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Albert_W.
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algebra problem
« on: Mar 26th, 2008, 8:11pm » |
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Why are there few references/results about infinite matrices and not much research done in this area?
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Benny
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Re: algebra problem
« Reply #1 on: Mar 27th, 2008, 11:18pm » |
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Well, there's an extensive theory of Hilbert spaces, the infinite dimensional inner product spaces which are important in quantum mechanics. One can study linear operators on a Hilbert space and the eigenvalues of these operators without introducing infinite matrices.
the multi-volume work 'Linear Operators' by Dunford and Schwartz address that, and in volume 1, there is a deep theory of infinite dimensional vector spaces and their homomorphisms, but it doesn't for the most part use the language of infinite matrices.
Perhaps someone else on this site can confirm that and expand on my answer.
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Obob
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Re: algebra problem
« Reply #2 on: Mar 27th, 2008, 11:33pm » |
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First of all, infinite matrices do not fall in the realm of algebra, as your title suggests, but rather in analysis. This is because without some notion of convergence of series, multiplication of infinite matrices is not generally well-defined.
Matrices themselves are not as important as the linear transformations they represent, and the matrices are not as "natural" as the linear transformations. For in order to obtain a matrix, one has to specify a basis of the vector space on which the linear transformation acts. Infinite dimensional vector spaces with "nice" bases are not all that common. Moreover, the matrix associated to a linear transformation of infinite dimensional vector spaces has only finitely many nonzero entries in each column, so has a very particular form.
When an infinite dimensional vector space is equipped with a notion of convergence in such a way that we have a "Schauder basis," i.e. a countably infinite linearly independent subset of the vector space whose span is dense, we can talk about the matrix of a linear transformation in terms of this Schauder basis. In particular this is the case any time we have a separable Hilbert space as BenVitale talks about. But (as far as I am aware) relatively little insight is gained from viewing linear transformations of these spaces in this light, and it is usually just better to view the linear transformations as the transformations themselves.
Even in linear algebra, the use of matrices is more valuable as a computational tool than a theoretical one. So when we pass to infinite dimensions, and the computations involved become so much more complex, it stands to reason that the tool would be less useful.
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Christine
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Re: algebra problem
« Reply #3 on: May 13th, 2008, 4:35am » |
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How to show that the polynom f = x4 - 16x2 + 4 is reducible over any finite field K?
Could anyone offer a solution? Thanks.
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Christine
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Re: algebra problem
« Reply #4 on: May 13th, 2008, 9:37am » |
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we know that over Z/2Z, f is reducible, since
x4 + 1 = (x2 + 1)(x2 + 1) (mod 2).
In fact, it can be shown that f is reducible in every prime modulus.
How about f = x4 - 16x2 + 4 ?
Can we predict whether or this polynomial will have this property?
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| « Last Edit: May 13th, 2008, 9:38am by Christine » |
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Benny
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Re: algebra problem
« Reply #5 on: May 13th, 2008, 12:33pm » |
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I'll take a shot at this problem.
You need to show that if f is irreducible and L/K is the corresponding field extension by the root of f, then its Galois group must be the Klein group Z/2 X Z/2. Since the Galois group of an extension of finite fields must be cyclic, you obtain a contradiction.
Can someone confirm this?
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Eigenray
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Re: algebra problem
« Reply #6 on: May 13th, 2008, 1:34pm » |
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Yes, it is true. For x4+1, we can try to factor it in 3 different ways:
x4+1 = (x2+a)(x2-a) mod p,
x4+1 = (x2+ax+1)(x2-ax+1) mod p, or
x4+1 = (x2+ax-1)(x2-ax-1) mod p.
For each congruence, determine when it has a solution with a  Z/pZ, and then show that for any p, at least one case holds. The same idea works for any polynomial of the form x4 + Ax2 + B2.
The deeper reason is that the roots of such a polynomial give a biquadratic extension, with Galois group (Z/2Z)2, and that this group contains no element of order 4. In general, suppose f is irreducible of degree n. For each prime p (not dividing the discriminant), you can look at how f factors mod p, which you can think of as a cycle type of a permutation on n elements, i.e., a conjugacy class in Sn. In fact, it will always correspond to a conjugacy class of the Galois group G, and moreover, the primes will be equally distributed in each conjugacy class of G.
In your example, G = <(12)(34), (13)(24)> is the Klein group. It has one conjugacy class consisting of the identity, and one class consisting of 3 permutations of the form (..)(..). So for 25% of the primes, f will split into 4 linear factors. The other 75% of the time, it will be a product of two quadratics. Since G contains no 4 cycle, f will never be irreducible mod p.
On the other hand, for a polynomial like f = x4+x2+2, the Galois group is the Quaternion group of order 8. We find that out of the first 1000 primes (other than 2 and 7),
-f splits into linear factors 11.7% of the time [e.g., f = (x+3)(x-3)(x+6)(x-6) mod 23].
-f is irreducible 25.7% of the time, e.g., p=3.
-f factors into two quadratics 37.6% of the time [e.g., f = (x2+2x-6)(x2-2x-6) mod 17].
-f factors into a quadratic and 2 linear factors 25.0% of the time [e.g., f = (x2+5)(x+2)(x-2) mod 11].
If we think of G = (Z/4Z) x| (Z/2) = <(1234),(24)> as a subgroup of S4, the cycle types are:
{(1)(2)(3)(4)} : 12.5% of the group
{(1234), (1432)} : 25.0% of the group
{(13)(24), (12)(34), (14)(23)}: 37.5% of the group
{(13)(2)(4), (24)(1)(3)} : 25.0% of the group.
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| « Last Edit: May 13th, 2008, 1:41pm by Eigenray » |
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Christine
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Re: algebra problem
« Reply #7 on: May 14th, 2008, 9:58am » |
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Wow...thanks, I just have to think about the deeper reason, i could solve the problem by showing that at least one of the elements 3,5,3*5 must be a square in K. So one can always factor f.
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Benny
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Re: algebra problem
« Reply #8 on: May 17th, 2008, 4:04am » |
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I want to prove the following claim: Let K be a field with characteristic p>0 and f be a polynomial in Z[X] (with Z being the set of integers). Then f(yp)=f(y)p for any y in K.
My consideration is this:
f = a0 + a1 x + a2 x2 + ... + an xn
Using the binomial theorem it easily follows that
f(y)p = (a0)p + (a1 y)p + (a2 y2)p + ... + (an yn)p = a0 + a1 yp + ... + an (yp)n = f(yp), as (ai)p = ai.
But this implication only holds if the coefficients are in Z/pZ which is not demanded.
So my questions: Is there a mistake in the description of the exercise? Or are the coefficients to be treated modulo p; if yes, why?
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Obob
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Re: algebra problem
« Reply #9 on: May 17th, 2008, 4:14pm » |
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There is a unique homomorphism from the integers to K which takes 1 to 1. The image of this homomorphism is the prime subfield Z/pZ of K. There are two possible interpretations of multiplying an element of K by an integer. One is that you add the element of K to itself repeatedly (or its additive inverse to itself repeatedly, if the integer is negative); the other is that you multiply the element of K by the image of the integer in K. These two notions are actually the same. Hence p times anything in K is zero. This means that multiplication in K by an element of Z is really the same thing as multiplication in K by the residue of the integer in Z/pZ.
The upshot is that you can in fact treat the coefficients as if they are integers mod p. Given any polynomial f in Z[x], you can reduce the coefficients mod p to get a polynomial g in (Z/pZ)[x]. If y is any element of K, then f(y)=g(y).
Alternately, in your proof, you needed to see that apy=ay for any integer a and any element y of K. But this is true since the image of ap-a in K is zero, exactly since ap = a (mod p).
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| « Last Edit: May 17th, 2008, 4:19pm by Obob » |
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Benny
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Re: algebra problem
« Reply #10 on: May 18th, 2008, 11:17am » |
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Yeah, thanks. I have just remembered the definition of how to multiply an element of the field with an integer and that was my only problem
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If we want to understand our world — or how to change it — we must first understand the rational choices that shape it.
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Albert_W.
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Re: algebra problem
« Reply #11 on: Jul 15th, 2008, 9:21pm » |
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Could you give examples of an Integral Domain that does not have any irreducible elements?
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Benny
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Re: algebra problem
« Reply #12 on: Jul 15th, 2008, 9:32pm » |
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on Jul 15th, 2008, 9:21pm, Albert_W. wrote:| Could you give examples of an Integral Domain that does not have any irreducible elements? |
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I believe UFD (unique factorization domain) answers to your question.
See examples and counterexamples at this linked document
http://en.wikipedia.org/wiki/Unique_factorization_domain
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Eigenray
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Re: algebra problem
« Reply #13 on: Jul 15th, 2008, 10:50pm » |
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on Jul 15th, 2008, 9:21pm, Albert_W. wrote:| Could you give examples of an Integral Domain that does not have any irreducible elements? |
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Any field will work, since every non-zero element is a unit. These are the only noetherian examples (so in particular the only UFD examples).
Some non-field examples are the ring of all algebraic integers, or the ring of fractional power series F[[x]][x1/2, x1/3, ...] over some field F.
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Albert_W.
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Re: algebra problem
« Reply #14 on: Jul 16th, 2008, 7:56pm » |
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Thanks. This question is in chapter 2 (Euclidean domains) in my textbook. Noetherian domains are in chapter 3, UFD comes later. So, how could i answer this question not knowing noetherian domain and UFD?
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Eigenray
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Re: algebra problem
« Reply #15 on: Jul 17th, 2008, 8:32am » |
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Try reading the first sentence of my post again.
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Albert_W.
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Re: algebra problem
« Reply #16 on: Jul 17th, 2008, 10:59am » |
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Oops! Sorry. I guess i missed that.
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william wu
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Re: algebra problem
« Reply #17 on: Oct 23rd, 2008, 6:52pm » |
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Hi guys,
Can someone verify for me if this thread naturally had multiple topics unrelated to one another (e.g., infinite matrices, and then, an algebra problem)? I would be concerned if the threads on this board are actually merging together due to some kind of nasty database bug.
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| « Last Edit: Oct 23rd, 2008, 8:42pm by william wu » |
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[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
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Obob
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Re: algebra problem
« Reply #18 on: Oct 23rd, 2008, 7:49pm » |
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Nothing is wrong with this thread. Albert just decided to ask another question in this thread rather than start a new thread.
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