show that x^p - x + a is irreducible over GF(p) if a is different from 0

**oblixps** · March 3rd 2012, 06:38 PM

Show that x^p - x + a is irreducible over GF(p) if a is different from 0.

I assumed for contradiction that x^p - x + a is reducible. Then it must split into irreducible factors (none of them linear) over GF(p) or just Z_p. Say one of the irreducible factors has degree m < p. Then the size of its splitting field must be p^m. The splitting field of x^p - x + a consists of all the roots of that polynomial. Let b be a root of that polynomial. Then since the multiplicative group of the splitting field is cyclic, all roots of that polynomial can be expressed as some power of b. So I need to find the minimum polynomial of b. If the minimum polynomial of b is just x^p - x + a then I am done. So I assume the minimum polynomial of b is the irreducible factor with degree m < p that I mentioned above. Then the splitting field of the irreducible factor of degree m must be the same as the splitting field of x^p - x + a since they both contain all the roots of x^p - x + a.

I am having trouble finding a contradiction in this argument. Can someone help me with this proof? Thanks.

**Deveno** · March 4th 2012, 08:15 AM

suppose b is a root of $x^p - x + a$ .

let k be any element of $\mathbb{Z}_p$ .

then $(b+k)^p - (b+k) + a = b^p + k^p - b - k + a = (b^p - b + a) + (k^p - k) = 0 + 0 = 0$ .

we then have p distinct roots of $x^p - x + a \in \mathbb{Z}_p(b)$ , so this IS the splitting field of $x^p - x + a$ .

now consider $\tau_k \in \text{Aut}_{\mathbb{Z}_p}(\mathbb{Z}_p(b))$ given by $\tau_k(b) = b+k$ .

the assignment $\tau_k \to k$ is a group homomorphism $\theta:\text{Aut}_{\mathbb{Z}_p}(\mathbb{Z}_p(b)) \to \mathbb{Z}_p$ .

since $\text{ker}(\theta) = \text{id}_{\mathbb{Z}_p(b)}$ , we know that $\text{Im}(\theta) = \{0\}$ or:

$\text{Im}(\theta) = \mathbb{Z}_p$ . the former implies that $[\mathbb{Z}_p(b):\mathbb{Z}_p] = 1$ ,

that is, that b is in $\mathbb{Z}_p$ . since a ≠ 0, this cannot be true, so $[\mathbb{Z}_p(b):\mathbb{Z}_p] = p$ ,

which in turn means $x^p - x + a$ is irreducible over $\mathbb{Z}_p$ .

**oblixps** · March 4th 2012, 12:21 PM

why is $\text{Im}(\theta) = \{0\}$ or $\text{Im}(\theta) = \mathbb{Z}_p$ ? also, could you please explain your motivation for defining the group homomorphism from $\text{Aut}(\mathbb{Z}_p(b))$ to $\mathbb{Z}_p$ ? i don't understand the reasoning for doing this.

**Deveno** · March 4th 2012, 03:49 PM

because $\theta$ is a homomorphsim, so it's image is a subgroup of the additive group of $\mathbb{Z}_p$ . but $\mathbb{Z}_p$ has prime order, so...

my motivation was to use the galois correspondence: $[\mathbb{Z}_p(b):\mathbb{Z}_p] = |\text{Aut}_{\mathbb{Z}_p}(\mathbb{Z}_p(b))|$ .

**oblixps** · March 4th 2012, 04:04 PM

thanks for the reply. i understand that the image must be {0} or $\mathbb{Z}_p$ but what I am still trying to figure out is why does $\text{Im}(\theta) = \{0\}$ imply that $[\mathbb{Z}_p(b): \mathbb{Z}_p] = 1$ ? I am having trouble seeing the connection between the degree of the field extension and the image of the map $\theta$ .

sorry i haven't learned galois theory yet so i was confused for a bit. but now that you posted the galois correspondence, it all makes sense now. thanks!

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