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| serpicojr |
 Posted: Tue Jun 17, 2008 5:59 pm Post subject: Algebraic Number Theory |
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Alright, I'm going to start a new thread on algebraic number theory. This is my area of research, and so I have a lot of passion for the subject. I'll be working out of Neukirch's text on the subject.
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So algebraic number theory is "algebraic" for two main reasons. First, it studies so-called algebraic numbers, which I'll define shortly. Second, many of the tools used in it come from algebra: at first linear algebra and Galois theory play a major role, with representation theory and group cohomology becoming the main tools of study as we progress (although I'm not so sure we'd make it so far in this thread). However, there is a lot of geometry, analysis, and topology floating around, too.
Our studies will look like very simple commutative algebra at first. This is because we're mostly dealing with rings which have very nice properties--fields (and a pretty restrictive set of fields, at that) and nice subrings of fields. But at some point we have to start using the special properties of the fields we're looking at, and this is where a lot of the meat of the theory and some of the less algebraic methods start popping up.
What do you, the reader, have to know? Linear algebra, basic abstract algebra, and basic Galois theory will be assumed, as will be some basic facts about the topology of the real and complex numbers and metric spaces. Unfortunately, you'll be left behind if you don't know this material.
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Now let me develop some of the basic questions we'll be asking. An algebraic number is a number which satisfies a polynomial with coefficients in Q, the rationals. 21/2 and i are some famous algebraic numbers, the former satisfying x2-2, the latter x2+1.
Any algebraic number α (that's supposed to be an alpha) determines a finite field extension of Q, the field K = Q(α). In general, we'll call a finite extension of Q a number field. I'm assuming you've seen this before if you're reading this, but let's recall that this ring is the same thing as:
-the set of numbers f(α)/g(α), where f(x), g(x) are polynomials with coefficients in Q, g(α) ≠ 0
-the set of numbers f(α), where f(x) is a polynomial with coefficients in Q
-the quotient ring Q[x]/(h(x)), where h(x) is a minimal polynomial for α over Q (and h(x) is unique if we stipulate that h(x) is monic, i.e. its leading coefficient is 1)
In a sense, we can reduce the study of α to the study of the number field K. Let's recall here that the K is a vector space of dimension n = deg(h(x)) over Q, and so we define the degree of the extension K over Q to be [K:Q] = n.
Note that any algebraic satisfies a polynomial with coefficients in Z. Indeed, if α satisfies h(x), let c be the least common multiple of the denominators of the coefficients of h(x). Then ch(x) has integer coefficients. If in fact α satisfies a monic polynomial with integer coefficients, we will call α an algebraic integer. We will learn shortly that the set of algebraic integers contained in K is a ring, and we'll call this the ring of integers of K. The ring of integers contains all of the information that we need to study K, and the fact that it doesn't have inverses makes some of the algebraic properties of K more evident. So we'll restrict a lot of our attention to this ring.
When K = Q, we naturally have that the ring of integers is just... the ring of integers, Z. One of the main results in elementary number theory is the Fundamental Theorem of Arithmetic--namely, given any nonzero number n ∈ Z, we can write uniquely (up to reordering of the factors):
n = ep1r1...pkrk
where e = ±1, the pi are distinct prime numbers, and ri ≥ 1. So a natural question is... what is the analog of FTA for rings of integers of number fields?
The first problem is that unique prime factorization of elements does not hold. However, using the theory of ideals, we can come up with a suitable approximation to unique factorization. Our first question is then:
1. How can we approximate unique factorization, and how bad does unique factorization fail?
It turns out we can quantize this via a gadget called the class group.
The next problem is that if two numbers have the same factorization in unspecified sense that I suggest above, their quotient may not be ±1. This is because there will generally be a great many algebraic integers u such that u-1 is also an algebraic integers. These algebraic integers will be called units, and they form a multiplicative group, the group of units of K. Our next question is then:
2. What is the structure of the group of units of K?
This question is answered by Dirichlet's Unit Theorem, which tells us quite a bit about the structure of the group.
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Let me know if you're interested and I'll keep going! |
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| river_rat |
| Posted: Wed Jun 18, 2008 1:10 am Post subject: |
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Count me interested - I can barely remember most of this.
Just a quick question, the unique factorization is for elements of the ring of integers in our field i assume? I know this question came about in relation to Fermat's last theorem, do you know the details?
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As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong. |
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| Guitarist |
| Posted: Wed Jun 18, 2008 3:13 am Post subject: |
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| As I am sure it will be good for me, count me also as a reluctant assenter! |
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| serpicojr |
| Posted: Wed Jun 18, 2008 7:20 am Post subject: |
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| river_rat wrote: |
| Just a quick question, the unique factorization is for elements of the ring of integers in our field i assume? |
Indeed!
| Quote: |
| I know this question came about in relation to Fermat's last theorem, do you know the details? |
This is true, and I should know the details, but unfortunately I don't. I believe it has something to do with factorization in cyclotomic fields. |
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| sunshinewarrior |
| Posted: Wed Jun 18, 2008 7:52 am Post subject: |
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On this stuff I don't have enough maths to follow it, so I shall just sit on the sidelines and watch. Fascinating stuff, though, and thanks for doing it. Perhaps some day I can do notes on the short stories of Rudyard Kipling or some such...  |
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| serpicojr |
| Posted: Wed Jun 18, 2008 12:03 pm Post subject: |
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Let me start with an example. Let let's K = Q(i). I claim the ring of integers is Z[i]. Indeed, to see this, suppose z = x+iy is an algebraic integer. If y = 0, then x is a rational number, and so x must be a rational integer. (Exercise 1: prove this!) So let's assume y≠0. Note that z is a root of the polynomial:
h(X) = X2-2xX+x2+y2
This polynomial clearly has rational coefficients. Since z is irrational, h(X) must be irreducible, and so it's the (monic) minimal polynomial of z. z is also an algebraic integer, so z satisfies some monic polynomial f(X) with integer coefficients. h(X) must divide f(X) due to minimality. But there is a theorem (due to Gauss?) that says a monic polynomial with coefficients in an integral domain factors over its field of fractions iff it factors over the domain. Thus h(X) must have rational integer coefficients.
Now we know 2x and x2+y2 are integers, and we wish to show that x and y are integers. There are a lot of ways to do this, and they're all pretty meandering. You just have to play around and find something that works. So x is either an integer or it's half an (odd) integer. If it's half an integer, then for x2+y2 to be an integer, y must also be half an integer. So there are integers m, n such that:
x = (2m+1)/2 = m+1/2
y = (2n+1)/2 = n+1/2
Note that:
x2 = m2+m+1/4
y2 = n2+n+1/4
x2+y2 = m2+m+n2+n+1/2
But this contradicts that x2+y2 is an integer. So x must be an integer. This implies that y2 and hence y are integers.
Z[i] is called the ring of Gaussian integers. Unlike most rings of integers, the Gaussian integers are pretty simple. They satisfy unique factorization, and there are only finitely many units. So they don't exhibit any of the weird behavior I described before, but this is all good and well, as it allows me to introduce some other gadgets we'll be using in a more general setting.
Let's introduce one right now. The norm of a Gaussian integer z = x+iy is the nonnegative real number N(z) = (x+iy)(x-iy) = x2+y2. This is the square of the usual complex modulus, and we've already seen this number as the constant term in the minimal polynomial for z. Note that N is multiplicative--letting * denote complex conjugation, we have:
N(zw) = (zw)(zw)* = zz*ww* = N(z)N(w)
We can immediately use the norm to tell us about the units of Z[i]. Recall that an element u of a (commutative) ring (with identity) R is a unit it has a multiplicative inverse in R. The following is a specific instance of a more general fact.
Theorem: u is a Gaussian unit iff N(u) = 1.
Proof: If u is a Gaussian unit, then u-1 is also a Gaussian integer. So N(u) and N(u-1) must be nonnegative integers. But N(u)N(u-1) = N(uu-1) = N(1) = 1, and so we must have N(u) = 1.
Conversely, if N(u) = 1, then if we let * denote complex conjugation, we have uu* = 1. u* = u-1, and since u is in Z[i], we must have u* is, too. Thus u is a unit. ☐
We can now completely determine the units of Z[i]. I leave the determination of this finite set as Exercise 2. |
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| river_rat |
| Posted: Wed Jun 18, 2008 11:00 pm Post subject: |
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| Quote: |
| But there is a theorem (due to Gauss?) that says a monic polynomial with coefficients in an integral domain factors over its field of fractions iff it factors over the domain. |
Yep, Gauss's polynomial lemma i think its called.
| Quote: |
| This is true, and I should know the details, but unfortunately I don't. I believe it has something to do with factorization in cyclotomic fields. |
I'll consider it homework 
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As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong. |
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| serpicojr |
| Posted: Thu Jun 19, 2008 8:11 am Post subject: |
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Alright, here's a long one. Let me know if it's too much for one sitting and I'll try to keep things shorter in the future.
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It's important to think of the norm as representing a size. This is useful in both geometric and algebraic senses. For example, in the special case of the Gaussian integers we have the
Proposition (Division Algorithm): Let z and w be Gaussian integers, w ≠ 0. Then there exist Gaussian integers q and r with N(r) < N(w) and z = wq+r.
Note: When I apply this proposition, I'll often say I'm dividing z by w. I'll call q the quotient, r the remainder.
Proof: Let v = z/w. v is some number in Q(i). Now picture the complex plane, and imagine the Gaussian integers sitting in the complex plane as the square lattice, i.e. all of the points in the plain with integer coordinates. Any complex number is within sqrt(2)/2 of a Gaussian integer. (Note that the center of each square has maximum distance from the vertices, and this distance is sqrt(2)/2.) Let q be such an integer. Then since the distance between v and q is less that sqrt(2)/2, N(v-q) < 1/2 (here we take the natural extension of N to all elements of K). Multiplying by N(w), we have N(z-wq) = N(wv-wq) = N(w)N(v-q) < N(w)/2 < N(w). Then let r = z-wq.☐
Let's recall some algebraic terminology:
-An integral domain (or simply a domain) is a commutative ring with identity and no zero divisors.
-A principal ideal domain (or a PID) is a domain in which all ideals are principal.
-Let D be a domain.
--An element d of D is irreducible if whenever you write d = ab as a product of elements a, b of D, then either a or b is a unit (equivalently, if a is not a unit, then b is).
--Two elements d and d' are associate to each other if there is a unit u such that d = ud' (equivalently, if d and d' divide each other).
--An element p of D is prime if it is nonzero, not a unit, and whenever p divides a product ab of elements a, b of D, then p divides a or p divides b (equivalently, if p doesn't divide a, then p divides b).
-A unique factorization domain (or a UFD) is a domain in which every element can be written uniquely as a product of irreducible elements, where uniqueness is in the sense that between any two factorizations there is a bijection between the terms of each factorization such that corresponding terms are associate to each other.
And let's recall some facts:
-Every PID is a UFD.
-In a UFD, primes and irreducible elements are the same.
So the Division Algorithm implies that the Gaussian integers are a special kind of domain called a Euclidean domain. I'd rather not spend time talking about such rings, since most rings of integers are not Euclidean. In fact, the property of being Euclidean is more stringent than that of being a PID:
Proposition: Any Euclidean domain is a PID and hence a UFD.
Proof: I'll just prove this for the Euclidean domain Z[i]. The general proof is virtually the same and really only requires the definition of Euclidean domain.
So let I be an ideal of Z[i]. If I = (0), I is prinipal, so assume I ≠ (0). Choose an element w of I of minimal norm greater than 0, which exists because I ≠ (0). I claim that I = (w), the ideal generated by w. If not, then there is an element z in I that is not in (w). Dividing z by w, we have that there exist q and r with N(r) < N(w) and z = wq+r. wq is in I since w is in I, and so r = z-wq is in I. But r has smaller norm than w, contradicting that w has minimal norm. So I = (w). Thus I is a principal ideal domain. ☐
So now we know that the Gaussian integers have unique factorization, so let's figure out what the prime elements are. The first thing to notice is that if π is a prime element, the N(π) is a rational integer divisible by π. Thus to find all prime elements, it suffices to seek prime divisors of nonzero rational integers. Now if a Gaussian prime divides a nonzero rational integer n, then by factoring n into rational primes and applying the definition of a prime element of the Gaussian integers, we see that π must divide some rational prime p. Thus N(π) divides N(p) = p2. Since π is not a unit, we must have N(π) = p or p2.
Conversely, suppose that z is a Gaussian integer and N(z) = p is a rational prime. We'll show that z is irreducible and hence prime. Indeed, if z = wv, w and v Gaussian integers, then N(w)N(v) = N(z) = p. So N(w) = 1 or N(w) = p. If w is not a unit, then N(w) = p, and hence N(zv) = 1. Thus v is a unit, and hence z is irreducible and prime.
Now we will characterize all Gaussian primes by factoring rational primes p into Gaussian primes.
Case I: Assume there exists π with N(π) = p. Then π is prime. Note that N(π*) = p so that π*, too, is prime. There are two possibilities: π and π* are associate, or they're not (which we'll call distinct).
Let π = x+iy. If π and π* are associate, then we must have that π* = x-iy is equal to one of x+iy, -x-iy, -y+ix, y-ix. The first possibility implies y = 0, while the second implies x = 0, either of which implies that p is a rational square, a contradiction to p being a rational prime. The third possibility implies x = -y, while the fourth implies x = y; in either case, we have p = 2x2, and so we must have x = ±1 and N(π) = 2. In particular, note that 2 = (1+i)(1-i) is the product of associate primes. Otherwise, p = ππ* is the product of distinct primes.
Case II: Assume there does not exist π with N(π) = p. p is divisible by some Gaussian prime π, and so N(π) = p2. But you'll show in Exercise 3 that this implies π and p are associate, so that π = ±p or ±ip. Thus p is a prime.
We can reformulate these results in terms of ideal-theoretic language. In a ring R, the product of the ideal I and J is the ideal (Exercise 4):
If R is a PID, then I = (a), J = (b) for some a, b in R, and you can show that (a)(b) = (ab) for Exercise 5. Thus we have:
IA: (2) = (1+i)2 ramifies, being the square of a prime.
IB: If N(π) = p for some π, then (p) = (π)(π*) splits into distinct primes.
II: Otherwise, (p) remains inert, i.e. it's still prime.
Here I use the word "prime" loosely to describe an ideal generated by a prime element, but indeed we have the general notion of a prime ideal. An ideal P is prime if IJ ⊂ P implies I ⊂ P or J ⊂ P (equivalently, I ⊄ P implies J ⊂ P). Then ideals generated by prime elements in a UFD are prime in this sense. Note that, in a PID, unique factorization implies that all ideals factor into a unique (up to order) product of prime ideals. We will find that all rings of integers of number fields satisfy this property. |
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| serpicojr |
| Posted: Fri Jun 20, 2008 9:05 am Post subject: |
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One natural thing to ask is what we can say about a prime given its factorization in the Gaussian integers. So suppose p is a norm, i.e. p = x2+y2. If we assume x, y nonnegative, then we have 0 < x, y < p1/2, so in particular p divides neither x nor y. So it's natural to look at this modulo p:
x2+y2 = 0 (mod p)
x2 = -y2 (mod p)
(x/y)2 = -1 (mod p)
So x/y is a square root of -1 modulo p. This is interesting. This means that if a prime splits in the Gaussian integers, then the polynomial X2+1 splits modulo p as (X-x/y)(X+x/y).
Conversely, suppose -1 is a square modulo p. Then there is 0 < k < p such that k2 = -1 (mod p), i.e. k2 + 1. We may assume that 0 < k ≤ (p-1)/2 by replacing k by -k modulo p. Then note that 1 < k2 + 1 < p. So p divides k2+1 exactly (i.e., no higher power of p divides k2+1). Any prime π dividing p must divide k2+1, and so either π or π* divides k+i. But then N(π) divides N(k+i) = k2+1, and so N(π) = p (as the other option is p2, which does not divide k2+1).
Note that this implies that if a prime p is inert in the Gaussian integers, then the polynomial X2+1 is irreducible modulo p!
Recall that 2 ramifies as the square of a Gaussian prime. Note that X2+1 = (X+1)2 modulo 2--ramification! This completes a demonstration of the
Fact: The factorization of a rational prime p in the Gaussian integers mimics the factorization of the polynomial X2+1 modulo p.
So when is -1 a square modulo p? Let's investigate the group structure of the group of units modulo p. This is a cyclic group of order p-1. Now the group of squares has order (p-1)/2, and so -1 being a square implies that 2 must divide (p-1)/2, i.e. 4 divides p-1, i.e. p = 1 (mod 4). Conversely, if p = 1 (mod 4), then (p-1)/2 is even, and hence the group of squares must have an element of order 2. -1 is the unique element of order 2 in the full group, so it must be the element of order 2 in the group of squares. This proves the
Lemma: -1 is a square modulo p iff p = 1 (mod 4).
Our inquiries culminate in the following rather succinct
Proposition: Let p be a rational prime. Then:
a. if p = 2, then 2 ramifies, 2 = -i(1+i)2, 1+i a Gaussian prime;
b. if p = 1 (mod 4), then p splits, p = ππ* for some Gaussian prime π; or
c. if p = 3 (mod 4), then p is inert, p is a Gaussian prime.
This motivates a third question:
3. Can we systematically determine the factorization of a rational prime in the ring of integers of a number field?
Yeah, this is a poorly defined question... but there is a detailed web of conjectures which describes what we believe the truth to be. We have solved some cases--for example, we can answer this question for abelian extensions of the rationals, i.e. extensions whose Galois groups are abelian groups. We almost certainly won't get to that result, although we will probably describe the case of quadratic extensions of the rationals in its entirety. |
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| Guitarist |
| Posted: Fri Jun 27, 2008 9:27 am Post subject: |
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serpicojr: Let me first thank you for starting this thread. It is a subject I avoided like the plague (whenever I could), but if you have a "passion for it" you should be able to awaken a glimmer in me (and others, I dare say).
And on a quick skim, I think, just think mind, I may have come partly on board.
After a trip away, I spent the last 15 min going from start to end; I will have some questions (actually, they are mainly to do with your notation, but some others too).
Plus I had an interesting thought (inspired by my Birkhoff & Mac Lane A survey of Modern Algebra) that connects your thread to mine.
But for now, it is Friday night here (UK). Traditionally we go get drunk and then beat our wives.
And why not?
Stay tuned for some really, and I mean really, dumb questions |
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| Guitarist |
| Posted: Sat Jun 28, 2008 6:45 am Post subject: |
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Hi serpicojr! Here are a few dumb questions for starters. I dare say you will be appalled by their dumbness, but there you go.
So, we have that,say Q[x] is the ring of polynomials in the indeterminate x with rational coefficients.
You then use the form Q(x). Am I to take this as the field of rational expressions in x?
Now we have that Q(α) is an extension of the field Q, where α is some number. The notation I have here is rather different, so let me just check this out;
I have that, if it is the case that the field K is a subfield of the field L, one says that L:K is field extension of K. Further, if it is the case that L = Q(α), then a) L:K is a simple extension of K and b) if α is an algebraic number, it is a simple algebraic extension, otherwise it is a simple transcendental extension. Is this right?
So, at some point you make the claim that Z[i] is the ring of integers. I don't follow this; Z[i], unless I have seriously misunderstood, is the polynomial ring in the indeterminant i with integer coefficients.
First, i is not an indeterminant, but even ignoring this, how can I recover a bog-standard integer from this? Or are you including non-real integers here?
Anyway, let's have some fun, which you did allude to. The theory of linear algebra offers us this fundamental theorem; any vector space Vn of dimension n over the field K is isomorphic to the field Kn.
Then provided only that dim(L:K) = dim(K), (um..do I need this assertion? Not sure) one may treat L:K as a vector space over K. But this is merely to define the forgetful functor U: Fld → K-Vec, that is the mapping from the category of fields to the category of vector spaces over K on the condition that this mapping forgets about the multiplicative unit in the field axioms.
What fun! |
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| serpicojr |
| Posted: Sat Jun 28, 2008 9:11 am Post subject: |
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I felt bad for letting this post fall by the wayside, but I'm glad that it actually gave you the opportunity to read and digest my posts! Let's get to work on these questions.
| Guitarist wrote: |
| I dare say you will be appalled by their dumbness, but there you go. |
Rule number 1 of the math forum: there are no dumb questions!
| Quote: |
So, we have that,say Q[x] is the ring of polynomials in the indeterminate x with rational coefficients.
You then use the form Q(x). Am I to take this as the field of rational expressions in x? |
Precisely.
| Quote: |
| I have that, if it is the case that the field K is a subfield of the field L, one says that L:K is field extension of K. |
I'll be using the notation L|K or L/K; I'll use [L:K] to denote the degree of the extension, i.e. [L:K] = dimK(L), thinking of L as a K-vector space.
| Quote: |
| Further, if it is the case that L = Q(α), then a) L:K is a simple extension of K and b) if α is an algebraic number, it is a simple algebraic extension, otherwise it is a simple transcendental extension. Is this right? |
Indeed. One wonderful fact is that every finite degree algebraic extension of Q is indeed simple (this is the primitive element theorem, I believe), although of course there are nonsimple transcendental extensions, e.g. Q(X,Y), X, Y indeterminates.
| Quote: |
| So, at some point you make the claim that Z[i] is the ring of integers. |
It is the ring of algebraic integers in the field Q(i). This is defined to be all elements of Q(i) which satisfy a monic polynomial with integer coefficients. The point here is that an element a+bi of Q(i), a and b rationals, is an algebraic integer iff a and b are rational integers.
| Quote: |
| Then provided only that dim(L:K) = dim(K), (um..do I need this assertion? Not sure) one may treat L:K as a vector space over K. |
Indeed, the definition of the degree of an extension is the dimension of L as a vector space over K. I think the only nontrivial thing to prove is that, if L is the root field of an irreducible polynomial f(X) in K[X], i.e. L = K(a) for some root a of f(X), or L = K[X]/(f(X)), then indeed [L:K] = deg(f(X)).
| Quote: |
| But this is merely to define the forgetful functor U: Fld → K-Vec, that is the mapping from the category of fields to the category of vector spaces over K on the condition that this mapping forgets about the multiplicative unit in the field axioms. |
An excellent connection between our two posts! |
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| Guitarist |
| Posted: Sun Jun 29, 2008 2:32 pm Post subject: |
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Ok, here's another elementary question.
Birkoff & Mac Lane are careful to distinguish between polynomial forms and polynomial functions. Specifically they state that any polynomial form
a0 + a1x + ....anxn
uniquely determines a polynomial function, and each polynomial function is determined by at least one such form.
This, as they remark, need not imply a one-to-one and onto map (a bijection) between such forms and such such functions, i.e. they are not isomorphic.
The example they give is from the quotient group Z3 = Z/3Z, i.e. the group of integers mod 3, one may have that the forms f(x) = x3 - x and g(x) = 0 determine the same function i.e the function that is identically zero.
They then make the astonishing remark, (noting that 3 is prime): "over any Zp equality of functions has an effectively different meaning than it does for forms" I like this a lot........
Just out of interest, do you agree with any of this (leaving aside all appeals to authority)? |
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| JaneBennet |
| Posted: Sun Jun 29, 2008 3:07 pm Post subject: |
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If you really want to be strict about things …
Polynomials over a ring R are actually sequences of elements of R. The polynomial
a0 + a1x + a2x2 + … + anxn
is defined as the sequence (a0, a1, a2, …, an, 0, 0, …) – i.e. (an)n=0∞ where ak = 0 for k > n.
On the other hand,
f(x) = a0 + a1x + a2x2 + … + anxn
is a function f : R → R.
Of course they are two totally different things. 
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| serpicojr |
| Posted: Sun Jun 29, 2008 5:30 pm Post subject: |
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| Guitarist wrote: |
| Just out of interest, do you agree with any of this (leaving aside all appeals to authority)? |
Indeed I do agree with this difference, and I would prefer we think of polynomials as forms (or sequences, as Jane puts it) as opposed to functions. This doesn't matter over infinite fields--a polynomial defines the constant function 0 iff it's the 0 polynomial, as it must have infinitely many roots and, for nonzero polynomials, the number of roots is bounded by the degree. But we'll be seeing that finite fields and their extensions play a very important role in our investigations. If you think of polynomials simply as functions and not as indeterminate algebraic expressions, then you can't talk about roots of your function which lie outside the field you're looking at. A function may have many extensions to an extension field. A polynomial only has one.
(Note: an abstract algebraic explanation of the difference of the two follows by considering quotients of the ring of polynomials of a finite field. For any q = pn, p prime, n ≥ 1, we have that the ideal (Xq-X) in Fq[X] is precisely the ideal of polynomials which define the constant function 0. Thus we have that polynomial functions over Fq are a quotient of polynomial forms over Fq, i.e. the ring Fq[X]/(Xq-X).)
(Fun stuff: the above discussion is an ingredient in the proof that the polynomial functions over Fq are precisely the functions p: Fq -> Fq. This follows from the Fundamental Theorem of Finitely Generated Modules over a PID.)
I'll begin discussing the general foundations of algebraic number theory in a short while. I'm trying to introduce concepts in an organic fashion, but I've hit a small snag that may compromise this plan.
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PS: Oh, and let me know what your background with finite fields is. I just need you to know that, for any prime power q = pn, there is a unique (up to isomorphism) finite field of order q, which I denote Fq. You may obtain this as the splitting field over Z/pZ of the polynomial Xq-X. For any m ≥ 1, Fqm is a simple Galois extension of Fq, and the Galois group is isomorphic to Z/mZ, with the map from the latter to the former given by sending k (mod m) to the automorphism of Fqm given by sending an element a to the element aqk. |
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