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I just realized that finding the splitting field of a polynomial over finite fields is not as "straightforward" as in $\mathbb{Q}$

I am struggling with the following problem:

"Find the splitting field of $f(x)= x^{15}-2$ over $\mathbb{Z}_7=\Bbb F_7$, the finite field of $7$ elements."

By direct computation, $f(x)$ has no roots on $\mathbb{Z}_7$ ; however, I do not how to prove that $f$ is actually irreducible.

I just found this lecture http://hyperelliptic.org/tanja/teaching/CCI11/online-ff.pdf

Using lemma 67, I can conclude that my polynomial is irreducible (although the proof seems a little weird)

Therefore, I think that the splitting field is $F= \mathbb{Z}_7(\alpha,\zeta)$ where $\alpha^{15} = 2$ and $\zeta$ is the $15th-$root of unity.

I want to describe $F$ as $\mathbb{F}_{7^n}$ for a suitable $n$.

Adam Hughes
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Blood Borne
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  • Do you mean $\Bbb Z_7$ as the $7$-adic integers or do you really mean $\Bbb Z/7\Bbb Z=\Bbb F_7$ the field with $7$ elements? – Adam Hughes Sep 29 '15 at 19:39
  • @AdamHughes: there is a clue in the title. – Rob Arthan Sep 29 '15 at 19:42
  • @RobArthan thanks for the pointer. I'll edit the question. – Adam Hughes Sep 29 '15 at 19:43
  • I mean the finite field with 7 elements. I will change to avoid such confusions. – Blood Borne Sep 29 '15 at 19:43
  • Similar question is this one, case c) and d). – Dietrich Burde Sep 29 '15 at 20:00
  • Dear Experts, Is this way correct: since for any $\alpha$ in $\mathbb{F}_7$, we have $\alpha^7=\alpha$, hence $\alpha^{15}-2=\alpha^7.\alpha^7.\alpha-2=\alpha^3-2$. As you have checked that the given polynomial has no root in $\mathbb{F}_7$, thus the polynomial $x^3-2$ is irreducible over $\mathbb{F}_7$. Can we conclude $x^{15}-2$ is also irreducible over $\mathbb{F}_7$? – Groups Sep 30 '15 at 05:19
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    @Groups: We cannot make such a conclusion. And this polynomial is not irreducible. See my answer for a factorization. – Jyrki Lahtonen Sep 30 '15 at 12:08

2 Answers2

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There is a typo in the statement of Lemma 67 in your source. The $n$th roots of unity are in $\Bbb{F}_p$ only if $n\mid p-1$ or, iff $p\equiv1\pmod n$ (not $n\equiv1\pmod p$ as is written there). Therefore that Lemma does not apply.

In fact, the polynomial $x^{15}-2$ is NOT irreducible in $\Bbb{F}_7[x]$. This follows trivially from the fact that $3^5=243\equiv-2\pmod 7$. Therefore $$ x^{15}-2=(x^3)^5+3^5=(x^3+3)(x^{12}-3x^9+3^2x^6-3^3x^3+3^4). $$

We immediately see that $x^3+3$ has no zeros in $\Bbb{F}_7$ (the cubes in that field are $0,\pm1$), so it is irreducible. Therefore the polynomial has a zero $\alpha$ in $\Bbb{F}_{7^3}$.

To get the splitting field of $x^{15}-2$ we need, as you observed, the primitive 15th roots of unity. We easily see that $$ 7^4=2401\equiv1\pmod{15}. $$ The multiplicative group of the field $\Bbb{F}_{7^4}$ is cyclic of order $7^4-1$, and thus it contains a primitive 15th root of unity $\zeta$.

A consequence of all this is that the splitting field of this polynomial is $$ \Bbb{F}_7[\alpha,\zeta]=\Bbb{F}_{7^{12}}. $$

Jyrki Lahtonen
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  • Raising to fifth power is a permutation of $\Bbb{F}_7$, so I was immediately confident that this polynomial is not irreducible. You don't need Mathematica/WA/CASofYourChoice to find that. To see that the degree 12 factor above is irreducible you need more precise information from Eric Wofsey's answer (+1): the zeros of this polynomial form a coset of the group of fifteenth roots of unity inside the bigger group of 45th roots of unity, and it is easy to see that the said coset then contains primitive 45th roots of unity. Those have minimal polynomials of degree 12. – Jyrki Lahtonen Sep 30 '15 at 12:05
  • Could you explain please the following: I see that $x^3+3$ has no zeros in $\mathbb{F}7$. How it follows that the polynomial has a zero $\alpha$ in $\mathbb{F}{7^3}$? – RFZ Mar 01 '19 at 16:20
  • @ZFR Do you see why $p(x)=x^3+3$ is irreducible in $\Bbb{F}7[x]$? It follows that one of the ways of constructing $\Bbb{F}{7^3}$ is to form the quotient ring $K=\Bbb{F}7[t]/\langle p(t)\rangle$. $K$ is a field because an irreducible polynomial generates a maximal ideal. And the coset $\alpha=t+\langle p(t)\rangle$ is then automatically a zero of $p$. Given that $|K|=7^3$ we can conclude that $K\simeq\Bbb{F}{7^3}$. Assuming you have proven uniqueness of a finite field (up to isomorphism) of a given cardinality. – Jyrki Lahtonen Mar 02 '19 at 06:20
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Here is a way to find the splitting field without having to factor the polynomial. Observe that $2$ is a primitive cube root of $1$ in $\mathbb{F}_7$, so $x^{15}-2$ splits completely in $\mathbb{F}_{7^n}$ iff there is a primitive $45$th root of $1$ in $\mathbb{F}_{7^n}$. It follows that the splitting field is $\mathbb{F}_{7^n}$ for the least $n$ such that $7^n-1$ is divisible by $45$. Doing some arithmetic mod $45$, it is not hard to compute that this $n$ is $12$.

Eric Wofsey
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