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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?

Community
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Lucas
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    Can you solve $x^2=7$ in your $\Bbb F_p$? – Angina Seng May 04 '18 at 06:41
  • 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. – Lucas May 04 '18 at 07:02
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    It may help to note that, modulo 11, $x^4-7=x^4+4=(x^2+2)^2-(2x)^2$. – Gerry Myerson May 04 '18 at 07:07

1 Answers1

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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.
Jyrki Lahtonen
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  • 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 $m=4\ell$. – Jyrki Lahtonen May 04 '18 at 07:53
  • @Jykri Lahtonen I am rather confused about this answer. Could you clarify a few things? 2. In finding the splitting field of $x^4-7$ over $\mathbb{F}5$ say. We would find that [7] has order 4 in $\mathcal{U}(\mathbb{F}_5)$. But there are no roots of this polynomial in $\mathbb{F}_5$ because x^4 can only ever be in [0] or [1] mod 5. 4. I am also confused because say we found that m=2 and $\mathbb{F}{25}$. How can this be the splitting field, because it has an entirely different operation from $\mathbb{F}_5$? What's more, nothing to the fourth power can be 2 or 7 mod 25 either... – Wyatt Kuehster Jun 15 '19 at 04:56
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    @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. – Jyrki Lahtonen Jun 17 '19 at 06:05
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    @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. – Jyrki Lahtonen Jun 17 '19 at 06:07
  • Thanks for fixing that big confusion! Sorry I am still on this topic, but I am still not sure why the fact that 7 has order 4 in $\mathbb{F}p^*$ means that we need 16th roots of unity to get the splitting field. I know we can factor out the principal ideal generated by $x^4-7$ from $\mathbb{Z}_5[x]$ which gives us a field isomorphic to $\mathbb{F}{625}$ containing a root of $t^4-7$ (but we don't know it's a splitting field). I can also see that $(x^4-7)^4 \equiv 1 \pmod 5$. But I'm not sure I'm getting a coherent picture of why this works. – Wyatt Kuehster Jun 25 '19 at 05:58
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    $7=2$ 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. – Jyrki Lahtonen Jun 25 '19 at 10:29
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    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 $\Bbb{F}_5$). – Jyrki Lahtonen Jun 25 '19 at 10:32
  • If $\alpha\in K=\Bbb{F}_{5^4}$ is one of the zeros, then $\alpha^5, \alpha^{25}$ and $\alpha^{125}$ are the others (Freshman's dream). They are all in the field $K$, because $\alpha$ is there (and hence its powers). – Jyrki Lahtonen Jun 25 '19 at 10:34
  • But, @Wyatt, I don't understand what you mean by $(x^4-7)^4\equiv1\pmod 5$? That cannot be right. $(x^4-7)^4$ is of degree sixteen so definitely not a constant. Are you possibly confused by the difference between a polynomial and a polynomial function? – Jyrki Lahtonen Jun 25 '19 at 10:37
  • Aha!! Thank you so much for the detailed explanation. I've got it now!!

    I was trying to say that you get 1 mod 5 for $x=0,1,2,3,4$, but I see why that was a dumb thing to say :)

     – Wyatt Kuehster Jun 25 '19 at 14:03