Finite Fields - Science topic

I am not an expert in logic so I am not sure how to answer your question, nor if what I am saying is relevant. However, there are algorithms for factoring polynomials over the integers, so it would seem to me that the problem is decidable.

Dear Rajen Kumar

There is no general method to generate a formula for a given sequence of data.

Each sequence has a different representation; some are linear with constant coefficients such as Fibonacci type sequences, others are linear with variable coefficients beside the higher-order sequences.

Show the data, and then one can search the best sequence, and we are lucky if a closed formula exists.

Regards

Dear Arash delfan

These articles might be resourceful, have a look:

1. https://tel.archives-ouvertes.fr/tel-02124550/document

2.

3.

Kind Regards

Qamar Ul Islam

can you using turbo pascal of simple programming..if have a problem can calling me or sending a message to help

http://arabteam2000-forum.com/index.php?/profile/228541-z3r0n3/&do=content&type=forums_topic_post&page=4

To make the "projecto-morphism" statement correct, the name "division algebra" has to be interpreted as what is also called a "skew field" or "division ring".

There is a huge literature on the connection between geometrical and algebraic properties. See e.g., Ruth Moufang, Marshall Hall, the book "Latin Squares and their Applications" by Denes and Keedwell and its second edition by Keedwell and Denes.

Consider the linear map T(X):=AX-XB. The original equation has exactly one solution for each C iff T is injective, iff ker T=0, iff the only solution for AX-XB=0 is X=0. So suppose AX-XB=0, i.e., AX=XB for a fixed solution X. Then A^2X=AAX=AXB=XB^2, and by induction, for any polynomial p(x) we get p(A)X=Xp(B).

Now consider the characteristic polynomial p of B. Then p(A)X=Xp(B)=0 since p(B)=0. Thus p(A)X=0 as a matrix equation. Either p(A) is nonsingular and there is only one solution, or p(A) is singular and there are multiple solutions. But p is the characteristic polynomial of B, which splits over the algebraic closure of the base field as

p(x)=(x-L1·I)...(x-Ln·I),

where I is the identity matrix and the Lk are the eigenvalues of B.

If p(A)=(A-L1·I)...(A-Ln·I) is nonsingular, then A-Lk·I is nonsingular for each k, hence no Lk is an eigenvalue of A. Therefore the minimal polynomials of A and B must be coprime (otherwise they would share at least a linear factor for some eigenvalue over the algebraic closure).

If p(A) is singular, we must have A-Lk·I singular for some k, hence Lk is an eigenvalue of A for some k, so the linear factor x-Lk divides the gcd of the minimal polynomials of A and B over the algebraic closure, therefore the minimal polynomials over the original field have nontrivial gcd.

in theses "Computational Aspects of Discrete Logarithms" by Rob Lambert,

Lanczos-type finite termination methods are considered for the solution

of linear systems over finite fields. Maybe it may help.

There are no challenges. An 8-bit finite field multiplier is about 200 gates. I think 65 nm has a gate density of about 400 kgates/mm2. That is 2000 8x8 multiplier per square mm. What are you after ?. What do you want to do with these multipliers ?. AES ?, BCH ECC ?. Reed-Solomon ECC ?

Dear Munir !

Thank you for your answer. The question is algorithmic: you are given a set (m matrices) of nXn matrices and you need to create a linear combination out of the given matrices such that the resulting matrix has the greatest rank that can be achieved in this way. The question is how to do it efficiently.

All The Best !

Any system of polynomial equations can be reduced to a system of quadratic equations in more variables. (This is well known.) For example, if a monomial m of degree d>=3 appears in an equation, write m=m_1*m_2

where m_1, m_2 are monomials of degree <d. Then add the variables z_1,z_2, replace the monomial m by the quadratic monomial z_1*z_2

and add the equations z_1=m_1 and z_2=m_2 of degree <d. By repeating this procedure we can get rid of all monomials of degree >=3 that appear in the equations.

So this shows that the fact that the equations are quadratic and not higher degree doesn't really help.

The 3-SAT problem could be expressed as a system of polynomial equations. This shows that solving a system of polynomial equations over a finite field is NP-complete. So one cannot expect a fast algorithm that works in general.

One approach would be using Buchberger's algorithm. If we are only interested in solutions in the finite field F_q (where q is a prime power)

then we can add the equations x_1^q-x_1,...,x_n^q-x_n where x_1,...,x_n are the variables. Of course this could be very slow. This could be doubly exponential in time.

Another approach could be just to check all possible vectors in F_q^n. This would be exponential in time.

Or, one could use a hybrid approach. Use Grobner basis with a lexicographic ordering, but compute a truncated version of it. With that I mean that we only do the computations in low degree and do not reduce S-polynomials whose degree is too high. This way, we might find some polynomials in fewer variables, say, the first k variables instead of all n variables. Then we can check for solutions in F_q^n using a back-tracking approach.

Dear Elahe,

Are you asking if there are types of footsteps which are considered normal and if you can diagnose conditions by measuring these parameters? In which case gait is very well studied and used for analysis of animal models with gait problems and also in human patients- briefly described here: https://emedicine.medscape.com/article/320160-overview#a1

@Aparna Sathya Murthy, what do you mean by "laws for large #s "? What is "#s" stands for? Please tell me little more details.

Recall the following fact.  Let A be a linear operator on a vector space V (over any field), and let v be a nonzero vector in V. Then A-cyclic subspace L_v is the smallest A-invariant subspace that contains v.

Deterministic equation solving over finite fields

Dear, please check this link,may help you:

There is no need to appeal to Galois theory. Basic considerations on the separability of the finite extension E/F are sufficient . The proof of the so called "primitive element" theorem  (see any textbook) uses the following lemma : E is monogeneous, i.e. of the form F(a), if and only if  E/F admits only a finite number of intermediary subextensions. Then :

1) If E/F is separable,  E is monogeneous (the above lemma is actually a key step in the proof of this theorem)

2) If E/F is not separable, the number of subextensions can be infinite. Classical example: take a field k of non null characteristic p, E = k(X, Y) (rational fractions on 2 variables over k), F = k(X^p, Y^p). It is immediate that E/F has degree p^2. And it is an exercise to show that E/F admits an infinite number of  subextensions  (show that E/F is not monogeneous, or exhibit directly an infinite number of subextensions) 

NB : This is probably the exercise in Lang's "Algebra" pointed out  by Giulio Peruginelli.

Not every finite ring is a field. Finite rings with zero divisors are not fields.

Every finie field is a subring of an infinite field: if F is a fileld, it is a subfield of F(x), the field of rational functions over F. It is also a subfield of its own algebraic closure.

Thanks for the 19_4 information. The question is for geometries over any field.  Type (0) is for general fields commutative of characteristic 0, type (1) is for general non-commutative fields (including the commutative ones), type (p) is for characteristic p commutative fields.  I do not know of examples of such n_k ctheorems that restrict to more than this: e.g. just to rationals.  Fano (or its matroid dual 7_4) is the only one I know that works only for a certain characteristic (2).  Instead of searching for n_k with n large and k fixed it could be more likely to find new ctheorems for k about n/2.  Maybe looking at the (p) case is easiest.  Perhaps Fano is the only such ctheorem.

Better news. The article:

by Zoe Chatzidakis if freely downloadable. At page 99 it is proven that there is a decision procedure if some sentence is true in all Z/pZ or not (without powers p^m!), and there they show also how to compute the exceptional primes for general formal sentences. It is this way which should be followed!

You should add to your question that t+k less than or equal to n. since it t+k>n, then the intersection not zero. 

Well known combinations are the Joux protocol or identity based encryption. 

You may read this.

Thank You Thong sir for providing valuable suggestion of  this question.

If you want the answer in term of coordinates or in term of polynomial, you must fix the irreducible polynomial which defines the field $F_16$ over $F_2$. If you take the polynomial $x^4+x+1$  (which is primitive) I agree that x^13 (the inverse of x^2) can be written $1+x^2+x^3$, but if we take the  irreducible polynomial $x^4+x^3+1$, then $x^13$ can be written $x+x^2$. I agree that the Euclidean algorithm is better than the "power" algorithm. For example in $Z/pZ$ where $p$ is a prime, you can compute $a^{-1}$  by the Euclidean Algorithm or by the power algorithm (with the square and multiply algorithm) both are linear in the size of $a$ but  the first runs about four times faster than the second (with the first, the data size decreases for each iteration, which is not the case with the second).

In general, Extended Euclid between the generator (viewed as a polynomial in x) and the polynomial defining the field.

I suggest you to take a look at the following book:

R. Lidl, H. Niederreiter, Introduction to finite fields and their applications, Cambridge University Press, Cambridge, 1986.

Please see the attachment in the following:

Regarding your question I found this material concise and easy to follow.

www.staff.uni-mainz.de/pommeren/MathMisc/QuGlChar2.pdf.

I doubt RS over GF(256) can be decoded mentally. However there are well known papers of Vera Pless for instance about decoding Golay code by hand.

Hello everyone. Unfortunately, Bruce´s examples are incorrect. The pair (y,z) indeed is not an orthogonal basis, but the span (y,z) has other bases that are: (y,z+2y) in the first example and (z+y,z+2y) in the second.

In fact, as Maximiliano pointed out, you can use the Gram-Schmidt process (provided the characteristic is odd), as follows. Take any ordered basis of your space: v_1,...,v_n. Assuming v_1,...,v_{i-1} are already orthogonal to all basis vectors, consider v_i. If it is singular, try to add a linear combination of v_{i+1},...,v_n to it to make it nonsingular; if impossible, then v_i is already orthogonal to the whole space, move it to the end, and continue with v_{i+1}. If, on the contrary, v_i is nonsingular, make v_{i+1},...,v_n orthogonal to it by subtracting suitable multiples of v_i from them. Thus, we end up with an orthogonal basis, with a basis for the singular subspace (if any) at the end.

In char. 2, the correspondence between bilinear forms and quadratic forms breaks down, and this doesn`t work, as Patrick´s example shows.

It depends of the matrix C. Over F_2, the equation is equivalent to X²+X+C=0. By multiplying with X from right and from left X²=X+C yelds the necessary condition XC = CX (they must commute!). This is already a system of linear equations with n²   equations and n² unknowns, and if it has rank d, we make X already dependent of only n² - d parameters. Now if you go back to the original equation, you get a system of quadratic equations where at most n² - d many unknowns are really squared. Finally, we can replace the squared variables with new variables, solve againt the linear system (the solution is again dependent of some parameters) and finally replace the new unknowns with squared variables. Now it will become clear if the particular instance of the equation has solutions or not in M.

Actually, you want to avoid these fuller finite matrix groups; For instance, a lot of the PSL(n,q) groups are simple! (Exceptions can occur for small n, and small finite fields q.)

Note that the group PSL(n,q) is just the quotient of SL(n,q) by it center (the diagonal matrices in SL(n,q)), so therefore the groups SL(n,q) are not far off from being simple.

Bogopolsky's book does a very nice job explaining in general terms how simplicity arises, and here are some hallmarks:

1) your group is highly transitive (say, at least two-transitive, for example) in its action of the vector space of the appropriate size,

2) your group has many elements of ``small support'' (given a point in the space you are acting on, there are elements in the group which only move this point and a few others).

3) your group is equal to its commutator subgroup.

In fact, these hallmarks were recognised by Higman in a paper going back to the 1950's, and then again in a well written short paper by D.B.A. Epstein (``The simplicity of certain groups of homeomorphisms'' Compositio Mathematica, 1970, pp 165--173), where in both cases, these hallmarks were looked for in infinite groups.

So, as any finite group can be found as a subgroup of permutation groups, and permutation groups are linear (embed any permutation group in a group of permutations of the major axes, which is a linear group). We see that there are lots of linear groups with lots of normal subgroups; just not really the ones you mentioned.

Thanks Jeremy

@Rama: in particular one nice property of a Galois ring is to be local, which would not be the case with a reducible polynomial

Hi Selda,

thanks for the tip. Part of the difficulty is that I want to count all polys not just the irreducible ones...

I will provide you wit a paper to discuss your inquiry