Determine the splitting field of $x^4 - 7$ over

(a) $\mathbb{Q}$

(b) $\mathbb{F}_{5}$

(c) $\mathbb{F}_{11}$

For (a): $x^4 - 7 = (x-\sqrt[4]{7})(x+\sqrt[4]{7})(x-i\sqrt[4]{7})(x+i\sqrt[4]{7})$. The splitting field of $x^4 - 7$ is $\mathbb{Q}(\sqrt[4]{7},i)$.

For (b) and (c): I want to determine the splitting field over $\mathbb{F}_{p}$ (for $p \neq 7$, of course). How can I determine this? Is possible?

  • 1
    $\begingroup$ Can you solve $x^2=7$ in your $\Bbb F_p$? $\endgroup$ – Angina Seng May 4 '18 at 6:41
  • $\begingroup$ If $p=5$, we have $1^2 = 1, 2^2 = 4, 3^2 = 4, 4^2 = 1$. If $p = 11$, we have $1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 5, 5^2 = 3, 6^2 = 4, 7^2 = 5, 8^2 = 9, 9^2 = 4, 10^2 = 1$. In both cases, has no solutions. For any $p \neq 7$, I don't know how to calculate. I couldn't see yet how this helps. $\endgroup$ – Corrêa May 4 '18 at 7:02
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    $\begingroup$ It may help to note that, modulo 11, $x^4-7=x^4+4=(x^2+2)^2-(2x)^2$. $\endgroup$ – Gerry Myerson May 4 '18 at 7:07

My algorithm for the fields $\Bbb{F}_p$:

  1. Determine the order of $7$ as a root of unity, i.e. its order in the multiplicative group.
  2. Determine the orders of the roots of the polynomial in the multiplicative group of the splitting field, call the largest of them $m$ (they may not all be equal, but the others are factors of the largest).
  3. Find the smallest exponent $n$ such that $m\mid p^n-1$.
  4. The field $\Bbb{F}_{p^n}$ is the smallest containing $m$th roots of unity. As it contains all of them it must be the splitting field.
  • $\begingroup$ In step 2 the extra clause is to cater for the following. Let's say that it turns out that $7$ is of odd order $\ell$. In that case there will be one solution of $x^4=7$ in the cyclic group $\langle 7\rangle$ (raising to fourth power is a bijection in a cyclic grop of an odd order). The other roots will have order $m=4\ell$. $\endgroup$ – Jyrki Lahtonen May 4 '18 at 7:53
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    $\begingroup$ @Wyatt True, there are no zeros of $x^4-7$. Why would that be a surpries? Anyway, you see that $7$ has order four. Thereafter it is easy to show that the roots of $x^4-7$ in some extension field must have order $16$. The smallles extension field of $\Bbb{F}_5$ that has roots of unity of order sixteen is $\Bbb{F}_{625}$ which then must be the splitting field. $\endgroup$ – Jyrki Lahtonen Jun 17 '19 at 6:05
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    $\begingroup$ @Wyat you also seem to be confused about fiinite fields. $\Bbb{F}_{25}$ has nothing to do with modulo $25$. It has characteristic five, so you do arithmetic with coefficients modulo five only. $\endgroup$ – Jyrki Lahtonen Jun 17 '19 at 6:07
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    $\begingroup$ $7=2$ has order four in $\Bbb{F}_5$. If $\alpha$ is a zero (in some extension field), then $\alpha^4=7$ implying that $\alpha^{16}=7^4=1$. Therefore the order of $\alpha$ is a factor of $16$. But $\alpha^8=7^2=-1$, so the order is not a factor of $8$. Ergo, the order is exactly sixteen. $\endgroup$ – Jyrki Lahtonen Jun 25 '19 at 10:29
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    $\begingroup$ Then, none of $\Bbb{F}_{5^\ell},\ell=1,2,3,$ have elements of order $16$, because $16$ is not a factor of $5^\ell-1$. Therefore $\alpha\in\Bbb{F}_{5^4}$ and $x^4-7$ is its minimal polynomial (so irreducible over $\Bbb{F}_5$). $\endgroup$ – Jyrki Lahtonen Jun 25 '19 at 10:32

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