• Support PF! Buy your school textbooks, materials and every day products Here!

Questions concerning Finite Fields (Basic Abstract Algebra)

  • Thread starter elias001
  • Start date
  • #1
12
0
Hello, everyone, i am a newbie here. I am currently taking a modern linear algebra course that also focus on vector spaces over the fields of Zp and complex numbers.
Since i am not familiar with typing up mathematics using tex or anything so that i can post on the forums, i will use the following notations for ease of readings.

Let F^M denotes F with a superscript M
Let Fp^M denotes F^M with suscript p, where p is an odd prime.

There is one part to my problem sets which i am having difficultty constructing examples. There are 3 parts to the question but i can't figure out the last part. Here it is:

Let F be a field. Suppose that M is a nonzero element of F. Let F^M={(a,b) such that a, b both belongs to F}. Define

(a1, b1)*(a2, b2)=(a1b2+b1b2M, a2b1+a1b2)

(a1, b1)+(a2, b2)=(a1+a2, b1+b2), a1, b1, a2, b2 are all elements of F.

a) Suppose that (a^2)-M is not zero for all a belonging to F. Then F^M is a field. Prove the following field axioms hold for F^M

associativity of multiplication
existence of multiplicative identity
existence of multiplicative inverse for nonzero elements

b) suppose that a^2=M for some a in F. Prove that F^M is not a field by demonstrating how one axiom in the definition of field fails to hold.

c) Let p be an odd prime. Prove that there exists a finite field that contains p^2 elements. (Hint: first, show that there exists M in Fp such that (a^2)-M is nonzero for all a in Fp. according to part a), Fp^M is a field. Show that Fp^M contains p^2 elements.


Part a) i solved, for multiplicative identity, (a1, b1) has to multiplied by (1,0) to work

For the mulitplicative inverse, M=1

For part b) if M does not equal to 1, then the axiom for the existence of multiplicative inverse fails

for part c) i do not know how to construct practical examples to show me what is actually going on. to show that (a^2)-M is nonzero in Fp, do i have to take into account of how the addition and multiplication operations in F^M are defined. And in
Fp, how can it have p^2 elements? For the last part of part c) how do i take into account that Fp^M is in (mod p) with the predefined arithimetic operations above. I mean how do i carry modular arithimetic with such messay mulitplications, and then how can the p^2 elements be listed?

If any of this is not clear, the link to the problem set is here:

http://www.math.utoronto.ca/murnaghan/courses/mat240/ps1.pdf

it is question 9 (c)

I changed alpha to M here.

I am not asking for a solution, but rather how to construct examples so that i can see what is going on in order to solve the question. Thanks everyone for any assistance/suggestions you can give me.
 

Answers and Replies

  • #2
matt grime
Science Advisor
Homework Helper
9,395
3
I think you ought to retry the existence of mult inverse if M is not 1; they do exist, indeed the complex numbers are an example of a field like this with F=R, M=-1, and we know they are a field, don't we... Indeed M=1 will not have inverses since x^2=1 has two roots in any of the fields you're looking at (i notice you are not allowing p=2).

If it helps, all you're doing is considering the set of symbols of the form a+db

where a and b are in your field, and d is a symbol that satsifies d^2=m. we define addition and multiplication in the obvious generalization from considering C as an extension of R.

try to think about complex conjugates as well x+iy--->x-iy, and what is the inverse of x+iy in terms of its complex conjugate? Now do you see why a^2-M must have no roots in F?
 
Last edited:
  • #3
shmoe
Science Advisor
Homework Helper
1,992
1
a) You have the correct identity, did you manage associativity?

M does not have to be 1 for F^M to have inverses, it just can't be a square like their condition asks (<--keep this in mind for part c)). (which M=1 actually is)

To show you have inverses, start with a non-zero element (a,b), meaning both a and b are not zero. Then find an element (c,d) where (a,b)(c,d)=(1,0) by going back to the definition of multiplication (you have to solve a linear system to find c and d and you may need a few cases)


b) If you can manage the above, it should give an idea on how to find an element without an inverse given an element a in F where a^2=M


c) You don't have to work with the operations in F^M at all here. Once you find a suitable M, you get a field by the construction above. That you have p^2 elements should be clear by the way F^M is constructed.

To get an idea how to prove a suitable M exists, try looking at the fields with 3, 5, 7, 11,... (how ever many you need) elements, they're small enough to be easy to work with. Which M's are suitable in these fields? More importantly, how many suitable M's are in the field with p elements? (alternatively, how many don't[/ii] work, and how can you find them?)
 
Last edited:
  • #4
12
0
yeah, i misinterpreted the multiplicative inverse as follows,

(a1, b1)*(x, y)=(1,1) instead, it should equal to (1,0). so the multiplicative inverse should be -a1/(some expression) and b1/(some expression) the "over some expression" means the same in both cases. I will tried you guys suggestions and see what i can figure out. If i am still stuck, i will come back and ask for clarification. Thank you for the help. :)
 
  • #5
12
0
Shome, sorry i forgot to ask what did you mean by

"You don't have to work with the operations in F^M at all here. Once you find a suitable M, you get a field by the construction above. That you have p^2 elements should be clear by the way F^M is constructed."

did you mean because the arithmetic are defined using order pairs and it is now in a finite field with p as prime, then there would be p square elements, because there are p choices for the first coordinate in the order pair and there are also p choices for the second coordinate in the second coordinate in the order pair.

Or did you mean that i can list the elements in Fp and count the direct product. Fp*Fp.
 
  • #6
shmoe
Science Advisor
Homework Helper
1,992
1
elias001 said:
"You don't have to work with the operations in F^M at all here. Once you find a suitable M, you get a field by the construction above. That you have p^2 elements should be clear by the way F^M is constructed."

did you mean because the arithmetic are defined using order pairs and it is now in a finite field with p as prime, then there would be p square elements, because there are p choices for the first coordinate in the order pair and there are also p choices for the second coordinate in the second coordinate in the order pair.
Correct on the number of elements.

About the operations, for part c), you don't have to perform any operations on the elements of F^M at all. It's not hard though, you're just working with a finite field of prime order so the operations are just the usual mod p.
 
  • #7
12
0
What i am still confused about is the following,

I know how to do modular arithmetic operation for one elemnts. a = b (mod p), or a+c = b (mod p)

but for order pairs, i am plain confused.

the arithimetic operations in the question is defined as

(a1, b1)+(a2, b2)=(a1+a2, b1+b2)

but the order paris are defined in Fp^M bit the addition operation defined above is in Fp

So if i would like to check the axiom (a1, b1) + (a1, b1)+(a3, c3)=(a1, b1) +((a2, b2)+(a3, b3))

for the left hand side, when i compute (a1, b1)+(a3, c3), do i just let it equal to (a1+b1, a2+b2) or do i have to account for mod p and hence take remainder.

I am confused as to if one has order pairs, or order n-tuples which are defined in some finite field p, where p could either be prime or non prime, should i not perform mathematical operation accourding to the laws of modular arithmetic. as an example, for p=3

like (2,3)+(4,6)=(6,9) (*)

Or is it (2,3)+(4,6)=((6 (mod 3)), (9 (mod 3)))=(0,0) (**)

so for arbitrarylly (a1,b1)+(a2,b2) (***)

Do i let it equal to

(a1+a2, b1+b2) or (a1+a2 (mod p), b1+b2 (mod p)) (****)
 
  • #8
matt grime
Science Advisor
Homework Helper
9,395
3
(a,b) + (c,d) = (a+c mod p, b+d mod p) all operations are done inside the brackets and are the mod p operations from the underlying field; there are no other operations possible.

It might help if you started to think of 4+2 =1 mod 5 directly and not think of it as 4+2=6 (ie adding in the integers first) then taking remainders mod 5. After all 6 is not an element of F_5 = {0,1,2,3,4} is it? It is an abuse of notation to write 0,1,2,3,4 instead of [0],[1],[2],[3],[4] for the equivalence classes, by picking out a distinguished element in each equivalence class.
 
Last edited:
  • #9
shmoe
Science Advisor
Homework Helper
1,992
1
elias001 said:
I am confused as to if one has order pairs, or order n-tuples which are defined in some finite field p, where p could either be prime or non prime, should i not perform mathematical operation accourding to the laws of modular arithmetic.
I just want to point out that the finite field with p elements is the integers mod p only when p is a prime. When p is a prime power, it's not so neat. This is the whole point of part c), you are trying to prove you have a field of order a prime squared.
 

Related Threads on Questions concerning Finite Fields (Basic Abstract Algebra)

  • Last Post
Replies
16
Views
2K
Replies
9
Views
3K
  • Last Post
Replies
1
Views
950
Replies
7
Views
770
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
976
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
888
Top