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I'm studying algebra by Herstein's Topics in Algebra, 2ed, and stuck at Ex.5.1.6(b):

In $\mathbb Q(\sqrt2,\sqrt[3]5)$ characterize all the elements $w$ such that $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt[3]5)$.

My approach is that since $\mathbb Q(\sqrt2,\sqrt[3]5)=6$, $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt[3]5)$ if and only if $w$ is not of degree 6, hence $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt[3]5)$ if and only if $w$ is of degree 1,2, or 3.

I get a hint from the answer to question Finding all reals such that two field extensions are equal. that $w$ is of degree 2 (resp. 3) if and only if it is in form of $a+b\sqrt2$ (resp. $a+b\sqrt[3]5+c(\sqrt[3]5)^2$). It is easy to show the if part is valid, but how to prove that $w$ is of degree 2 (resp. 3) only if it is in form of $a+b\sqrt2$ (resp. $a+b\sqrt[3]5+c(\sqrt[3]5)^2$)? Thanks.

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    $\begingroup$ Do you know Galois theory? This could help. $\endgroup$
    – Martin Brandenburg
    Jul 15, 2014 at 6:59
  • $\begingroup$ No, I don't. But Galois theory is not introduced yet where the exercise appears, so the exercise is to be solved without using Galois theory. $\endgroup$
    – pngleo
    Jul 15, 2014 at 7:27

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Your hypothesis is correct. Let $w=\sqrt{2}+a\sqrt[3]{5}+b\sqrt[3]{25}$ and assume that $w$ has degree $3$, then we have $\sqrt{2} \in \mathbb{Q}(\sqrt[3]{5}, w)$ and $$[ \mathbb{Q}(\sqrt[3]{5}, w): \mathbb{Q}]=[ \mathbb{Q}(\sqrt[3]{5}, w): \mathbb{Q}(w)][ \mathbb{Q}(w): \mathbb{Q}]=3 \ \text{or} \ 9$$ but this is impossible since $[ \mathbb{Q}(\sqrt[3]{5}, w): \mathbb{Q}]=[ \mathbb{Q}(\sqrt[3]{5}, w): \mathbb{Q}(\sqrt{2})][ \mathbb{Q}(\sqrt{2}): \mathbb{Q}]$ is even.

The other case can be solved similarly.

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  • $\begingroup$ Thanks for your reply. But why can't $[\mathbb Q(\sqrt[3]5,w):\mathbb Q(w)]$ be 2? $\endgroup$
    – pngleo
    Jul 16, 2014 at 6:01
  • $\begingroup$ You are adding the root of $x^3-5$, if this is irreducible over $\mathbb{Q}(w)$ then the degree is $3$. If its reducible then since one root is real and the other two are complex, and the fact that $\mathbb{Q}(w)$ is real the root must be $\sqrt[3]{5}$ and the degree would be $1$. $\endgroup$
    – Rene Schipperus
    Jul 16, 2014 at 12:55
  • $\begingroup$ Got it. But it's still unclear for me how to do it in a similar way for degree 2. Could you give some hint? My idea is, since $aw^2+bw+c=\sqrt2\alpha+\beta$, where $\alpha=2a\gamma+b,\beta=a\gamma^2+b\gamma+2a+c$ and $\gamma=a\sqrt[3]5+b\sqrt[3]{25}$ and $\alpha,\beta,\gamma\in\mathbb Q(\sqrt[3]5)$, we have $aw^2+bw+c=0$ if and only if $\alpha=\beta=0$. And $\alpha=0$ implies $\gamma=-b/(2a)$, so $2ab\sqrt[3]{25}+2a^2\sqrt[3]5+b=0$, a contradiction with the fact that the degree of $\sqrt[3]5$ over $\mathbb Q$ is 3. $\endgroup$
    – pngleo
    Jul 17, 2014 at 9:38

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