Haven't seen this posted here before. This is another old prelim problem. My feeling is that no such field extension exists. Why I think this is that $\mathbb{Q}(x)$ has no index 2 subgroups, but I'm not sure how to show this.
2026-04-25 23:10:46.1777158646
Does there exists $F\subseteq K$ with $[K:F]=2$ where $F\simeq K\simeq \mathbb{Q}(x)$
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Let the fields $F,K$ be given by $F=Q(x^2),\;K=Q(x)$.