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Let $K = \mathbb{Q}(\sqrt{2},\sqrt{5},\sqrt{7})$: I know that $K/F$, with $F =\mathbb{Q}$, is Galois and has degree 8. By the primitive element theorem there is a a $\theta$ such that $K = F(\theta)$, in this case it could be a linear combination given by $\sqrt{2}+\sqrt{5}+\sqrt{7}$. My question is: What is the degree $d$ of the minimal polynomial for $F(\theta)$. Show that $d$ is independent of $\theta$.

The minimal polynomial should have degree 8 as $[F(\theta):\mathbb{Q}] = deg(m_{\theta}(x))$. How do I prove the independence?

Thanks

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    $\begingroup$ Note that you didn't define $F$. $\endgroup$ – user21820 May 16 '14 at 6:34
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Since you have $d = \deg(m_θ) = [\mathbb{Q}(θ):\mathbb{Q}] = [K:\mathbb{Q}] = 8$, you already have that $d$ is independent of the choice of $θ$ as long as it is a primitive element for $K/\mathbb{Q}$. But I'm not so sure you understand the second equality, and how did you know that $a = \sqrt{2}+\sqrt{5}+\sqrt{7}$ is a primitive element?

From your comment I see that you don't actually know how to prove that $a$ is a primitive element. Jyrki gave one hint for proving it, which is to observe that any element in $Gal(K/F)$ must preserve the roots of $(t^2-2)$ and $(t^2-7)$ and $(t^2-5)$ as polynomials in $t$, and so it can only flip the signs of the radicals. Since the only element of $Gal(K/F)$ that fixes $a$ is the identity automorphism, $Gal(K/F(a))$ has size $1$ and so $[K:F(a)] = 1$, and hence $K = F(a)$. However, this method is curiously specific to this particular primitive element, and cannot be used to prove that some other choices of signs also gives a primitive element.

Here is a generic method that works for many other fields as well. Let $(r,s,t)=(\sqrt{2},\sqrt{5},\sqrt{7})$. Then we know that the subextensions of $K/F$ correspond to the subgroups of $Gal(K/F)$, which is itself a subgroup of ${S_2}^3$ because any element in $Gal(K/F)$ must preserve each of $\{r,-r\}$ and $\{s,-s\}$ and $\{t,-t\}$, and hence must be ${S_2}^3$ because $|Gal(K/F)| = [K:F] = 8$. Now consider the eight subextensions $F,F(r),F(s),F(t),F(r,s),F(r,t),F(s,t),F(r,s,t)$. They are all distinct because if any two are the same then $F(r,s,t)$ collapses to a smaller field. For example, if $F(r) = F(s)$ then $F(r,s,t) = F(r,t)$ which has too low extension degree. Therefore these are all the subextensions of $K/F$. If $F(r+s+t)$ is equal to any of them except $K$, then again $F(r,s,t)$ collapses. For example, if $F(r+s+t) = F(r,s)$ then $F(r,s,t) = F(r,s,r+s+t) = F(r,s)$. Therefore $F(r+s+t) = F(r,s,t)$ and likewise this shows that in fact any non-zero coefficients for a linear combination of $r,s,t$ would give a primitive element. (Note that this is not true in general!)

Nevertheless in this particular problem the crucial point is that $[K:F] = 8$, which is most easily proved by proving the independence of $r,s,t$, and so the above method is redundant! Do you really have a correct proof of $[K:F] = 8$?

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  • $\begingroup$ $\sqrt{2}+\sqrt{5}+\sqrt{7}$ is a primitive element by the proof of the primitive element theorem. As the proof of the primitive element indicates, a primitive element for an extension can be obtained as a simple linear combination of the generators for the extension. $\endgroup$ – Leonhard Leibniz May 16 '14 at 6:35
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    $\begingroup$ @LeonhardLeibniz: Yes, but the theorem does not tell you which linear combination works, so your claim does not follow. $\endgroup$ – user21820 May 16 '14 at 6:38
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    $\begingroup$ If you know that $K/F$ is Galois over of degree 8, and that the Galois group consists of mappings determined by all sign combinations of $\pm\sqrt2$, $\pm\sqrt5$ and $\pm\sqrt7$, then the primitivity of $\alpha=\sqrt2+\sqrt5+\sqrt7$ follows from the fact that all its conjugates are strictly less than $\alpha$. $\endgroup$ – Jyrki Lahtonen May 16 '14 at 13:23
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    $\begingroup$ @JyrkiLahtonen: My point was that Leonhard Leibniz did not know that the primitive element theorem does not give what he claims at all. $\endgroup$ – user21820 May 16 '14 at 13:53
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    $\begingroup$ @JyrkiLahtonen: It's okay I was just making clear that it's an unproven claim that is not trivial. I've added an explanation of your comment, in case it is useful to Leonhard Leibniz. =) $\endgroup$ – user21820 May 16 '14 at 14:50

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