# Irreducibility over Integers mod P

apalmer3

## 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)$$\in$$ 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...

## 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!

Homework Helper
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??

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...

Homework Helper
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.

apalmer3
Well... 2+3 = 5. So it's a partition of 5...?

Homework Helper
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.

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!)