENT OF THE

REQUIREMENTSFORTHEDEGREEOF

MASTERSIN SCIENCE

(MATHEMATICS)

at the

CALIFORNIA STATE UNIVERSITY - CHANNEL ISLANDS

2011 © 2011

James M. McDonough

ALL RIGHTS RESERVED

Dedication

To my friends and family, and especially my thesis advisor Dr. Jesse Elliott, who have supported me throughout this endeavor.

iv Acknowledgements

This was done with the help of my thesis advisor, Dr. Jesse Elliott, my commit- tee members, Dr. Ivona Grzegorzyck and Dr. Brian Sittinger, and technology advisor Dr. Jorge Garcia, all from California State University, Channel Islands.

Camarillo, California

August 17, 2011

v Abstract

Using techniques of and p-adic analysis, we classify all integral domains arising as quotient rings of Z[[x]].

1 Table of Contents

1 Introduction 3

2 Background on p-adic Numbers and Power Series Rings 6

2.1 p-adic Numbers ...... 6

2.2 Background on Theory ...... 18

2.3 Power Series Rings ...... 20

3 Proofs and Examples 28

3.1 Proof of Main Theorem ...... 28

3.2 Examples ...... 38

3.3 Further Directions ...... 41

Bibliography 42

2 CHAPTER

1 Introduction

While comparing the ring Z[x] of polynomials with coefficients with

Z[[x]] its x-adic completion, the ring of with integer co-

efficients, an interesting dilemma arises. While they do have some things

in common, for instance, they are both Noetherian unique factorization do-

mains (UFDs) of Krull dimension 2, one major difference between them has

to do with known irreducibility criteria for their elements. For Z[x], there

are known irreducibility criteria, such as Eisenstein’s criterion and Gauss’s

Lemma. No such criteria are known for Z[[x]]. A 2008 paper by Daniel Bir-

majer and Juan Gil (see [1]) gives examples of polynomials that are irreducible

over Z[x], but reducible over Z[[x]], and vice versa. For example they show

that 6 x x2 is irreducible over Z[x] but reducible over Z[[x]], and 2 7x 3x2 + + + +

3 CHAPTER 1. INTRODUCTION is irreducible over Z[[x]] but reducible over Z[x].

These examples lead to the following question. How do the quotient rings differ or compare between Z[x] and Z[[x]] when factoring out by one of these irreducible elements? The situation for Z[x] is well understood already. When factoring out Z[x] by an nonconstant , the resulting quotient ring is isomorphic to Z[α], where α C is any root of the polyno- ∈ 2 h 1 p 23 i mial. For instance Z[x]/(6 x x ) Z − + − . What can we say about + + =∼ 2 Z[[x]]/(f ), where f is irreducible? This answer to this question is not provided by any of the standard algebra texts. We will show, for example, that the ring

Z[[x]]/(2 7x 3x2) is in fact isomorphic to the ring of 2-adic Z . Even + + 2 when a polynomial is irreducible over both Z[x] and Z[[x]], the quotient rings are generally not isomorphic. An example are the quotient rings Z[x]/(2 x2) − and Z[[x]]/(2 x2), which are isomorphic to Z[p2] and Z [p2] respectively. − 2 It is believed that a better understanding of the quotient rings of Z[[x]] may lead to clues about irreducibility criteria in Z[[x]]. Since Z[[x]] is a UFD, its prime elements and irreducible elements are the same; so when Z[[x]] is factored out by an generated by an irreducible element the resulting quotient ring is an . Therefore, if one can classify the quo- tient rings that are integral domains, then the irreducibility characteristics of

4 CHAPTER 1. INTRODUCTION power series in Z[[x]] may be more apparent.

In this paper, all integral domains arising as quotient rings of Z[[x]] are classified. First a brief introduction to p-adic numbers will be given, followed by some properties of power series. In particular, the irreducibility of ele- ments in Z[[x]] and the prime ideals in Z[[x]] will be discussed. Using this information, as well as the p-adic Weierstrass Preparation Theorem, the fol- lowing theorem will be proved.

Main Theorem. The integral domains arising as a quotient rings of Z[[x]] are, up to isomorphism, precisely the following:

Z[[x]],(Z/pZ)[[x]], Z, Z/pZ, Zp [α] where p is prime and α is any element of Q with α 1. Moreover, Z [α] p | |p < p is a local ring with maximal ideal (p,α) and is a DVR if and only if (p,α) is principal.

In addition, it is also shown that for any irreducible power series f Z[[x]], ∈ excluding associates of p and x, the quotient ring Z[[x]]/(f ) is isomorphic to

Z [α] for some prime p, where Z is the ring of p-adic integers and α Q is p p ∈ p any root of f .

5 CHAPTER

2 Background on p-adic

Numbers and Power Series

Rings 2.1 p-adic Numbers

Most mathematicians are familiar with Q and its completion R with respect to

the usual . An absolute value on a field is defined as follows.

Definition 2.1.1. An absolute value on a field K is a function from K to R 0 |·| ≥ satisfying the following properties for any x and y in K .

1. x 0 iff x 0 | | = =

2. x y x y | | = | || |

3. x y x y | + | ≤ | | + | | 6 2.1. P-ADIC NUMBERS

Moreover, is non-archimedean if x y max( x , y ) for all x, y K.A | · | | + | ≤ | | | | ∈ normed field is a field K together with a norm on K . | · |

We now introduce the p-adic absolute value.

Definition 2.1.2. Let p be prime. Then is the p-adic absolute value, where | |p for all a Q, we have a is 1 divided by the power of p in the prime factoriza- b ∈ | b |p a tion of b .

It is not difficult to show that this definition does indeed define an abso- lute value on Q, and it is non-archimedean (see [3] or [4]).

Example 2.1.3. For any prime p we have the following.

1. pn 1 | |p = pn

2. 1 pn | pn |p =

3. 16 1 | 17 |2 = 24

4. 16 33 | 27 |3 =

We now give the definition of a complete normed field (see[5, 1.4-3 Defi- nition]) and the completion of a normed field (see[3, pp. 49-59]).

7 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

Definition 2.1.4. A normed field is said to be complete if every sequence that is Cauchy with respect to the norm converges in the field. The completion | · | of a normed field K with respect to the norm is a complete normed field | · | containing K as a dense normed subfield and is unique up to isomorphism of normed fields.

We now introduce the p-adic rational numbers.

Definition 2.1.5. Let p be prime. The normed field of p-adic rational num- bers, Qp , is the completion of Q with respect to the p-adic absolute value.

Note that R is the completion of Q with respect to the usual absolute value.

Remark 2.1.6. In contrast to R, the series

1 p p2 + + + ···

converges in Qp . Similarly, the series

1 1 1 + p + p2 + ···

converges in R but diverges in Qp .

One might ask: what are all the completions of Q with respect to its abso- lute values? The next theorem answers this question.

8 2.1. P-ADIC NUMBERS

Theorem 2.1.7 (Ostrowski). Any completion of Q with respect to an absolute value is isomorphic as a normed field to R or Qp for some prime p.

To help visualize elements of Q , it is useful to know that every x Q has p ∈ p a unique p-adic expansion by [3, Corollary 3.3.12]. For instance every x Q ∈ p can be written as follows:

n0 n0 1 X n x bn0 p bn0 1p + bn p + = + + ··· = n n0 ≥ where n Z and where 0 b p 1 for all n n . Notice that in this form, 0 ∈ ≤ n ≤ − ≥ 0 it is easy to factor out the largest power of p that divides the expansion and

n0 compute the p-adic absolute value of x: if b 0 then x p− . n0 6= | | =

Definition 2.1.8. Z is the set of all x Q such that x 1. p ∈ p | |p ≤

Alternatively, Zp is the set of expansions of the following form:

2 n X∞ n b0 b1p b2p bn p bn p + + + ··· + + ··· = n 0 = where 0 b p 1 for all n by [3, Corollary 3.3.11]. Note that Z is a ≤ n ≤ − p of Qp .

Now that we have Qp , we want to find its analog to C, where C is the normed field of complex numbers. This is not as easy as going from R to C, which is a finite extension.

9 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

Definition 2.1.9. Qp is the algebraic closure of Qp .

The degree of Qp over Qp is infinite, and unfortunately the algebraic clo- sure Qp is not complete by [3, pp. 176]. The good news is that its completion is algebraically closed.

To complete Qp , an extended absolute value must be defined. In order to do so, we need to use algebraic norms on finite extension fields of Qp . An algebraic norm of an extension field K of a field F is a function

N : K F. K /F →

The following are two ways to define the norm function:

1. Let α K and let F (α) be the smallest sub-extension containing F and α. ∈ n n 1 Let f (x) x an 1x − a1x a0 F [x] be the minimal irreducible = + − +···+ + ∈ polynomial of α over F . Then N (α) ( 1)sn as , where s [K : F (α)] K /F = − 0 = is the degree K over F (α) (see[3, pp. 147-148]).

2. Suppose K F (α) is a normal extension of F , and let f (x) xn = = + n 1 an 1x − a1x a0 F [x] be the minimal irreducible polynomial − + ··· + + ∈

of α over F . Let αi be the roots of f ; these are called the conjugates of

Qn α. Define NK /F (α) i 0 αi (see[4, pp. 60]). = =

10 2.1. P-ADIC NUMBERS

Now we define the extended p-adic absolute value for a finite extension of Qp , which also extends to Qp by [3, Theorem 5.3.5].

Definition 2.1.10. Let x Q , and let n [Q (α): Q ]. The extended p-adic ∈ p = p p absolute value is defined as follows:

q n x NQ Q (x) . | |p = | p (α)/ p |p

Now, completing Qp with respect to the extended p-adic absolute value, we get the following definition (see[3, Proposition 5.7.6]).

Definition 2.1.11. Cp is the completion of Qp with respect to the the extended p-adic absolute value.

Finally, Cp is both complete and algebraically closed by [3, Proposition

5.7.6 and Proposition 5.7.8].

Next, we will focus on the convergence properties of sequences and series in a complete non-archimedean normed field K of zero. The

first two lemmas and proofs follow from Gouvêa (see [3, Lemma 4.1.1 and

Corollary 4.1.2]).

Lemma 2.1.12. Let (an) be a sequence in K . Then (an) is Cauchy and therefore convergent iff lim an 1 an p 0. n + →∞ | − | =

11 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

Proof. Suppose (a ) is Cauchy. Then for every ² 0 there exists an inte- n > ger M 0 such for all n,m M, a a ². Now for any n,m M > ≥ | n − m|p < ≥ we get an am p an an 1 an 1 am 1 am 1 am p max( an | − | = | − − + − − ··· − + + + − | ≤ | − an 1 p , an 1 an 2 p ,..., am 2 am 1 p , am 1 am p ) ². Now as ² 0, − | | − − − | | + − + | | + − | ≤ →

M , and therefore lim an 1 an p 0 for any n M. → ∞ M | + − | = ≥ →∞

Now assume that lim an 1 an p 0 implies that for every ² 0 there n + →∞ | − | = > exists an integer N 0 such that for all n0 N, an 1 an p ². Let s r > ≥ | 0+ − 0 | < ≥ ≥

N. Then as ar p as as 1 as 1 ar 1 ar 1 ar p max( as | − | = | − − + − − ··· − + + + − | ≤ | − as 1 p , as 1 as 2 p ,..., ar 2 ar 1 p , ar 1 ar p ). However, since each index − | | − − − | | + − + | | + − | is greater than or equal to r , every ar i 1 ar i p ², so the max is less than | + + − + | < ² and the sequence is Cauchy. It immediately follows in a complete field that a Cauchy sequence converges.

Next we will show when a series converges.

P P Lemma 2.1.13. Let ak be a series in K . Then the series ak is convergent iff

lim an p 0. n →∞ | | =

P Proof. Suppose ak is convergent and therefore Cauchy. By the pre-

P vious lemma the series ak is Cauchy and convergent if and only if ¯ n n 1 ¯ ¯ X X− ¯ lim ¯ ak ak ¯ lim an p 0. n ¯ ¯ n →∞ ¯k 0 − k 0 ¯ = →∞ | | = = = p

12 2.1. P-ADIC NUMBERS

The next lemmas and proofs deal with the convergence of sums and prod-

ucts of convergent series. These proofs follow as in the real case and similar

proofs can be found in [8, Theorem 3.47 and Theorem 3.50].

Lemma 2.1.14. Suppose Pa a and Pb b both converge in K . Let c k = k = k = a b . Then Pc converges to a b. k + k k +

n n n n X X ³ ´ X X Proof. First note ck ak bk ak bk . Now by taking the k 0 = k 0 + = k 0 + k 0 = n = n = = n n P X X X X limit we get ck lim ck lim (ak bk ) lim ak lim bk n n n n = →∞ k 0 = →∞ k 0 + = →∞ k 0 + →∞ k 0 = X X = = = = a b a b. k + k = +

Lemma 2.1.15. Suppose Pa a and Pb b both converge in K . Let c k = k = k = k X P ai bk i . Then ck converges to ab. i 0 − = n n n X X X Proof. First set An ak , Bn bk , Cn ck and βn Bn b. Then = k 0 = k 0 = k 0 = − = = = Cn a0b0 (a0b1 a1b0) (a0bn a1bn 1 anb0) a0Bn a1Bn 1 = + + +···+ + − +···+ = + − +

anB0 a0(b βn) a1(b βn 1) an(b β0) Anb a0βn a1βn 1 ···+ = + + + − +···+ + = + + − +···+

anβ0. Now lim Anb ab; so we must show that the leftover terms go to zero. n →∞ = n n n ¡ Note that a0βn a1βn 1 anβ0 a0βn a1βn 1 a β 2 | + − +···+ | ≤ | + − +···+ b 2 c d 2 e|+| d e− n ¢ n n n 2 a β an 1β1 anβ0 max a j max( βn ,..., β ) max b c · d 2 e b 2 c + ··· + − + | ≤ 0 j | | · | | | d 2 e| + 0 j ≤ ≤∞ ≤ ≤∞ ¡ n n ¢ b j max( a n ,..., an ). Now taking the limit as n yields lim 2 2 2 n | |· d e−b c ·| d e| | | → ∞ →∞

13 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

n n n ¡ ¢ n max a j max( βn ,..., β ) max b j max( 2 2 a ,..., an ) 0 0 j | |· | | | d 2 e| +0 j | | · d e−b c ·| d 2 e| | | = ≤ ≤∞ ≤ ≤∞ since lim βn 0 and lim an 0 by Lemma 2.1.13. n n →∞ | | = →∞ | | =

The next theorem is an important tool for determining if a polynomial is reducible over Zp [x] (see [3, Theorem 3.4.6]).

Theorem 2.1.16 (Hensel’s Lemma). Let f ,g ,h Z [x] such that f 1 1 ∈ p ≡ g1h1 (mod p) where g1 is monic and g1 and h1 are relatively prime in

(Z/pZ)[x]. Then there exist polynomials g,h Z [x] such that f gh where g ∈ p = is monic and g g (mod p) and h h (mod p). ≡ 1 ≡ 1

The final topic discussed in this section is the p-adic Weierstrass Prepa- ration Theorem. This theorem has many versions, but all show the existence of a factorization of a p-adic power series into a polynomial and a unit power series. Power series are discussed in greater detail in Section 2.3. The version that follows is from [7, pp 115-116 Exercises 8 and 9] and [9, Proposition 7.2 and Theorem 7.3]. Before we prove the theorem, we prove that Zp [[x]] has the following remainder theorem.

X∞ v Proposition 2.1.17 (Washington). Let f ,g Zp [[x]] with f av x such ∈ = v 0 = that p a for 0 v n 1 but p a . Then g can be written uniquely as | v ≤ ≤ − 6 | n

g q f r = +

14 2.1. P-ADIC NUMBERS where q Z [[x]] and r Z [x] with deg r n 1. ∈ p ∈ p ≤ −

Proof. First we show existence. To do so, define the following Zp -linear oper- ator τn on Zp [[x]]:

³ X∞ v ´ X∞ v n τn bv x bv x − . v 0 = v n = = n 1 X− v n Let U τn(f ) and P (1/p) av x such that f pP x U. Note U is a unit = = v 0 = + = since p a . Now define q to be 6 | n

i 1 X∞ i i ³ 1´ q U − ( 1) p τn PU − (τn(g)) = i 0 − ◦ = 1 ³ ¡ 1 ¢ 2 ³ 1 ¡ 1 ¢´ ´ U − τ (g) pτ PU − τ (g) p τ PU − τ PU − τ (g) . = n − n n + n n n − ···

We claim τ (f q) τ (g). Set n = n

i i i 1 1³ 1´ k ( 1) p + PU − τ PU − (τ (g)) i = − n ◦ n and

h k xnτ (k ). i = i − n i

i 1 X∞ Each hi is a polynomial of degree n 1 in Zp [x] divisible by p + so hi con- − i 0 =

15 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS verges to a polynomial of degree n 1 in Z [x]. Then − p

i n ³ 1 X∞ i i ³ 1´ ´ f q (pP x U) U − ( 1) p τn PU − (τn(g)) = + i 0 − ◦ = µ i ¶ X∞ n i i ³ 1´ ´ ki x ( 1) p τn PU − (τn(g)) = i 0 + − ◦ = µ i ¶ n X∞ n i i ³ 1´ ´ k0 x τn(g) ki x ( 1) p τn PU − (τn(g)) = + + i 1 + − ◦ = n X∞ ¡ n ¢ x τn(g) k0 ki x τn(ki 1) = + + i 1 − − = n X∞ x τn(g) hi . = + i 0 =

Therefore

³ n X∞ ´ n τn(f q) τn x τn(g) hi τn(x τn(g)) τn(g). = + j 0 = = =

It now follows that g f q is a polynomial of degree n 1. Set r g f q. Now − − = − f q r f q g f q g. + = + − = We show uniqueness by contradiction: suppose to obtain a contradiction that g f q r and g f q r , where q q or r r and deg r ,r = 1 + 1 = 2 + 2 1 6= 2 1 6= 2 1 2 ≤ n 1. Then f (q q ) (r r ) 0. Let pk be the largest power of p that − 1 − 2 + 1 − 2 = divides (q q ) and (r r ). Then f (q q ) (r r ) 0 can be rewritten 1 − 2 1 − 2 1 − 2 + 1 − 2 = k as p (f q0 r 0) 0, hence f q0 r 0 0, where deg r 0 n 1 and p q0 or p r 0. + = + = ≤ − 6 | 6 | n n It follows that f q0 x τ (f q0) r 0 0, but p f q0 x τ (f q0) since p a for − n + = | − n | v

0 v n 1 so p r 0. By reducing f q0 r 0 modulo p it follows that p f q0, but ≤ ≤ − | + |

16 2.1. P-ADIC NUMBERS

p f by hypothesis so p q0. Since also p r 0, this is a contradiction. Therefore 6 | | | q0 r 0 0 so q q and r r . = = 1 = 2 1 = 2

Theorem 2.1.18 (p-adic Weierstrass Preparation Theorem). Every nonzero power

X∞ v series f av x Zp [[x]] admits a unique representation = v 0 ∈ =

f pµPU, =

where µ Z 0, U is a unit in Zp [[x]] and P Zp [x] is a monic polynomial ∈ ≥ ∈ satisfying P xdeg P (mod p). ≡

µ µ Proof. Let f be a power series in Z [[x]] such that f p f 0, where p is the p = X∞ v largest power of p dividing f and f 0 av0 x , where p ai0 for i 0,...,n 1 = v 0 | = − = n and p a0 . Then x f 0q r by Proposition 2.1.17, where r is a polynomial 6 | n = + with deg r n 1 in Z [x]. One then sees that ≤ − p

n ³ X∞ v ´ x r f 0q av0 x q (mod p) − = ≡ v n = so we must have r 0 (mod p). Setting P xn r , we have P xdeg P (mod p). ≡ = − ≡ n Let q0 be the constant term of q. Now the coefficient of x is 1 and we have

1 1 q a0 (mod p). This implies q is a unit in Z [[x]]. Therefore setting U q− ≡ 0 n p = 1 n µ we get f 0 q− (x r ) UP and conclude that f p UP as desired. Since = − = =

17 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

1 n n f 0 q− (x r ) with deg r n 1, this equation can be rearranged as x = − ≤ − = f 0q r , so the uniqueness of P and U now follow from Proposition 2.1.17. +

2.2 Background on

In this section we will provide the necessary background in the theory of com- mutative rings. All rings are assumed commutative with identity.

Definition 2.2.1. An integral domain is a ring that is a subring of some field.

Equivalently, an integral domain is a non trivial with no nonzero divisors of 0 in R [10, Definition 3.4 and Definition 6.3]

Two elements of a ring R are called associates if one is a unit times the other. This means x and y are associates in R if and only if y ux for some = unit u of R. Irreducibility is now defined as follows (see [10, Definition 9.1]).

Definition 2.2.2. A nonzero element x of an integral domain R is irreducible if it is not a unit, and its only divisors are units and associates of x.

We now give the definition of unique factorization domain (UFD), princi- pal ideal domain (PID) and discrete valuation domain (DVR) (see [10, Defini- tion 9.4 and Definition 10.1] and [2, Theorem 16.2.7]).

18 2.2. BACKGROUND ON RING THEORY

Definition 2.2.3. An integral domain R in which every nonzero element has a unique factorization into irreducible elements in R is called a unique factor- ization domain. An integral domain is a if every ideal is principal. A discrete valuation domain is a PID with a unique .

We now define the prime ideals of a ring.

Definition 2.2.4. Let R be a commutative ring and let Q be a proper ideal in R.

Then Q is prime if and only if R/Q is an integral domain.

Now the definition of height of a is given (see [10, 14.2]).

Definition 2.2.5. Let Q be a prime ideal of a ring R. The height of Q is the largest non-negative integer k such that there is a strictly descending chain

Q Q Q Q Q = 0 ⊃ 1 ⊃ 2 ⊃ ··· ⊃ k

of prime ideals Qi of R.

Next we define the Krull dimension of a ring (see [10, 14.1]).

Definition 2.2.6. The (Krull) dimension of a ring R is the supremum of the heights of the prime ideals of R. We write dim R for the dimension of R.

19 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

2.3 Power Series Rings

This section discusses power series rings, in particular, Z[[x]]. Some basic properties of irreducibility and reducibility will be discussed for Z[[x]] as well as the prime spectrum for Z[[x]]. The discussion of irreducibility in Z[[x]] starts with a discussion of the units in Z[[x]] and the units of power series rings in general. The following result is well known.

Proposition 2.3.1. Let R be a commutative ring, and let Pa xn R[[x]]. Then n ∈ P n an x is a unit in R[[x]] if and only if a0 is a unit in R.

P n Proof. Suppose an x is a unit in R[[x]]. Then, there exists a power series

Pb xn in R[[x]] such that (Pa xn)(Pb xn) 1. This implies a b 1 so a n n n = 0 · 0 = 0 is a unit.

Now suppose that a0 is a unit in R. We inductively define the multiplica-

P n P n tive inverse bn x of an x .

1 b a− 0 = 0

1 b a− (a b ) 1 = − 0 1 0 P Now assume that the first n terms b0,b1,...,bn 1 of the inverse of an are − defined in a similar way. Define bn as follows:

1 bn a0− (bn 1a1 bn 2a2 b0an). = − − + − + ··· +

20 2.3. POWER SERIES RINGS

Then one checks that (Pb xn)(Pa xn) 1. n n =

Since the only units in Z are 1, we have the following corollary. ±

Corollary 2.3.2. Let Pa xn Z[[x]]. Then, Pa xn is a unit if and only a n ∈ n 0 = 1. ±

Now we have the following from [10, Theorem 13.6].

Theorem 2.3.3. If R is a principal ideal domain, then the power series ring

R[[x]] is a UFD.

The following corollary follows immediately from the theorem since Z is a PID.

Corollary 2.3.4. The power series ring Z[[x]] is a UFD.

Next we present some known criteria for indicating whether or not power series in Z[[x]] are irreducible.

Lemma 2.3.5. Let Pa xn Z[[x]] such that a is a prime integer. Then Pa xn n ∈ 0 n is irreducible.

Proof. Suppose Pa xn (Pb xn)(Pc xn) in Z[[x]]. Then b c p. Since p n = n n 0 0 = is prime either b 1 or c 1, and one of the factors of Pa xn must be a 0 = ± 0 = ± n P n unit. Therefore an x is irreducible by Proposition 2.3.1.

21 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

The next proposition shows that if a power series in Z[[x]] has a nonzero constant term whose absolute value is not a prime power, then the power se- ries is reducible. This follows from [1, Proposition 3.4].

Proposition 2.3.6. Let Pa xn Z[[x]]. Then if a 0,1 and a is not a n ∈ | 0| 6= | 0| P n prime power then an x is reducible.

P n Proof. It will be shown that the power series an x can be factored as

(Pb xn)(Pc xn). If a 0,1 and a is not a prime power then a can be n n | 0| 6= | 0| 0 factored into two relatively prime nonunit factors b0 and c0. Since b0 and c0 are relatively prime, there exist α and β in Z such that b α c β 1. It now 0 + 0 = follows from here that a b αa c βa . Now setting b βa and c αa 1 = 0 1 + 0 1 1 = 1 1 = 1 we get a b c c b . Now assume all the b ’s and c ’s have been found up 1 = 0 1 + 0 1 i i Ãn 1 ! X− to i n 1. Then we require an b0cn bi cn i bnc0, or = − = + i 1 − + = n 1 X− an bi cn i b0cn bnc0, − i 1 − = + = which follows if we set

n 1 n 1 ³ X− ´ ³ X− ´ bn β an bi cn i and cn α an bi cn i . = − i 1 − = − i 1 − = =

Therefore by induction Pa xn (Pb xn)(Pc xn). n = n n

We now have the following corollary.

22 2.3. POWER SERIES RINGS

Corollary 2.3.7. Every irreducible power series in Z[[x]] besides any associate of x has constant term pn for some positive integer n. ±

The next proposition shows a sufficient condition for irreducibility (see

[1, pp 546]).

Proposition 2.3.8. Let Pa xn Z[[x]] such that a pk for some prime p n ∈ 0 = ± and k 2. If p a , then Pa is irreducible in Z[[x]]. ≥ 6 | 1 n

Proof. Let p a , and suppose to the contrary that Pa xn (Pb xn)(Pc xn) 6 | 1 n = n n where neither factor is a unit. Since a pk and a b c it follows that 0 = ± 0 = 0 · 0 b ps and c pt , where s t k and s,t 1. Now a b c b c 0 = ± 0 = ± + = ≥ 1 = 0 1 + 1 0 = psc pt c implies that p a , a contradiction. 1 + 0 | 1

This condition is only sufficient, not necessary. For instance the polyno- mial 4 2x 3x2 is irreducible in Z[[x]] and 2 divides a (see [1, pp 546]). At + + 1 this point, no criteria has been established to handle all cases of irreducibility.

Birmajer and Gil have the following additional criteria (see [1, Lemma 4.1 and

Theorem 4.3]).

Lemma 2.3.9 (Birmajer and Gil). Let Pa xn Z[[x]] with a p2 for some n ∈ 0 = prime p and assume that p a . If a αβ (mod p) for every α,β Z/pZ such | 1 2 6≡ ∈ that a /p α β (mod p), then Pa xn is irreducible. 1 ≡ + n

23 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

Theorem 2.3.10 (Birmajer and Gil). Let Pa xn with a p2 for some prime n 0 = p. Assume that for some m we have p2 a for j 1,...,m, and p a for every | j = | j even number j m 1,...,2m. If p does not divide a2m 1, or if p a2m 1 but ∈ + + | + a2m 2 αβ (mod p) for every α,β Z/pZ such that am 1/p α β (mod p), + 6≡ ∈ + ≡ + P n then an x is irreducible.

Now I prove the following proposition, which classifies the prime ideals of

Z[[x]].

Proposition 2.3.11. Let Z[[x]] be the ring of formal power series in x with co- efficients in Z. Then the following can be said about the prime ideals of Z[[x]]:

1. (0) is the unique minimal prime ideal.

2. All maximal ideals are height 2 of the form (p,x), where p is prime in Z.

3. There is a unique principal prime ideal, namely (x), contained in every

maximal ideal.

4. All height 1 prime ideals are principal, generated by an irreducible power

series.

24 2.3. POWER SERIES RINGS

5. Every maximal ideal (p,x) contains its own unique uncountable set κp

of height one prime ideals, where each ideal is generated by an irreducible

power series f with constant term pk for some k.

6. Every maximal chain of prime ideals has one of the following forms:

(p,x) (x) (0) or (p,x) (f ) (0), ⊃ ⊃ ⊃ ⊃

where (f ) κ . ∈ p

7. dimZ[[x]] 2 =

Proof. Statement 1 is trivial. To show statement 2 we have from [10, Theorem

12.4] that the maximal ideals of a power series R[[x]] are of the form M (x), + where M is a maximal ideal in R. It immediately follows since the maximal ideals of Z are those generated by a prime that the maximal ideals of Z[[x]] are (p,x) where p is a prime in Z. This implies that (p,x) (x) (0), so (p,x) ⊃ ⊃ has height greater than or equal to 2. Now from [10, Theorem 14.10] if a ring

R is Noetherian then

dimR[[x]] dimR 1. = +

Any chain of prime ideals in Z has the following form: (p) (0), where p is ⊃ a prime integer. This implies Z is Noetherian and dimZ 1 so dimZ[[x]] = =

25 CHAPTER 2. BACKGROUND ON P-ADIC NUMBERS AND POWER SERIES RINGS

2. Therefore, all maximal ideals in Z[[x]] have height at most 2, hence have height exactly 2.

\ Statement 3 follows since (p,x) (x). p = To get statement 4 we have that Z[[x]] is a UFD by Corollary 2.3.4 and by

[6, Theorem 20.1] all height one prime ideals in UFD are principal. Moreover in a UFD an element is prime if and only if the element is irreducible [10, p.

92]. Therefore all height one prime ideals must be generated by an irreducible power series.

For statement 5, we claim that for each (p,x) the set κp is the set of all prime ideals generated by irreducible power series with constant term pk for some k. By Corollary 2.3.7 all irreducible elements besides associates of x have a prime power as the absolute value of its constant term. Since Z[[x]] is a UFD an element is prime if and only if it is irreducible. This means all height

1 prime ideals, except (x), will be generated by an irreducible power series f with constant term pk . To show uniqueness, we show any irreducible power series f pk a x must be in (p,x). Suppose f pk a x (q,x), = + 1 + ··· = + 1 + ··· ∈ where q is prime; then f can be written as r q sx where r,s Z[[x]]. This + ∈ means pk (q) and a x (x). If pk (q) then p (q) and q p. However ∈ 1 + ··· ∈ ∈ ∈ | p and q are prime; so q must be an associate of p, meaning q p. One can = ±

26 2.3. POWER SERIES RINGS conclude (q,x) (p,x) since the ideal (u p) (p) for any unit u. The fact that = · = the set κp is uncountable follows from [11, Theorem 4.3].

Statements 6 and 7 follow from combining the information from 1 to 5.

The following prime , similar to those found in [11, p. 16], helps explain the previous proposition.

(2,x) (3,x) (5,x) ··· OO oOO ggOO ddd OO oo OOgggg ddOdOdddd OoOOoo gggggOdOdOdddd OOO oo ggOdgOdgddddd OO OO dogodgodgdgdgdd OO OO OO (x) κ κ κ 2 3 5 ··· OO o gg ddd OO oo gggg ddddddd OOO ooo gggggdddddd OO oo ggdgdgddddd OOdogodgodgdgdgdd (x)

Now that we know what the prime ideals of Z[[x]] are, we can use these to classify all the integral domains that arise as quotient rings of Z[[x]]. We will classify them in the next Chapter.

27 CHAPTER

3 Proofs and Examples

3.1 Proof of Main Theorem

In this section we present a complete proof of the Main Theorem. To begin

this section the Main Theorem will be restated followed by subsequent propo-

sitions and finally the proof of the Main Theorem.

Theorem 3.1.1 (Main Theorem). The integral domains arising as a quotient

rings of Z[[x]] are, up to isomorphism, precisely the following:

Z[[x]],(Z/pZ)[[x]], Z, Z/pZ, Zp [α]

where p is prime and α is any element of Q with α 1. Moreover, Z [α] p | |p < p is a local ring with maximal ideal (p,α) and is a DVR if and only if (p,α) is

principal.

28 3.1. PROOF OF MAIN THEOREM

Before the Main Theorem is proved, we will discuss the case, where the quotient ring is isomorphic to Zp [α]. This is the most difficult and most in- teresting case. First it will be shown that for every irreducible power series f pk a x in Z[[x]] not associate to p there exists an α as described in = + 1 + ··· the theorem.

Proposition 3.1.2. Let f Z[[x]] be a nonzero power series. Write f pµPU ∈ = as in the Weierstrass Preparation Theorem. Let α C be a root of f . Then the ∈ p following are true.

1. α 1 if and only if α is a root of P. | |p <

2. α 1 if and only if α is a root of U. | |p ≥

Moreover, if P is not constant then P has a root in Cp and therefore f has a root

β C with β 1. ∈ p | |p <

Proof. We first show if α is a root of f then α is a root of P or α is a root of U.

Since P is clearly a unit in Q ((x)) ¡Q [[x]]¢[1/x] (the ring of formal Laurent p = p series over Q ), one has U f /pµP in Q ((x)). If α is not a root of P, then p = p P(α) 0, so U(α) converges in C and equals f (α)/pµP(α) 0. 6= p = Let P xn P x P in Z [x]. Suppose α is a root of P and α 1. = + ··· + 1 + 0 p | |p ≥ Then α n P α i for all i 0,...,n. Since α is a root of P it follows that | |p > | i |p | |p =

29 CHAPTER 3. PROOFS AND EXAMPLES

n n n 1 P(α) p 0. However P(α) p α P1α P0 p max( α p , Pn 1α − | | = | | = | +···+ + | ≤ | | | − + P α P ). Now by the non-archimedean property αn P α P ···+ 1 + 0|p | +···+ 1 + 0|p = n n 1 n n 1 max( α p , Pn 1α − P1α P0 p ) when α p Pn 1α − P1α | | | − + ··· + + | | | 6= | − + ··· + + n 1 i P0 p . This is the case since Pn 1α − P1α P0 p max Pi α p | | − + ··· + + | ≤ | | = max P α i αn α n . Therefore 0 αn P α P α n 1 | i |p | |p < | |p = | |p = | + ··· + 1 + 0|p = | |p ≥ a contradiction. Therefore every root α of P has α 1 and every root α of f | |p < with α 1 must be a root of U. | |p ≥ Let U u u x . Now suppose that α is a root of U and α 1. = 0 + 1 + ··· | |p < It follows that u u α u α2 α( u u α ) so 1 α u 0 = − 1 − 2 − ··· = − 1 − 2 − ··· = | |p | − 1 − 2 n u2α p α p max(1, α p , α p ,...) max α p α p 1, a contradiction. − ···| ≤ | | | | | | = n 1 | | = | | < ≥ Hence every root α of U has α 1 and every root of f with α 1 must be | |p ≥ | |p < a root of P.

Finally, the last statement of the proposition is clear.

Now we have the following proposition.

Proposition 3.1.3. Let α C be a p-adic root of an irreducible power series ∈ p f pk a x Z[[x]] with α 1. Then, the function φ defined by = + 1 + ··· ∈ | |p <

φ : Z[[x]] C , → p

30 3.1. PROOF OF MAIN THEOREM where

h h(α) 7→ for all h Z[[x]], is a well-defined with equal to ∈

(f ) and image equal to Zp [α]. Therefore Z[[x]]/(f ) Zp [α]. =∼

Proof. First we show that this is a well-defined mapping, or equivalently all power series in Z[[x]] when evaluated at α will converge in Cp . To begin, all numbers in Z have p-adic absolute value less than or equal to 1. So taking a power series Pa xn in Z[[x]] and evaluating it at α we see that a αn n | n |p ≤ n n n α p 1 for all n. Now we have that lim α p lim α 0, and we can n n p | | < →∞ | | = →∞ | | = conclude by Lemma 2.1.13 that the series converges in Cp . This shows the mapping is well-defined.

Next we show this mapping is indeed a homomorphism. Note that 0 and 1 in Z[[x]] trivially map to 0 and 1 in Cp . To show addition is pre-

P n P n served, let an x and bn x be in Z[[x]]. Evaluating φ at each power ³ ´ ³ ´ series and adding we see that φ Pa xn φ Pb xn Pa αn Pb αn, n + n = n + n ³ ´ which by Lemma 2.1.14 converges and is equal to P a αn b αn P(a n + n = n + ³ ´ ³ ´ b )αn φ P(a b )xn φ Pa xn Pb xn . Hence φ preserves addition. n = n + n = n + n Now, we show multiplication is preserved using the same arbitrary power se- ³ ´ ³ ´ ³ ´ ³ ´ ries. We get φ Pa xn φ Pb xn Pa αn Pb αn which by Lemma n · n = n · n

31 CHAPTER 3. PROOFS AND EXAMPLES

n n X∞ X i n i X∞ ³X ´ n 2.1.15 converges and is equal to ai α bn i α − ai bn i α n 0 i 0 − = n 0 i 0 − = n n = = = = ³ X∞ ³X ´ n´ ³ X∞ X i n i ´ ³X n X n´ φ ai bn i x φ ai x bn i x − φ an x bn x . Hence n 0 i 0 − = n 0 i 0 − = · = = = = multiplication is also preserved. Thus φ is a well-defined homomorphism.

Next we will prove that the image of this homomorphism is Zp [α].

To show this we will first show that Z [α] im φ and then we will show p ⊂ that im φ Z [α]. ⊂ p X∞ i k 2 Let ci p Zp . Note that f (α) 0 implies that p a1α a2α . i 0 ∈ = = − − − ··· = 2 X∞ n Set g a1x a2x . Let s bn g , where bn cnk cnk 1p = − − − ··· = n 0 = + + + = k 1 Pk 1 j c(n 1)k 1p − j −0 cnk j p . Since φ is continuous, evaluating at ··· + + − = = + µ ¶ k 1 X∞ n X∞ n X∞ nk X∞ ³ X− j ´ nk φ gives φ bn g bn g(α) bn p cnk j p p n 0 = n 0 = n 0 = n 0 j 0 + ======k 1 X∞ X− nk j cnk j p + . Since every nonnegative integer can be written uniquely n 0 j 0 + = = k 1 X∞ X− nk j in the form nk j for some n and j, we have, φ(s) cnk j p + + = n 0 j 0 + = = = X∞ i ci p . Therefore Zp im φ. Finally, since α im φ, we have Zp [α] im φ. i 0 ⊂ ∈ ⊂ = Now we show im φ Z [α]. By the p-adic Weierstrass Preparation The- ⊂ p orem f pµPU, where U is a unit in Z [[x]] and P is a monic polynomial in = p Z [x] with P xn (mod p) and α is a root of P by Proposition 3.1.2. If P pb p ≡ = 0+ n n n 1 pb1x x then α pb0 pb1α pbn 1α − p( b0 b1α + ··· + = − − − ··· − − = − − − ··· − n 1 n 1 n P i bn 1α − ). Set β b0 b1α bn 1α − so α pβ. Let ci x Z[[x]]. − = − − − ··· − − = ∈

32 3.1. PROOF OF MAIN THEOREM

n 1 n 1 ³X i ´ X i X∞ X− in j X∞ X− i i j It follows that φ ci x ci α cin j α + cin j p β α . = = i 0 j 0 + = i 0 j 0 + = = = = n 1 i j i j X− k Since cin j Z and β ,α Zp [α] it follows that cin j β α di,j,k α , + ∈ ∈ + = k 0 n 1 n 1 = ³X i ´ X∞ X− X− k i where di,j,k Zp for all i, j,k. We then have φ ci x di,j,k α p . ∈ = i 0 j 0 k 0 = = = n 1 Ã Ãn 1 ! ! X− X∞ X− i k These sums can be rearranged as follows: di,j,k p α . The sum k 0 i 0 j 0 = = = n 1 X− di,j,k has a finite number of terms in Zp so lies in Zp . The infinite sum j 0 = Ãn 1 ! X∞ X− i i di,j,k p converges in Zp since lim p p 0. Therefore, the finite sum i 0 j 0 i | | = = = →∞ n 1 Ã Ãn 1 ! ! X− X∞ X− i k di,j,k p α lies in Zp [α]. Therefore im φ Zp [α], and this shows k 0 i 0 j 0 ⊂ = = = im φ Z [α]. = p Finally we show ker φ (f ). Since Z[[x]]/ker φ im φ, is an integral do- = =∼ main, ker φ is a prime ideal. Now φ(f ) f (α) 0 which implies f ker φ. = = ∈ Since f is irreducible we have the following chain of prime ideals: (0) (f ) ⊂ ⊂ ker φ. By Proposition 2.3.11 dim Z[[x]] 2. If ker φ (f ) then ker φ must = 6= be maximal. However, this is a contradiction, since Zp [α] is not a field. This shows that (f ) ker φ. =

To prove the converse of Proposition 3.1.3 we need the following lemmas.

X∞ ki i Lemma 3.1.4. Let f p vi x be a power series in Zp [[x]] with a nonzero = i 0 = constant term, where each vi is a unit in Zp . Then there exists a unit power se-

33 CHAPTER 3. PROOFS AND EXAMPLES

X∞ i X∞ i riesU ui x in Zp [[x]] such that g U f gi x is in Z[[x]] with constant = i 0 = = i 0 = = term pk0 .

1 k0 1 k0 Proof. Set u v− . Then g p v v− p and U once defined will be a 0 = 0 0 = 0 0 = j X ki unit in Zp [[x]]. We will now define u j such that g j p vi u j i is in Z. To do = i 0 − j j = j X ki k0 X ki X ki t j so note g j p vi u j i p v0u j p vi u j i . Set p vi u j i p w j = i 0 − = + i 1 − i 1 − = = = = X∞ i t j where w j w ji p − is in Zp . To define each u j there are two cases: k0 t j = i t > = j 1 X∞ i k0 and k t . For case 1, k t so set u v− w p − . It then fol- 0 ≤ j 0 > j j = − 0 ji i k0 = k0 t k0 1 X∞ i k0 t X∞ i t lows that g p v u p j w p v v− w p − p j w p − j j = 0 j + j = − 0 0 ji + ji = i k0 i t j = = k0 1 − t j X i t j 1 t j k0 p w ji p − which is in Z. For case 2 k0 t j so set u j v0− p − w j . i t ≤ = − = j k0 t k0 1 t k0 t Then g p v u p j w p v v− p j − w p j w 0 which is also in j = 0 j + j = − 0 0 j + j = Z.

Lemma 3.1.5. Let α C . Then α Q with α 1 if and only if α is a root of ∈ p ∈ p | |p < a polynomial in Z [x] such that P xdegP (mod p). p ≡

Proof. First note α is a root of a minimal irreducible polynomial P xn = +···+ P x P in Q [x]. It follows that α pn P 1 by the Definition 2.1.10. 1 + 0 p | |p = | 0|p < One easily sees from this that P 1 so P pZ . Now by [3, Lemma 5.3.6] | 0|p < 0 ∈ p all the coefficients Pn 1,...,P1,P0 are in Zp . −

34 3.1. PROOF OF MAIN THEOREM

Now assume some of the coefficients Pn 1,...,P1 are not divisible by p. − Reducing modulo p, we get P xn P xk (mod p), where k is the least ≡ + ··· + k positive integer such that Pk is not divisible by p. Factoring modulo p yields

n k k P (x − P )x (mod p). One can now apply Hensel’s Lemma because ≡ + ··· + k n k k x − P and x are relatively prime modulo p. Therefore P is reducible + ··· + k in Z [x], which is a contradiction. Hence P xdegP (mod p). The converse p ≡ follows from the proof of Proposition 3.1.2.

Proposition 3.1.6. Let α Q with α 1. Then the integral domain Z [α] ∈ p | |p < p is isomorphic to a quotient ring of Z[[x]].

Proof. By Lemma 3.1.5, α is a root of a polynomial P such that P ≡ xdegP (mod p). Now by Lemma 3.1.4 there exist a power series f in Z[[x]] such that f UP with constant term equal to pk , where U is a unit in Z [[x]]. = p

Since α p 1 Proposition 3.1.3 applies. If f is irreducible then Z[[x]]/(f ) | | < =∼

Zp [α]. If f is reducible then f can be factored into irreducible factors as f (pk1 a x ) (pkn a x ). Suppose without loss of generality = ± + 11 + ··· ··· + n1 + ···

k1 that α is a root of g p a11x . Then Z[[x]]/(g) Zp [α] by Proposition = + + ··· =∼

3.1.3. Therefore Zp [α] is isomorphic to a quotient ring of Z[[x]].

We are now ready to prove the Main Theorem.

35 CHAPTER 3. PROOFS AND EXAMPLES

Proof of Main Theorem. By Proposition 2.3.11 and Proposition 3.1.3 we get the following exhaustive table of prime ideals Q in Z[[x]] and their associated quotient ring Z[[x]]/Q:

Q Z[[x]]/Q

(0) Z[[x]]

(p,x) Z/pZ

(x) Z

(p) (Z/pZ)[[x]]

(f ) Zp [α] where f is any irreducible power series in Z[[x]] not associate to p or x and

α Q is any root of f with α 1. (Note that f must have at least one such ∈ p | |p < root by Proposition 3.1.2.) Moreover, by Proposition 3.1.6 every ring of the form Z [α] with α Q and α 1 is isomorphic to a quotient ring of Z[[x]]. p ∈ p | |p <

Thus the only statement in the theorem to prove is that Zp [α] is a local ring with maximal ideal (p,α) and is a DVR if and only if (p,α) is principal.

According to [6, Theorem 11.2], a ring is a DVR if and only if the ring is a

Noetherian local ring with Krull dimension greater than zero and has a unique maximal ideal that is principal. We will show that Zp [α] is a Noetherian local ring with dimension greater than zero.

36 3.1. PROOF OF MAIN THEOREM

It follows that Zp [α] is Noetherian since Z[[x]]/(f ) Zp [α], because any =∼ quotient ring of a is Noetherian. By [2, Theorem 7.3.8], there is a bijection between the maximal ideals of Z[[x]] that contain (f ) and the maximal ideals of Z[[x]]/(f ) Zp [α]. By Proposition 2.3.11, the maximal ide- =∼ als of Z[[x]] have the form (p,x) and every ideal (f ), generated by an irre- ducible power series, is contained in exactly one maximal ideal. This implies that there is only one maximal ideal containing (f ) and only one maximal ideal in Z[[x]]/(f ) Zp [α] equal to (p,α). This shows Zp [α] is local and has =∼ dimension greater than zero. Now it follows from [6, Theorem 11.2] that Zp [α] is a DVR if and only if (p,α) is principal, thus completing the proof.

The results in this section yield the following proposition.

Proposition 3.1.7. Let α C with α 1. Then α Q if and only if α is a ∈ p | |p < ∈ p root of some nonzero power series f Z[[x]]. ∈

Proof. First suppose that α Q . We may assume α 0. By Lemma 3.1.5 α ∈ p 6= is a root of a polynomial P Z [x] such that P xdeg P (mod p). By factoring ∈ p ≡ out a power of x, we may assume P has a nonzero constant term. Then by

Lemma 3.1.4 there exists a unit power series U Z [[x]] such that g UP is ∈ p = in Z[[x]]. Therefore α is a root of the nonzero power series g in Z[[x]].

37 CHAPTER 3. PROOFS AND EXAMPLES

Conversely assume α is a root of some nonzero power series f Z[[x]]. ∈ f pµPU as in the p-adic Weierstrass Preparation Theorem. It follows from = Proposition 3.1.2 that α must be a root of P since α 1. Therefore α | |p < ∈

Qp .

3.2 Examples

Example 3.2.1. For any prime p one has Z[[x]]/(p x) Zp . − =∼

Proof. Since p is prime the power series p x is irreducible by Lemma 2.3.5, − with p as a root and p 1 1. Therefore by the Main Theorem Z[[x]]/(p | |p = p < − x) Zp [α] Zp . =∼ =

2 Example 3.2.2. Z[[x]]/(2 7x 3x ) Z2. + + =∼

Proof. 2 7x 3x2 is irreducible by Lemma 2.3.5, since 2 is prime. However + + 2 7x 3x2 factors as (2 x)(1 3x) where 1 3x is a unit in Z[[x]]. Thus 2 is + + + + + − a root of 2 x with 2 1 1. Therefore by the Main Theorem Z[[x]]/(2 + | − |p = 2 < + 2 7x 3x ) Z2[ 2] Z2. + =∼ − =

Example 3.2.3. For any prime p and any positive integer n one has Z[[x]]/(pn − x) Zp . =∼

38 3.2. EXAMPLES

Proof. It follows from Proposition 2.3.8 that pn x is irreducible. It has a root − n n 1 n α p where p p n 1. Now by the Main Theorem, Z[[x]]/(p x) = − |− | = p < − =∼ Z [ pn] Z . p − = p

Example 3.2.4. For any prime p and any positive integer n one has Z[[x]]/(p −

n n x ) Zp [pp]. =∼

Proof. First, note p xn is irreducible by Lemma 2.3.5. Now we get pn p n − | |p =

1 n 1 p , so pp n 1. Therefore, the Main Theorem applies and | |p = p | |p = pp <

n n Z[[x]]/(p x ) Zp [pp]. − =∼

n n 1 Note that pp Q if n 1 because pp n , which is irrational, but 6∈ p > | |p = pp

n for any r in Qp , r p is rational. Thus Zp [pp] Zp . | | =6∼

3 2 Example 3.2.5. For any prime p one has Z[[x]]/(p x ) Zp [ppp]. − =∼

Proof. To show p3 x2 is irreducible, suppose to the contrary that p3 x2 − − = P i 2 P i (p i∞ 1 ai x )(p i∞ 1 bi x ) for some ai ,bi Z. The following equations + = + = ∈ must be satisfied: p2a pb 0 and p2a a b pb 1. Solving the 1 + 1 = 2 + 1 1 + 2 = − first for b gives b pa . Substituting for b in the second equation gives 1 1 = − 1 1 p2a pa2 pb 1 and implies p 1, a contradiction. 2 − 1 + 2 = − | −

39 CHAPTER 3. PROOFS AND EXAMPLES

The roots of p3 x2 are ppp. Let α ppp. It follows that ppp 2 − ± = | |p = p3 1 , so ppp 1 1. Now the Main Theorem applies and | |p = p3 | |p = p3/2 < 3 2 Z[[x]]/(p x ) Zp [ppp]. − =∼

Example 3.2.6. 6 x can be factored as (2 a x )(3 b x ) Z[[x]]. − + 1 + ··· + 1 + ··· ∈

Moreover Z[[x]]/(2 a1x ) Z2, Z[[x]]/(3 b1x ) Z3 and Z[[x]]/(6 + + ··· =∼ + + ··· =∼ − x) Z2 Z3. =∼ ×

Proof. 6 x factors in Z[[x]], because 6 is not a prime power and can be fac- − tored as (2 a x )(3 b x ) by the proof of Proposition 2.3.6. Notice + 1 + ··· + 1 + ··· that 2 a x and 3 b x are irreducible since 2 and 3 are primes. It fol- + 1 +··· + 1 +··· lows that 6 is a 2-adic root of 2 a x and a 3-adic root of 3 b x since + 1 +··· + 1 +··· 3 b x maps to a 2-adic unit and 2 a x maps to a 3-adic unit when + 1 +··· + 1 +··· they are evaluated at 6. Therefore by the Main Theorem, Z[[x]]/(2 a1x ) + +··· =∼

Z2 and Z[[x]]/(3 b1x ) Z3. To show the last isomorphism, note that + + ··· =∼ (2 a x ) (3 b x ) 1 c x which is a unit. Therefore 1 is in + 1 + ··· − + 1 + ··· = − + 1 + ··· ¡(2 a x ),(3 b x )¢ and ¡(2 a x ),(3 b x )¢ Z[[x]]. It now + 1 +··· + 1 +··· + 1 +··· + 1 +··· = follows from the Chinese Remainder Theorem (see [2, Proposition 16.3.18]) that Z[[x]]/(6 x) Z2 Z3. − =∼ ×

40 3.3. FURTHER DIRECTIONS

3.3 Further Directions

We have shown that Z[[x]]/(f ), where f is irreducible and not associate to p or x, is isomorphic to Z [α], where α is a root of a monic polynomial P p ≡ xdegP (mod p), has p-adic absolute value less than 1, and is algebraic over

Qp . In addition we proved the converse: Zp [α], where α has the properties described above, is isomorphic to a quotient ring of Z[[x]].

There are still no irreducibility criteria for Z[[x]]. The p-adic Weierstrass

Preparation Theorem shows any power series in Z[[x]] can be factored as pµPU, where P is a monic polynomial with coefficients in Zp . The irreducibility of a power series might be able to be determined by looking at this monic polyno- mial P and then using known irreducibility criteria over Zp [x].

What other types of quotient rings arise in Z[[x]]? This paper only focused on the quotient rings that are integral domains. Classifying all the quotient rings of Z[[x]] is not yet solved. The research done in this paper might be useful in classifying other quotient rings. For instance, if one was interested in classifying some non-integral domains then Example 3.2.6 is a simple place to start. Lastly, someone may want to try to generalize the results for R[[x]], where R is a PID, or more generally a UFD.

41 Bibliography

[1] D.BIRMAJERAND J.B.GIL, Arithmetic in the ring of formal power series

with integer coefficients, American Mathematical Monthly, 115 (2008).

[2] D.S.DUMMITAND R.M.FOOTE, , John Wiley and Sons

Inc, New Jersey, 2004.

[3] F.Q.GOUVÊA, p-adic Numbers, Springer-Verlag, Berlin, 1997.

[4]N.K OBLITZ, p-adic Numbers, p-adic Analysis and Zeta-Functions,

Springer-Verlag, New York, 1984.

[5]E.K REYSZIG, Introductory Functional Analysis With Applications, John

Wiley and Sons. Inc., 1978.

42 3.3. FURTHER DIRECTIONS

[6]H.M ATSUMURA, Commutative ring theory, Cambridge University Press,

Cambridge, 1992.

[7]J.N EUKIRCH, , Springer-Verlag, Berlin, 1999.

[8] W. RUDIN, Principles of Mathematical Analysis, McGraw-Hill, Inc, Singa-

pore, 1976.

[9]L.W ASHINGTON, Introduction to Cyclotomic Fields, Springer-Verlag, New

York, 1982.

[10]J.W ATKINS, Topics In Commutative Ring Theory, Princeton University

Press, New Jersey, 2007.

[11]R.W IEGANDAND S.WIEGAND, Prime ideals in notherian rings: A survey,

(2009).

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