# Characterize elements in $\mathbb Q(\sqrt2,\sqrt5)$

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

In $\mathbb Q(\sqrt2,\sqrt5)$ characterize all the elements $w$ such that $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt5)$.

My approach is that since $\mathbb Q(\sqrt2,\sqrt5)=6$, $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt5)$ if and only if $w$ is not of degree 6, hence $\mathbb Q(w)\ne \mathbb Q(\sqrt2,\sqrt5)$ 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. $a+b\sqrt5+c(\sqrt5)^2$). 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. $a+b\sqrt5+c(\sqrt5)^2$)? Thanks.

• Do you know Galois theory? This could help. Jul 15, 2014 at 6:59
• 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. Jul 15, 2014 at 7:27

Your hypothesis is correct. Let $w=\sqrt{2}+a\sqrt{5}+b\sqrt{25}$ and assume that $w$ has degree $3$, then we have $\sqrt{2} \in \mathbb{Q}(\sqrt{5}, w)$ and $$[ \mathbb{Q}(\sqrt{5}, w): \mathbb{Q}]=[ \mathbb{Q}(\sqrt{5}, w): \mathbb{Q}(w)][ \mathbb{Q}(w): \mathbb{Q}]=3 \ \text{or} \ 9$$ but this is impossible since $[ \mathbb{Q}(\sqrt{5}, w): \mathbb{Q}]=[ \mathbb{Q}(\sqrt{5}, w): \mathbb{Q}(\sqrt{2})][ \mathbb{Q}(\sqrt{2}): \mathbb{Q}]$ is even.
• Thanks for your reply. But why can't $[\mathbb Q(\sqrt5,w):\mathbb Q(w)]$ be 2? Jul 16, 2014 at 6:01
• 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 $\sqrt{5}$ and the degree would be $1$. Jul 16, 2014 at 12:55
• 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 $\gamma=a\sqrt5+b\sqrt{25}$ and $\alpha,\beta,\gamma\in\mathbb Q(\sqrt5)$, we have $aw^2+bw+c=0$ if and only if $\alpha=\beta=0$. And $\alpha=0$ implies $\gamma=-b/(2a)$, so $2ab\sqrt{25}+2a^2\sqrt5+b=0$, a contradiction with the fact that the degree of $\sqrt5$ over $\mathbb Q$ is 3. Jul 17, 2014 at 9:38