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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:

http://mathforum.org/library/drmath/view/69482.html

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

```

```
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)

Therefore,

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
http://mathforum.org/dr.math/

```

```
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
again.

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,

Sincerely,

Fab
```
Associated Topics:
College Modern Algebra

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