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Let $L = \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ be a field extensions of $\mathbb{Q}$. Show that if $f(x) \in \mathbb{Q}[x]$ is a monic irreducible polynomial of degree $3$, then $f$ has no roots in $L$.
I already showed that $L/\mathbb{Q}$ is Galois so, in particular, $L/\mathbb{Q}$ is normal. Thus, if $f$ has a root in $L$ then $f$ splits on $L$. But, I don't know how to use this to get a contradiction. Can someone help me? Thanks for the advance!
You don't need that $L/\mathbb{Q}$ is Galois, or anything sophisticated at all. Indeed, it is easy to see that the only prime divisor of $[L:\mathbb{Q}]$ is $2$. (With a little more work, one can show that $[L:\mathbb{Q}] = 8$, but this is not necessary). Any root $\alpha$ of $f$ generates a degree $3$ extension of $\mathbb{Q}$, since $f$ is irreducible of degree $3$. If $\alpha \in L$, then $L$ would have a subfield of degree $3$ over $\mathbb{Q}$, which is a contradiction, since this would imply $3$ divides $[L:\mathbb{Q}]$.
For the chain of fields $$ \mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(\sqrt{2},\sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}) $$ each successive extension is generated by an element of degree $2$ over $\mathbb{Q}$, hence degree at most $2$ (in fact, exactly $2$ but all we need is at most $2$) over the prior field in the chain.
Thus, each of the integers \begin{align*} &[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}(\sqrt{2},\sqrt{3})]\\[4pt] &[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]\\[4pt] &[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]\\[4pt] \end{align*} is either $1$ or $2$, hence $[L:Q]$ is a divisor of $8$.
It follows that $L$ has no subfield whose degree over $\mathbb{Q}$ is $3$.
The Galois group consists (quite obviously) of maps that send $\sqrt2\mapsto\pm\sqrt 2$, $\sqrt3\mapsto\pm\sqrt 3$, and $\sqrt5\mapsto\pm\sqrt 5$. In other words, $G\cong (\Bbb Z/2\Bbb Z)^3$ and acts on the 3-element set of roots of $f$. The image of $G$ in $S_3$ is a $2$-group hence either trivial or consists of a 2-cycle and the identity. At any rate, it is not transitive, i.e., one of the roots of $f$ is $G$-invariant, i.e., $f$ has a rational root.
