13

I am currently studying for my Algebra Qualifying Exam. I came across the following problem and am stuck on where to even begin:

Prove that $x^5-2$ is irreducible over $\mathbb{Z}_{31}$.

I have found a similar question with $x^3-9;$ however, with degree $5$, proving there is no linear factor does not prove that $x^5-2$ is irreducible. Any hints or helpful insight would be nice. Thanks!

daOnlyBG
  • 2,711
user149593
  • 307
  • While excluding the linear factors does not finish the job, it is an excellent start. What else needs to be ruled out? – hardmath Feb 24 '15 at 15:37
  • 3
    We must also rule out quadratic factors, which will in turn rule out cubic factors, correct? – user149593 Feb 24 '15 at 15:39
  • Before tackling all possible quadratic divisors, let's look around to see if there are any shortcuts available. E.g. do we need all quadratics? – hardmath Feb 24 '15 at 15:40
  • 1
    We only need to consider monic quadratics factors. – user149593 Feb 24 '15 at 15:42
  • And only the irreducible ones. But before you start tabulating those (which sometimes is not a bad approach), have a look at this similar exercise. If the logic applies, it makes quick work of your problem. – hardmath Feb 24 '15 at 15:45
  • Okay, I will look into that. Thank you! Quick question, once we prove that $x^5-2$ does not have any linear or irreducible factors then we are done with the problem since those two cases, in turn, prove that there are no irreducible degree 3 or 4 factors? – user149593 Feb 24 '15 at 15:48
  • Yes, that is correct. The degree argument is analogous to the fact that a composite has a factor no bigger than its square root. – hardmath Feb 24 '15 at 15:49
  • Okay, perfect. I will look into those hints and see what I come up with. Thank you! – user149593 Feb 24 '15 at 15:53
  • Do you know Berlekamp's algorithm? It is very useful to prove that modular polynomials are irreducible. – Crostul Feb 24 '15 at 15:53
  • To prove that there is no irreducible quadratic factor using a similar proof in the link in the comment you left above, we consider if $x^5-2$ did have a quadratic factor. Then there would be a root $\alpha$ in the quadratic extension of $\mathbb{Z}{31}$. The proof then goes on to say that there are $960$ units of this quadratic extension and uses an order argument to finish the proof. I was first wondering, the quadratic extension of $\mathbb{Z}{31}$ is $\mathbb{Z}_{31^2}$ correct? Secondly, how are they calculating the number of units of this extension? – user149593 Feb 24 '15 at 23:53
  • A short answer is that all elements of a field are units with obvious exception of zero. – hardmath Feb 25 '15 at 00:00
  • That's what I thought at first however, don't you use Euler's phi function to calculate the number of units? In which case , $\mathbb{Z}_{31^2}$ would have $\phi(31^2)=(31)(30)$ units? What is the incorrect reasoning in my logic? – user149593 Feb 25 '15 at 00:04
  • I'll post an Answer, but to address this last point, the quadratic field extension of characteristic 31 (they are all isomorphic) has $31^2$ elements, but it is not $\mathbb{Z}_{31^2}$, which is a cyclic ring and not a field at all. – hardmath Feb 25 '15 at 14:43

2 Answers2

4

This has to do with the structure of finite fields and their Galois theory.

First look at $2$ as an element of $\Bbb F_{31}^*\,$: Since $2^5=1$ there, $2$ is a primitive fifth root of unity, and so you’re asking about the degree of the extension gotten by adjoining the twenty-fifth roots of unity.

Thus the question becomes just which extension of $\Bbb F_{31}$ finally has a number of elements congruent to $1$ modulo $25$. In other words, you’re now asking for the order of $31$ in $(\Bbb Z/25\Bbb Z)^*$, in other words the order of $6$ in that multiplicative group. You can calculate $6^5=7776$, but I would have hoped that you had seen that $1+p$ is a topological generator of $1+p\Bbb Z_p$, the principal units in the $p$-adic integers. Either way, the fifth power of $6$ is the first one that’s $\equiv1\pmod{25}$, so there you are.

Lubin
  • 62,818
4

The OP previously verified that $x^5 - 2$ has no root in $\mathbb{Z}_{31}$, equiv. no linear factor.

Lubin's Answer contains the key observation that any root $\alpha$ of $x^5 - 2$ in a field extension of $\mathbb{Z}_{31}$ would be a twenty-fifth root of unity, since $2$ is already a fifth root of unity in $\mathbb{Z}_{31}$. Indeed $\mathbb{Z}_{31}$ contains the five roots of $x^5-1$, namely $1,2,4,8,16$, as many as a field can have for a fifth degree polynomial (by Fund. Thm. of Algebra).

The OP has puzzled out an argument to finish this Qualifying Exam problem based on the order of the multiplicative group of any quadratic field extension.

If $\alpha$ satisfied a quadratic irreducible factor of $x^5 - 2$, say polynomial $p(x)$, then the simple field extension $\mathbb{Z}_{31}(\alpha)$ would have dimension two over $\mathbb{Z}_{31}$ as a vector space and contain $961 = 31^2$ elements.

The nonzero elements of this field extension are units and form a multiplicative group of order one less (omitting zero), $960$.

But the multiplicative order of $\alpha$ is $25$, just restating that it is a twenty-fifth root of unity but not a fifth root of unity. But if $\alpha$ is in the multiplicative group of order $960$, a consequence of Lagrange's Thm. is that $25$ divides $960$.

Since that just isn't true, $x^5 - 2$ has no quadratic irreducible factors over $\mathbb{Z}_{31}$. By eliminating any linear and quadratic factors of $x^5 -2$ we conclude this fifth degree polynomial is irreducible (if it factored, one factor would necessarily have degree two or less).

hardmath
  • 37,015