Let $F \subseteq E$ be a field extension. If $\sigma$ is an isomorphism between $E$ and $F$, does it follow that $E=F$?
This has maybe an obvious answer, but I couldn’t figure it out myself or find it anywhere (I’m taking an introductory course on abstract algebra). I’m guessing it’d depend on whether the degree of the extension is finite or not.
On
It doesn't have to be. For example let (R,+) the group of the real numbers with the addition operation, and let (R⁺,*) be the group of the positive real numbers and with the multiplicative operation. Then the isomorphism via the exponential and the logarithm.
https://en.wikipedia.org/wiki/Isomorphism
Sorry, I just realized you were talking about fields and not groups.
No. The extension $\mathbb{Q}(X)/\mathbb{Q}(X^2)$ has degree $2$, so the two fields are not equal, but they are clearly isomorphic.
Nevertheless, you have the following positive result: let $K$ be the prime subfield of $E$ (which is also the prime subfield of $F$). If $[E:K]$ is finite and $F\simeq E$, then $F=E$.
Indeed, a ring isomorphism is automatically an isomorphism of extensions of $K$ ($K$ is just the subring generated by $1$ !). In particular, $[E:K]=[F:K]$, which are both finite , which implies $E=F$ (two $K$-vector spaces of same finite dimension and containing each other are equal)