Is there an easier way to calculate the degree of extension of a splitting field for a polynomial like $$x^7-3\quad\text{over}\quad\Bbb Z_5\,?$$My approach for several of these have been to find all roots in the given field (in this case, I think $x=2$ is the only root), then I can factor it via long division. In this case, I get $$x^7-3=(x-2)(x^6+2x^5+4x^4+3x^3+x^2+2x+4).$$At this point, I would check that $2$ is not a repeated root, and here it is easy to check that this is not the case. Since I've checked all of the other elements, at this stage I would adjoin a root of this polynomial, use long division and get a $5^\text{th}$ degree polynomial. Now, the new field I'm working in would have degree $6$, since it is the root of a $6^\text{th}$ degree irreducible polynomial, right?

At this point, it begins to feel like I'm searching for a needle in a haystack; I have several more elements that I have to begin trying, and for this particular problem, that gets to be overwhelming.

At some point, I had thought the map $\alpha\mapsto \alpha^p$, where $p=5$ in this case, would work, but I had another problem where that wasn't the case (specifically, I tried to find the splitting field of $x^5+x+1$ over $\Bbb Z_2$. Here it was easy to see it was irreducible, so I adjoined a root, lets call it $\gamma$, and using the method above, I found $\gamma^2$ was a root, but not $\gamma^4$).

So my question is the following: is there a better approach than what I'm doing to factor these (and in the process, find the degree of extension)?

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    $\begingroup$ In general for the problem of factoring polynomials over finite fields there is Berlekamp's algorithm. In small cases such as this, experience and ad hoc techniques will usually be good enough. $\endgroup$ – Jyrki Lahtonen Feb 25 '13 at 15:05
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    $\begingroup$ But something went wrong in your binary example: $$x^5+x+1=(x^3+x^2+1)(x^2+x+1).$$ This is not irreducible! The squares of its roots (as well as fourth powers) are, of course, still also roots. $\endgroup$ – Jyrki Lahtonen Feb 25 '13 at 15:07
  • $\begingroup$ @JyrkiLahtonen: Thanks! I didn't realize that! $\endgroup$ – Clayton Feb 25 '13 at 23:42

Once you know one root of $x^7-3$, namely $2$, you get all others by multiplying with $7$-th roots of unity. Since $x^7-1$ and its derivative $7x^6$ have no common roots, the polynomial $x^7-1$ is separable, so $\overline{ \mathbb F}_5$ really contains $7$ different roots of unity, say $1,\zeta,\zeta^2,\ldots,\zeta^6$. Then $2,2\zeta,\ldots,2\zeta^6$ are the different roots of $x^7-3$.

Assume that the extension $\mathbb F_{5^n}$ of $\mathbb F_5$ contains $\zeta$. Then $\zeta$ generates a subgroup of $\mathbb F_{5^n}^\times$ of order $7$, so by Lagrange's theorem, $7|(5^n-1)$, or equivalently $5^n \equiv 1 \pmod 7$. Thus $n$ is a multiple of the order of $5$ modulo $7$.

Conversely, if $n \geq 1$ is such that $5^n \equiv 1 \pmod 7$, then $7$ divides $5^n-1$ and this implies that the polynomial $x^7-1$ divides $x^{5^n-1} - 1$. Since $\mathbb F_{5^n}$ is the splitting field of $x^{5^n-1} - 1$, we get $\zeta \in \mathbb F_{5^n}$.

This shows that $\mathbb F_5(\zeta) = \mathbb F_{5^n}$ where $n$ is the order of $5$ modulo $7$, which is $6$. Therefore, the splitting field of $x^7-3$ has degree $6$ over $\mathbb F_5$.

Of course these arguments only worked because we could reduce the problem to finding the degree of $\mathbb F_5(\zeta)$ over $\mathbb F_5$ where $\zeta$ is a primitive $7$-th root of unity. For general polynomials, things are probably more difficult.

  • $\begingroup$ The order of $5$ modulo $7$ must divide $7-1$, so it can't be $4$. $\endgroup$ – Gerry Myerson Feb 25 '13 at 6:13
  • $\begingroup$ Hm, you're right. I corrected that in my answer. $\endgroup$ – marlu Feb 25 '13 at 14:23
  • $\begingroup$ @marlu: For the converse statement, why is this true? The part that $7$ divides $5^n-1$ so $x^7-1$ divides $x^{5^n-1}-1$? $\endgroup$ – Clayton Feb 26 '13 at 2:07
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    $\begingroup$ @Clayton: In general, if $a|b$ then $x^a-1|x^b-1$. Write $b=ac$, then by the geometric summation formula, $\frac{x^{ac}-1}{x^a-1} = 1+x^a+\ldots+x^{(c-1)a}$. $\endgroup$ – marlu Feb 26 '13 at 15:25

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