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Using Galois theory to prove that x^4 +1 is reducible in Z_p[X] for all primes p

Date: 11/09/2008 at 16:10:42
From: fab
Subject: x^4 +1 reductible in F_p[X] for any prime p

Hi Dr. Jacques,

I am not familiar with the number theory concepts that you used in 
this proof:


How would you prove -- using Galois theory -- that the polynomial x^4 
+ 1 is reducible in F_p[x] for any prime p?

Thank you in advance.

Date: 11/10/2008 at 05:26:25
From: Doctor Jacques
Subject: Re: x^4 +1 reductible in F_p[X] for any prime p

Hi Fab,

If p = 2, then x^4 + 1 = (x + 1)^4, and we are done. So let us assume 
that p is an odd prime.

As x^8 - 1 = (x^4 + 1)(x^4 - 1) in any field, the roots of x^4 + 1 in 
a splitting field are eighth roots of unity.  They are actually the 
primitive 8th roots of unity.  We say that x^4 + 1 is the cyclotomic 
polynomial of order 8.

GF(p^n) is the finite field of p^n elements.  So consider now the 
extension field GF(p^2), which is an extension of degree 2 of GF(p).

If p is any odd integer (not necessarily prime), we have p^2 = 1 (mod 
8), which you may confirm for the 4 possible cases.

The multiplicative group of GF(p^2) contains p^2 - 1 elements, and is 
therefore a multiple of 8.  As the multiplicative group of any finite 
field is cyclic, the group in question contains a cyclic subgroup of 
order 8, and all the elements of that subgroup satisfy x^8 = 1.

This means that GF(p^2) contains all the 8th roots of unity, and, in 
particular, all the roots of x^4 + 1. In other words, the splitting 
field of x^4 + 1 is GF(p^2), an extension of degree 2.  However, if 
x^4 + 1 were irreducible, any root would generate an extension of 
degree 4, which is a contradiction.

You can use this kind of argument to produce many other examples. For 
example, the cyclotomic polynomial of order 15 is:

  f(x) = (x^15 - 1)(x - 1)/(x^3 - 1)(x^5 - 1)
       = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1

If p is any prime different from 3 and 5, we have:

   p^4 = 1 (mod 3)
   p^4 = 1 (mod 5)


   p^4 = 1 (mod 15)

Using the same kind of argument as before, we see that f(x) splits in 
GF(p^4), an extension of degree 4. This means that the irreducible 
factors of f(x) have degree at most 4. (Using more elaborate methods, 
you can even prove that these factors have the same degree.)

If p = 3, we note that, since x^15 - 1 = (x^5 - 1)^3, a root of f(x) 
must be a root of x^5 - 1, and therefore a root of the cyclotomic 
polynomial of order 5, namely:

  g(x) = x^4 + x^3 + x^2 + x + 1

Indeed, we have f(x) = (g(x))^2 over GF(3).  You can use a similar 
argument for p = 5, where you will find that:

  f(x) = (x^2 + x + 1)^4

With greater effort, you can prove that f(x) is irreducible over Q -- 
true for any cyclotomic polynomial.

Please feel free to write back if you want to discuss this further.

- Doctor Jacques, The Math Forum

Date: 11/10/2008 at 13:18:01
From: fab
Subject: Thank you (x^4 +1 reductible in F_p[X] for any prime p)

Thank you, Dr. Jacques,

I'll study this with attention and if I've any question, I'll ask you 

Thank you.

Date: 11/11/2008 at 16:42:10
From: fab
Subject: Thank you (x^4 +1 reductible in F_p[X] for any prime p)

Dear Dr. Jacques,

Thank you for your complete and easy-to-understand answer.  It was 
very helpful.


Associated Topics:
College Modern Algebra

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