Irreducibility over Integers mod P

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In summary, the polynomial x^2+1 is irreducible over the field F of integers mod 11, and x^2+x+4 is irreducible over the field F of integers mod 11.
  • #1
apalmer3
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Homework Statement



a. Prove that x^2+1 is irreducible over the field F of integers mod 11.
b. Prove that x^2+x+4 is irreducible over the field F of integers mod 11.
c. Prove that F[x]/(x^2+1) and F[x]/(x^2+x+4) are isomorphic.


Homework Equations


A polynomial p(x) in F[x] is said to be irreducible over F if whenever p(x)=a(x)b(x) with
a(x),b(x)[tex]\in[/tex] F[x], then one of a(x) or b(x) has degree 0 (i.e. constant).

I was also told by somebody it's sufficient to show that there aren't any zeros...


The Attempt at a Solution


a. The zeros of x^2+1 are + and - i. Therefore, it is irreducible over F.
b. The zeros of x^2+x+4 are also imaginary (-.5 + or - 1.93649167 i), and it is therefore irreducible over F.
c. Each field has 121 elements, so they're isomorphic?

Thanks in advance for the help!
 
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  • #2
x^2+1 is reducible mod 5. It's equal to (x+2)(x+3). How can that be when it's roots are imaginary??
 
  • #3
I don't know. I don't know how to prove that they're irreducible, really. I'm just following what somebody said about it being sufficient to show that it has no zeros...
 
  • #4
Apparently, the fact that a polynomial doesn't have real roots doesn't imply that it's irreducible mod n. You agree with that, right? So throw that out. How do you think I found that x=2 and x=3 solve x^2+1 mod 5? Just think about it before I tell you.
 
  • #5
Well... 2+3 = 5. So it's a partition of 5...?
 
  • #6
5+6 is a partition of 11. But (x+5)(x+6) doesn't work. Ok, I'll tell you. I guessed. x^2+1=0 mod 5 is the same as x^2=4 mod 5. 2 clearly works, so -2=3 mod 5 also works. For mod 11, you need to solve x^2=10 mod 11. You can prove it has no solution mod 11 by just trying all of the numbers {0,1,2,...10} for x. There may be a more systematic way to do this by using quadratic residues or something, but I'm not an expert in that. I just wanted to let you know that the real root argument was going nowhere.
 
  • #7
Okay. I think I understand that now.

(i.e. x^3-9 is irreducible over integers mod 31 cause no x makes x^3=9mod31 true. But It is reducible over integers mod 11... (x-4)(x^2+4x+5).)

But how do I prove part c? (Sorry I'm asking so many questions. I appreciate the help!)
 
  • #8
I'm the one who is sorry, like I said, this is a bit out of my field and I'd have to study up on it. In short, I don't know. Reply to this so it appears at the top of this list without my name as the last person to reply and maybe someone who know Galois type stuff can help you.
 
  • #9
Okay. Thanks Dick. Your help is very much appreciated.

To everyone else... help with part c?
 
  • #10
Finite fields of the same order are isomorphic. Equivalently, for all primes p and all positive integers n, there is a unique field up to isomorphism of order p^n.

The proof of this is beyond me to explain. Maybe the question can be answered without relying on this knowledge, by using irreducibility or extension fields.
 

1. What does "irreducibility over integers mod P" mean?

Irreducibility over integers mod P refers to a mathematical concept in which a polynomial cannot be factored into smaller polynomials with integer coefficients, when considered modulo a prime number P. In other words, it is a way of determining whether a polynomial is "prime" in the ring of polynomials with coefficients mod P.

2. How is irreducibility over integers mod P different from irreducibility over the real numbers?

Irreducibility over the real numbers refers to a polynomial that cannot be factored into smaller polynomials with real coefficients. However, in the context of integers mod P, the coefficients are restricted to a specific set of numbers (integers mod P), which can change the nature of the irreducibility of a polynomial.

3. What is the significance of irreducibility over integers mod P?

Irreducibility over integers mod P has important applications in number theory, algebra, and cryptography. It is commonly used in determining the structure of finite fields and in the construction of error-correcting codes. It also plays a crucial role in the study of prime numbers and their distribution.

4. How is irreducibility over integers mod P determined?

One method of determining irreducibility over integers mod P is by using the Eisenstein criterion, which states that if a polynomial has integer coefficients and can be written as a product of two polynomials with smaller degree and one of the factors has a leading coefficient that is not divisible by P, then the polynomial is irreducible over integers mod P.

5. Can a polynomial be irreducible over integers mod P but reducible over the real numbers?

Yes, a polynomial can be irreducible over integers mod P but reducible over the real numbers. For example, the polynomial x^2 + 1 is irreducible over integers mod 3 (since it cannot be factored into two polynomials with integer coefficients), but it is reducible over the real numbers (x^2 + 1 = (x + i)(x - i)).

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