Prove that the polynomial $$f (t) = t^7 + 10t^2 − 5$$ has no roots in the field $\mathbb Q(\sqrt[3] 5, \sqrt[5] 5)$.
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1$\begingroup$ do you mean cubic and quintic roots i.e. $\mathbb{Q}(\sqrt[3]5,\sqrt[5]5)$? $\endgroup$– obatakuJan 5, 2014 at 9:55
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$\begingroup$ @oldrinb sorry not good at English.. Yep it has been edited already. $\endgroup$– JammyJan 5, 2014 at 9:57
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$\begingroup$ For me this question does not seem to be natural... what is the background for this question... $\endgroup$– user87543Jan 5, 2014 at 10:09
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2 Answers
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Hint 1. Find the degree of field extension $\mathbb Q\subset\mathbb Q(\sqrt[3] 5, \sqrt[5] 5)$.
Hint 2. Prove that $f (t) = t^7 + 10t^2 − 5$ is irreducible over $\mathbb Q$.
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The extensions $\mathbb Q(\sqrt[3] 5)$ and $\mathbb Q( \sqrt[5] 5)$ have degrees $3$ and $5$ over $\mathbb Q$.
Since $3$ and $5$ are relatively prime the extension $\mathbb Q(\sqrt[3] 5, \sqrt[5] 5)$ has degree $15$ over $\mathbb Q$.
On the other hand, the polynomial $f(t)$ is irreducible by Eisenstein.
So if it had a root $a\in \mathbb Q(\sqrt[3] 5, \sqrt[5] 5)$, we would have $[\mathbb Q(a):\mathbb Q]=7$.
But then we would have (by the multiplicativity of degrees in towers of extensions) $$15=[\mathbb Q(\sqrt[3] 5, \sqrt[5] 5):\mathbb Q]=[\mathbb Q(\sqrt[3] 5, \sqrt[5] 5):\mathbb Q(a)] \cdot [\mathbb Q(a):\mathbb Q] = [\mathbb Q(\sqrt[3] 5, \sqrt[5] 5):\mathbb Q(a)] \cdot 7 $$ an impossibility.
So indeed no root $a$ of $f(t)$ can exist in $\mathbb Q(\sqrt[3] 5, \sqrt[5] 5)$ .
answered Jan 5, 2014 at 10:33
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$\begingroup$ @user: 1) You claim without any argument that this a homework question and your 99% figure is drawn from thin air.2) If the OP appreciates my answer so much the better: I wanted to help him and not bow to your criteria for a good answer. 3) What I said suffices to prove that the extension has degree exactly $15$ to anyone who has a minimal understanding of elementary field theory. It is rather ironical that you reproach me both for giving "a too extended answer" and then for not giving a full proof of an essentially trivial result! $\endgroup$ Jan 5, 2014 at 11:04