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I was wondering if anyone could check to see if i have done the following problem correct.

I want to find the splitting Field for $f(x)$ over $\mathbb Q$, $E$ say, and also evaluate $[E:\mathbb Q]$

Let $f(x)=x^{7}+3$

EDIT:

The roots of $x^{7}+3$ are the seven, $7$-th roots of $-3$. Now since $7$ is odd, $\exists$ a real $7$-th root of $-3$, namely $-(\sqrt[7]{3})$. Let $\alpha=-(\sqrt[7]{3})$. Let $\omega=e^{\frac{2\pi i}{7}}$ (a primitive root $7$-th root of unity. Then the seven,$7$-th roots of $-3$ are $\alpha,\alpha\omega,\alpha\omega^{2},\alpha\omega^{3},\alpha\omega^{4}...\alpha\omega^{6}$. Thus $\mathbb E=\mathbb Q(\alpha,\omega)$

Now to evaluate $[\mathbb E:\mathbb Q]=[\mathbb E:\mathbb Q(\omega)][\mathbb Q(\omega):\mathbb Q]$

So first calculating $[\mathbb Q(\omega):\mathbb Q]$.

First note that $f(x)$ is monic and also irreducible using E.I.T with prime $3$.This shows that $[\mathbb Q(\omega):\mathbb Q]=\deg f(x)=7$

Now calculating $[\mathbb E:\mathbb Q(\omega)]$.

So, now note that $\mathbb E=\mathbb Q(\alpha,\omega)$ and $x^{7}-1$ is the minimum polynomial of $\omega$ over $\mathbb Q(\alpha)$. This is because $\mathbb Q(\alpha)$ contains only real numbers and $x^{7}-1$ has only non-real solutions. Thus $[\mathbb E:\mathbb Q(\omega)]=\deg(x^{7}-1)=\deg \Phi_{7}(x)=6$. Thus $[\mathbb E:\mathbb Q]=6*7=42$.

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  • $\begingroup$ Compare with the answer here, with $n=7$ and $a=-3$. $\endgroup$ – Dietrich Burde May 17 '18 at 19:02
  • $\begingroup$ From the comparisions made, i assume i have calculated $E$ incorrectly? $\endgroup$ – Gibberish May 17 '18 at 19:08
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You need the $\;7\,-$ th roots of $\;-3=3e^{\pi i+2k\pi i},\,\,k\in\Bbb Z\;$ , and these are the seven numbers

$$z_k=\sqrt[7]3\,e^{\frac{\pi i}7\left(1+2k\right)}\;,\;\;k=0,1,2,...,6$$

Observe that $\;\zeta_7:=e^{\frac{\pi i}7}\;$ is a primitive unit root of order $\;7\;$ , so your polynomial's roots are

$$z_k:=\sqrt[7]{-3}\,\zeta_7^k\;,\;\;k=0,1,...,6$$

and from here you get a rasonably easy way to express the splitting field:

$$\Bbb E=\Bbb Q\left(\sqrt[7]{-3}\,,\,\,\zeta_7\right)$$

and its $\;\Bbb Q\,-$ dimension is $\;[\Bbb E:\Bbb Q(\zeta_7)][\Bbb Q(\zeta_7):\Bbb Q]=7\cdot6=42\;$

Your calculations are fine up to the point when you try to determine the splitting field since

$$\;\color{red}{(-i\sqrt[7]3)^7}=-\left(i\sqrt[7]3\right)^7=-i^7\cdot3=-3i\color{red}{\neq-3}\,\ldots\;$$

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  • $\begingroup$ Should $\mathbb{E}=\mathbb Q(\sqrt [7]{-3},\zeta_{7})$. be $\mathbb{E}=\mathbb Q(-\sqrt [7]{3},\zeta_{7})$?, how would i fix my working? $\endgroup$ – Gibberish May 18 '18 at 17:38
  • $\begingroup$ No. $\;\sqrt[7]{-3}\;$ is fine, as when you rise that to the seventh power you get $\;-3\;$ , which is what we want: $$x^7+3=0\iff x^7=-3$$ Of course, as the root is an odd one, it is just the same $\;\sqrt[7]{-3}=-\sqrt[7]3\;$ ...:) $\endgroup$ – DonAntonio May 18 '18 at 17:44
  • $\begingroup$ Hi i edited my post to show my new solution, can you check if its correct? $\endgroup$ – Gibberish May 18 '18 at 18:20
  • $\begingroup$ @Gibberish I really am lazzy to read all that...but you still claim that $\;-i\sqrt[7]3\;$ is a root of your polynomial and I'd expect at least this is already clear from my post above that it is false ...so I don't think your stuff can be correct. $\endgroup$ – DonAntonio May 18 '18 at 18:23
  • $\begingroup$ I think you may have read the wrong part , I separated it from what I initially wrote by EDIT: $\endgroup$ – Gibberish May 18 '18 at 18:24

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