# Irreducibility over Integers mod P

1. Apr 24, 2008

### apalmer3

1. The problem statement, all variables and given/known data

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.

2. Relevant 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)$$\in$$ 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...

3. 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!

2. Apr 24, 2008

### Dick

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. Apr 24, 2008

### apalmer3

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. Apr 24, 2008

### Dick

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. Apr 24, 2008

### apalmer3

Well... 2+3 = 5. So it's a partition of 5...?

6. Apr 24, 2008

### Dick

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. Apr 24, 2008

### apalmer3

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. Apr 24, 2008

### Dick

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. Apr 24, 2008

### apalmer3

Okay. Thanks Dick. Your help is very much appreciated.

To everyone else... help with part c?

10. Apr 30, 2008

### mutton

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.