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

Solution 1:

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).

Solution 2:

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.