User p vanchinathan - MathOverflow

Answer by P Vanchinathan for Existence of nontrivial roots of a homogeneous polynomial over a finite field in extension fields

2013-05-17T05:44:19Z

For a prime $p\ne2$, take the homogeneous polynomial $\sum_{j=1}^{p^2-1} X_j^{p^2-1}$ of degree $p^2-1$ in as many variables. For any non-zero element of $\mathbf{F}_{p^2}$ each term evaluates to 1 and so the given polynomial is the constant function $-1$ and has no zero. The number of variable $p^2-1$ is a multiple of 2, and yet has no solution in the quadratic extension of the prime field, providing another counterexample.

Answer by P Vanchinathan for problem factoring a cubic polynomial

2013-05-16T02:39:47Z

You already know 2 is a root, so divide by $(x-2)$, which will give no remainder, and get a quadratic polynomial. Now you are left with a quadratic which should be easier to deal with. I suggest you ensure that questions you ask are at a level fit for this forum, and not at pre-college level.

Answer by P Vanchinathan for polynomial zero within a square

2013-05-14T10:20:23Z

Use Lagrange Interpolation Formula: Specify values at $1, i, 1+i$ etc to be numbers of very high absolute value and value at 0 to be of much smaller absolute value. By this formula we will get a polynomial. The problem could be it can have roots of small absolute value. To avoid this and meet your requirement, change the polynomial $f(z)$ obtained to $f(cz)$ with the scaling constant chosen to push out the zeros far away from 0. Note that $f(cz)$ has the same constant term as $f(z)$ and so your condition will be preserved.

Answer by P Vanchinathan for Does there exist a polar decomposition of matrices over finite fields?

2013-05-03T00:44:38Z

Orthogonal group consists of fixed-points of an involutary automorphism in $GL(n)$. There is a general theory of involutions and symmetric varieties. For an involution $\sigma$ of a semi-simple algebraic group $G$, the subvariety ${x\sigma(x)\mid x\in G}$ is affine generalizes the notion set of symmetric matrices, viz. $ = GL(n)/O(n)$.

See works of Deconcni-Procesi or T.A. Springer for stratifying $G/H$ $H$ being the fixed-point subgroup of $\sigma$.

Answer by P Vanchinathan for Generate a higher degree symmetric polynomial from an existing one

2013-05-01T08:28:49Z

Don't know if there is a name. Possibly this is known to Newton; the inductive proof of Newton's theorem on elementary symmetric polynomials goes along similar lines.

When we start with some polynomial and take the sum over its orbit under $S_{n+1}$ it will be invariant under $S_{n+1}$. You are starting with $u(x_{n+1}) p(x_1,x_2,\ldots, x_n)$, and summing it over the generating set of $n$ transpositions of $S_{n+1}$.

Irreducible Degrees and the Order of a Finite Group

2013-04-08T07:27:08Z

This is a question of aesthetics.

For a finite group of order $n$, the proof that the degree $d$ of a complex irreducible representation divides $n$ goes by showing that the rational number $n/d$ is an algebraic integer. As an application of the fact that $\mathbf Z$ is an integrally closed domain, this proof is really spectacular. But I feel that this is an indirect proof, not providing an insight into what is actually going on.

Being a statement of very basic nature perhaps there are other 'natural' or alternative ways of seeing why this happens. I would be grateful if experts here can point out other proofs or explain what is happening in the traditional proof.

Answer by P Vanchinathan for Combined standard deviation

2013-04-11T07:14:28Z

Queston not appropriate for this forum.

Answer by P Vanchinathan for Parabolic-type subgroups of GL(V)

2013-04-01T05:50:25Z

In Galois theory of algebraic number fields while discussing a prime lying above a prime of the base filed the inertial group is defined as one inducing identity at the residue field level. Your definition is analogous to that. SO inertial subgroup of the parabolic could be a candidate reasonable name.

Answer by P Vanchinathan for Decomposition of solvable Lie group

2013-03-26T03:34:04Z

In the case of connected linear algebraic groups it is true: Any inner automorphism of $G$ is an algebraic group automorphism of $R$. And so it carries all the unipotent elements of $R$ to unipotent elements, (See Section 19. Connected Solvable Groups in J.E. Humphreys textbook "Linear Algebraic Groups")

Answer by P Vanchinathan for 3D Rotation Representation for Multiple Turns

2013-03-21T03:41:29Z

The resolution in 2D that you suggested may also be viewed as going from the circle to its universal covering space: $\mathbb{R}\to S^1$.

SO the the same trick should work: take the universal cover of SO(3).

Answer by P Vanchinathan for Subfield of rational function field and which is not a rational function field

2013-03-15T07:45:36Z

A result of H.W.Lenstra (Inventiones Math., 1974): (a clearly written paper)

For a prime number $p$ let $K=Q(x_1,x_2,\ldots, x_p)$, be a pure transcendental extension over the rational numbers. Let $F$ be the subfield consisting of those elements of $K$ that are fixed by the cyclic permutation of the variables.

Then $F/Q$ is not purely transcendental for $p=47$ (Swan, 1969) and for infinitely many primes $p$.

The paper contains much more, about abelian group of permutations and their invariant subfields.

Answer by P Vanchinathan for The inverse Galois problem, what is it good for?

2013-01-12T06:31:15Z

Any branch of mathematics after the first few definitions will make everyone routinely ask themselves some basic questions. I consider Inverse Galois Problem is one such. If the question (i) is not highly technical, (ii) can be understood at very early stages and (iii) does not sound concocted then it justifies itself.

These are the natural questions the subject should attempt to answer. (It is irrelevant if solving them requires Fields medallists or undergraduates).

Let me list more questions in the same category (not necessarily of the same level of difficulty!)

Which divisors of $|G|$ are orders of subgroups of $G$?
Which connected open subsets of the complex plane are biholomorphic to the unit disc?
For which numbers $d$, is the ring $\mathbf{Z}[\sqrt d]$ a UFD?
Which finite groups occur as subgroups of $\mathbf{SO}(3)$?
Which integers are represented by an indefinite/definite integral quadratic form?
Which projective curves are subvarieties of the projective plane?

I have been under the impression that this is the way mathematicians think. If someone questions the relevance of the above questions it would be difficult for me to communicate with that person.

Answer by P Vanchinathan for Does every polynomial diophantine equation have solutions modulo p?

2013-01-06T10:18:50Z

Fermat's theorem that $a^p\equiv a\pmod p$ for any inegers $a,p$ with $p$ prime allows us to conclude that the equation $X^p-X+1=0$ for any prime number $p$ does not have any solution modulo that prime number $p$..

In general, as polynomial functions and polynomials are two different animals for finite fields we have many non-constant polynomials that are constant as functions. Actually a friend of mine used to joke that $\mathbf{Z}/2$ is an algebraically closed field: any non-constant polynomial function over $\mathbf{Z}/2$ should assume more than one value hence it has a zero!

Primitive Elements for $S_n$ Galois Extensions?

2013-01-03T04:26:06Z

This is an offshoot of my other question two days ago. http://mathoverflow.net/questions/117820/how-to-apply-hilberts-irreducibilty-theorem

But it is of independent interest.

Solutions of Inverse Galois Problem for a finite group $G$ give us polynomials whose splitting field has Galois group $G$. I would like to know if there is a way of getting the minimal polynomial (degree equal to order of $G$) for a primitive element of that splitting field.

If this is too cumbersome at least is there a reasonable description of a primitive element?

For example in the case of $S_n$ can we get a recursive description something like this? give two algebraic numbers $\alpha,\beta$ one of them of degree $(n-1)!$ such that $\alpha+\beta$ is of degree $n!$ and generates an $S_n$-Galois extension? (as $S_n$ is generated by an $n$-cycle and an involution, the number of degree $(n-1)!$ above could possibly be fixed by that $n$-cycle).

Known baby example: Considering that cube roots of one and two together generate a Galois extension with $S_3$ as Galois group it is straightforward to see $\omega +\sqrt[3]{2}$ as primitive element and write down its minimal polynomial. (It turns out to be $x^6 + 3x^5 + 6x^4 - 13x^3 - 24x^2 + 33x + 121$, quite a mouthful).

How to apply Hilbert's Irreducibilty theorem?

2013-01-02T00:50:21Z

I do not know how to correctly interpret Hilbert's Irreducibility theorem with Galois group as my aim.

Here $K$ is a number field (or simply $\mathbf{Q}$).

Scenario 1: Take a field $L$ that is a finite Galois extension of $K(t)$ ($t$ an indeterminate) with Galois group $G$. Writing $L=K(t)[X]/(f(t,X))$ for an irreducible polynomial $f(t,X)\in K(t)[X]$, and taking a specialization $t=a\in K$ guaranteed by Hilbert we can see the Galois group descends and we get a $G$-Galois extension over the number field $K$ as $K[X]/(f(a,X))$.

I understand this situation well.

Scenario 2: Instead of a $G$-Galois extension we are merely provided with an irreducible polynomial whose SPLITTING FIELD has $G$ as Galois group, so the {\it degree of the polynomial can be less than the order of $G$.}

I took the following example from Malle and Matzat's book on Inverse Galois Theory. (Page 88, attributed to Beckman). (Instead of a general degree $n$ I take $n=3)$.

He claims $f(t,X) = X^3-3tX +2t \in \mathbf{Q}(t)[X]$ is irreducible with $S_3$ as Galois group. (of its splitting field).

For the special value $t=4$ we get the irreducible polynomial $X^3-12X+8$, but the discriminant is a square (of 72) and we get a cubic number field as splitting field and not the expected $S_3$ extension of $\mathbf Q$.

What mistake am I making in this scenario?

Instead of giving a degree $n$-polynomial in $K(t)[X]$ with $S_n$ as Galois group I would be more comfortable with an irreducible polynomial of degree $n!$ with $S_n$ as Galois group so that I can specialize that polynomial. Perhaps it is expecting too much.

Answer by P Vanchinathan for Is there a "geometric" intuition underlying the notion of normal varieties?

2012-12-28T05:29:49Z

An excellent non-algebraic meaning (using analysis) of normality is found in Kollar's article in the Bulletin of AMS (1987).

Restrict to irreducible varieties $X$ so we can talk of function fields. A point $x_0\in X$ is considered normal whenever a rational function exhibits decent behaviour in a neighbourhood of $x_0$ then it finds a place in the local ring of $X$ at $x_0$.

Decent behaviour here is: If $f\in K(X)$ and if $|f(x)|$ remains a bounded function as $x$ approaches $x_0$ by paths lying in $X$, then $x_0$ should be good enough to admit $f$ in its local ring.

This survey article of Kollar is about Mori's Fields-medal winning work on 3-folds. But it starts from the scratch defining what an algebraic variety is. It is a great source to learn the meanings of fundamentals objects of algebraic geometry. (for example Kollar explains why we have to deal with line bundles when we study projective varieties).

Answer by P Vanchinathan for parabolic subgroup

2012-12-27T14:03:14Z

Example: Subgroups of $SL(n, \mathbf{C})$ which contain upper triangular subgroup, and closed in Zariski topology (i.e. matrices satisfying polynomial conditions on its entries) is a parabolic subgroup. (also their conjugate subgroups).

See the textbook by Humphreys Linear Algebriac Groups.

Answer by P Vanchinathan for Factoriality of cones

2012-12-17T14:49:54Z

A result of Popov and Vinberg gives a positive case. For a simply-connected semi-simple algebraic group $G$ in characteristic zero. they consider the orbits of the highest weight vectors under in irreducible representations.

Then the cones over these orbits are factorial varieties for fundamental representations.

Answer by P Vanchinathan for Properties of permutations with unknown pattern avoidance descriptions

2012-12-16T02:49:32Z

I hope I understood the question correctly. I have a feeling that questions on permutations of algebraic as opposed to combinatorial nature, could be candidates.

Lakshmibai and Sandhya's theorem is a geometric question and it is a significant theorem because it reduces geometry to combinatorics. With this understanding of your question let me attempt to give four examples:

(1) A permutation being of specific order $m$ .

Suppose we attempt pattern avoidance like: for any $k$ relatively prime to $m$ it should not have a length $k$ cycle. A permutation of order, for example $m^2$, will also satisfy that criterion and will be accepted wrongly.

(2) Permutation being even. (avoidance criterion may not work: because presence of an even number of cycles of any particular length, as opposed odd number of them, will be ok)

(3) Some irreducible character vanishing in it. This is conjugacy class question. Can be argued similarly

(4) Commuting with another specific permutation.

Answer by P Vanchinathan for What is the ideal corresponding to the Plücker embedding?

2012-12-13T08:36:09Z

References for this purpose are: A series of papers initiated by C S Seshadri, Lakshmibai, Musili develops "Standard Monomial Theory" to deal with this. It gives equations for Schubert varietes, describes their singular loci, proves many cohomology-vanishing theorems for line bundles on them.

V. Lakshmibai & K.N. Raghavan have written a book published by Springer (2008). Encyclopaedia of Mathematical Sciences, 137.

Answer by P Vanchinathan for Cyclic extensions

2012-12-12T08:16:54Z

Take $\zeta = e^{2\pi i/p}$ for a prime number $p\equiv1$ (mod 3), e.g. $p=7$. Then $Q(\zeta+\bar\zeta)$ is a totally real cyclic Galois extension of $\mathbf{Q}$ of degree a multiple of 3, hence contains a cubic extension $L$ that is Galois with cyclic Galois group. Being totally real it cannot be the splitting field of a polynomial of the form $X^3-a$. (or use David Loeffler's argument above).

Dirichlet's theorem on primes in arithmetic progressions assures us that we have infinitely many such examples over the base $\mathbf{Q}$.

Answer by P Vanchinathan for At what point does number theory stop playing with finite rings?

2012-12-10T04:28:56Z

Dirichlet's Theorem on Primes in Arithmetic Progressions, proved in 1837, needing real-analytic methods could possibly be the first major candidate for a number-theoretic result departing from finite methods. (This was proved 50 years earlier than Prime Number Theorem).

Answer by P Vanchinathan for Is there a high-concept explanation for why characteristic 2 is special?

2012-12-10T09:58:50Z

Here is my computational reason (instead of a high-concept explanation) why the primes 2 and 3 are special (hypotheses of many theorems on algebraic groups, linear or projective exclude both these primes).

The numbers 2 and 3 got into the Primes Club by 'dubious' means!

Given $p$, to certify it as a prime a number, we need to check no number $d $ with $1 < d \leq [ \sqrt p ]$ divides it. For 2 and 3 this condition is vacuously true as there are no integers in that interval, wheres 5 onwards they really needed to pass the test!

Answer by P Vanchinathan for Non-rigorous reasoning in rigorous mathematics

2012-12-05T00:24:42Z

Close to he requirement in the original question: Waring's problem which generalizes Lagranges's four-square theorem. Every positive integer can be expresses as a sum of 9 cubes, a sum of 19 fourth powers etc. For the $k$-th powers the number of summands required, denoted $g(k)$, was a heuristic guess, $g(k) = 2^k + [ (\frac32)^k ]- 2$ and some variations of this. Though Hilbert proved $g(k)$ is finite before 1910 actual specific values were proved decades later. The reason I know this is because one of the persons who 'nailed the last nail into the coffin" of this problem in 1980s was working where I started my PhD.

The number of summands is very high for low numbers because (heuristically) you have only 1 and $2^k$ to use. For 4-th powers 79 is the culprit, needing fifteen 1's and four 16's.

So this lead to related another natural question: as numbers needing that many summands are small in size they may be a finite number of exceptions. Define $G(k)$ as the number of summands needed for expressing every sufficiently large integer as sum of $k$-th powers (i.e. treat 79 as an exception for the case of fourth powers). $G(4)$ is known to be 16.

Are the canonical actions on Schubert Cells Linearizable?

2012-11-01T10:01:43Z

G. Schwarz constructed a (counter)example for an action of a simple algebraic group on an affine space that is not linearizable (i.e., it is not a representations).

Natural examples of affine spaces that are not readily vector spaces are Schubert cells. So it was tempting to look for reductive group actions on them and see if they can lead to more counter-examples.

For a parabolic subgroup $P$ of a linear algebraic group $G$, (say $G$ semi-simple) we can take a Schubert cell $C\subset G/P$. By definition $C$ is the orbit for a (maximal) solvable subgroup
which is far from a reductive group. However one can look at the largest parabolic subgroup of $G$ acting on $C$, and restrict the action to a Levi part $L$ and ask if the action of $L$ on $C$ is linearizable. (Easy to see examples where $L$ is more than a maximal torus of $G$).

In a 5-minute meeting with M. Brion I asked this question and he said `Yes, it would follow from the slice theorem'. Can any one elaborate on his brief answer?

Answer by P Vanchinathan for Generating finite simple groups with $2$ elements

2012-04-16T23:57:43Z

There is a paper in arxiv by Robert Guralnick and Gunter Malle that answers your question in a stronger way. Their aim is to prove existence of algebraic surfaces obtained in a specific way as a quotient of finite group actions on products of curves of genus > 1. They prove the existence of two conjugacy classes in a finite simple group with the property that picking one element each from these classes always generates the group.

Here is the link:

http://arxiv.org/abs/1009.6183

Comment by P Vanchinathan

2013-05-19T02:58:53Z

My answer given yesteray is wrong. One can choose $p$ variables to have non-zero value, and set all others to be zero to get a non-trivial solution. Should I delete it?

Comment by P Vanchinathan

2013-05-18T10:32:57Z

The above is wrong. One can choose $p$ variables to have non-zero value, and set all others to be zero to get a non-trivial solution. I am sorry to have rushed in without fully verifying.

Comment by P Vanchinathan

2013-05-15T00:26:03Z

Am I misreading the question? Do you want to know if $4/9$ is a RATIO of two Pythogorean fractions, as opposed to being a Pythagorean fraction by itself? The wording of your question means an expression for $4/9$ as $\frac{p_1q_1/(p_1^2-q_1^2)}{p_2q_2/(p_2^2-q_2^2)}$ is what is desired (using Barry Cipra's suggestion).

Comment by P Vanchinathan

2013-05-15T00:07:09Z

I see the snag. Can't we, with hindsight, specify values to be even higher at 1, $1+i$ so that scaling constant can't bring down. Of course this depends on the least absolute value of the zero. I'll try to rework again. Thanks, Johan!

Comment by P Vanchinathan

2013-05-14T06:19:38Z

Perhaps you have more info that could reduce the degree. For example, the fields generated individually by $p_{41}$ and $p_{21}$ could intersect in a field of degree 16. and similar things.Also possibly a a convenient automorphism (a Galois group element) might send your number $p_4$ to a number more amenable for computations.

Comment by P Vanchinathan

2013-04-29T08:22:06Z

Have no idea how to solve this problem. But do remember that M. Isaacs (author of a book on Character Theory) has written papers that have a lot to do with the set of character degrees.

Comment by P Vanchinathan

2013-04-11T09:10:36Z

Thanks, John for a clear writing.

Comment by P Vanchinathan

2013-04-11T00:34:33Z

More along this line: The cone over the Grassmannian for the Pl\"ucker embedding is even a factorial variety. (And quadratic relations among Pl\'ucker co-ordinates holds.) Book by Lakshmibai on 'Standard Monomial Theory' has more generalisations. Popov's generalisation: same is true for orbits of highest weight vectors in an irreducible representation of simply-connected semisimple algebraic group $G$ (over C?).

Comment by P Vanchinathan

2013-04-09T02:56:17Z

Correction to my earlier comment: I meant real two-dimensional representation for odd cyclic groups.

Comment by P Vanchinathan

2013-04-08T23:38:52Z

@David Speyer: Thanks for the detailed answer here and at SE. I am teaching rep theory without using group algebras and fumbling. In Lagrange's theorem we see cosets are of same cardinality and provide a partition of $G$. About degree of intermediate fields in finite extensions also we have a transparent proof. I am looking for such a simple underlying idea. As odd ordered cyclic groups have 2-dimensional irreps as symmetries of regular polygons we need to bring the dependence on complex numbers (via Schur's lemma). Not that easy perhaps. :-(

Comment by P Vanchinathan

2013-04-08T09:57:58Z

(TeX comment) Insert a missing dollar sign to make the question readable.

Comment by P Vanchinathan

2013-03-09T02:44:47Z

Dear Kali, This is an exercise question in a first course in Linear algebra and not suitable for this forum where people discuss more advanced questions.

Comment by P Vanchinathan

2013-01-20T00:51:09Z

The equality connecting arithmetic mean and geometric mean should be relevant here.

Comment by P Vanchinathan

2013-01-10T15:44:59Z

For $k=\mathbf{Q}$ work of Saltman,(retract rational) on Noether's problem could be relevant, and Swan, Lenstra for abelian groups. But all of them talk of higher dimensional varieties. Specifically cyclically permuting $n$ variables over $\mathbf{Q}$ does not give a fixed field that is purely transcendental for $n=47$ (Swan) and $n=8$ (Lenstra).

Comment by P Vanchinathan

2013-01-07T10:50:11Z

Sorry for the goof up. I read it as ideals. Now notice that question is about maximal ideals! SO my above answer is wrong. I'll see if it can be deleted.