Proving (1+a2)(1+a3)...(1+an)=0 for Any Finite Field F

In summary, for a finite field F with elements a1, a2, a3, ..., an, where n is odd, the product (1+a2)(1+a3)...(1+an) = 0 holds true. This is because at least one of the elements, say x, must have itself as its inverse and therefore x^2 = 1 which implies x = -1. For the case where n is even and none of the elements are inverses of themselves, it suffices to show that one of the terms in the product is 2, which is equivalent to 0 in a field of characteristic 2 and can be relabeled as -1 in any other field. This is true because
  • #1
mansi
61
0
let F= { 0=a1, a2,a3,a4...an} be a finite field. show that
(1+a2)(1+a3)...(1+an) = 0.

when n is odd, it's simple since 1 belongs to F. then odd number of elements are left( they're distinct from 0 and 1). at least one of them, say x,must have itself as its inverse. x^2 = 1 implies x=-1 ( since x != 1)...so the result is true.

for n even and the case where none of them are inverses of themselves, can somebody suggest a solution?
 
Physics news on Phys.org
  • #2
One of the a_r must be 1, which is distinct from 0, so one of the terms is 2, which is zero in char 2, and wlog a_2 = -1 in any other field.

It's a field. If elements multiply together to give 0 one of them must be zero, ie, 1+a_2=0 after relabelling. a_2= -1, the inverse of 1 which isn't zero and hence must be in the list you gave.
 
  • #3
matt grime said:
one of the terms is 2, which is zero in char 2, and wlog a_2 = -1 in any other field.

didn't get this...please elaborate
 
  • #4
You get that the product is zero if and only if one of the factors is zero, since it is a field?

So it suffices to show that one of the 1+a's is zero.

But this is true since 1 has an additive inverse, -1 (which equals 1 in a field of char 2)
 
  • #5
thanks...that makes it clear...
but here's another question :redface: does this mean i could have done this problem without considering separate cases?
 
  • #6
Yes, absolutely. In any field there is a non-zero element, x, such that 1+x=0. The result is also true in any finite ring with unit, any finite domain too.
 

FAQ: Proving (1+a2)(1+a3)...(1+an)=0 for Any Finite Field F

How do you prove that (1+a2)(1+a3)...(1+an)=0 for any finite field F?

To prove this statement, we can use mathematical induction. We first show that it holds for n=1, which is trivial. Then, we assume it holds for n=k and prove it for n=k+1. By expanding the product, we can see that it becomes a sum of terms, each of which has the form (1+ai) where i ranges from 2 to k+1. Since F is a finite field, all elements are invertible, so we can multiply each term by the inverse of ai, which results in a product of (1+1)=2. Since F is also finite, 2 is equal to 0, thus proving the statement for n=k+1. By the principle of mathematical induction, we can conclude that the statement holds for all finite values of n.

Why is it important to prove this statement for any finite field F?

Proving this statement demonstrates the fundamental properties of finite fields and their elements. It also allows us to make generalizations about finite fields and their operations, which can be applied in various areas of mathematics and computer science.

Can this statement be extended to infinite fields?

No, this statement does not hold for infinite fields. In infinite fields, the elements are not necessarily invertible, so we cannot use the same approach of multiplying by the inverse to prove the statement. Additionally, the concept of "finite" does not apply to infinite fields, so the statement itself would not make sense in that context.

How does this statement relate to polynomial roots in finite fields?

The statement (1+a2)(1+a3)...(1+an)=0 for any finite field F can be interpreted as saying that the polynomial (x-a2)(x-a3)...(x-an) has all of its roots in F. This is because if we substitute any element of F into the polynomial, it will result in 0, which satisfies the statement. This is a useful property in fields of cryptography, where we want to ensure that certain polynomials have roots in the field.

Are there any real-world applications of this statement?

Yes, this statement has many applications in cryptography and coding theory. It can be used to prove the existence of error-correcting codes in finite fields, which are essential in data transmission and storage. It also has applications in the design and analysis of secure cryptographic algorithms, as well as in the study of finite geometries.

Similar threads

Comp Sci An Introduction to Deep Learning and Modifying Code
Replies
7
Views
1K
I Proving inequality about dimension of quotient vector spaces
Replies
0
Views
833
I Existence and Uniqeness of Finite Fields ....
Replies
3
Views
2K
I Prove that elements of x^i 0<i<n-1 are distinct
Replies
3
Views
1K
MHB Exploring Finite Group Theory: Finding the Upper Bound of Groups of Order
Replies
2
Views
1K
MHB Existence and Uniqeness of Finite Fields .... Example from D&F ....
Replies
2
Views
2K
A Anti-dual numbers and what are their properties?
Replies
1
Views
1K
MHB Number of Elements in Finite Fields?
Replies
8
Views
2K
Orbital dynamics math review for a computer simulation
Replies
11
Views
1K
Why Does the Fourier Series of |sin(x)| Treat n=1 Differently?
Replies
3
Views
3K

Hot Threads

Recent Insights

Back
Top