Does $K \subseteq L$ and $K \cong L \implies$ $K=L$?

Question

Suppose $K,L$ are infinite fields with $K \subseteq L$ and $K \cong L.$ Does it imply necessarily that $K=L$?

This is certainly not true for infinite groups. For instance $2 \Bbb Z \subsetneq \Bbb Z$ but $\Bbb Z \cong 2 \Bbb Z.$ Is there any similar counter-example for infinite fields?

Any help in this regard will be highly appreciated. Thank you very much for your valuable time.

Answer 1

No.

Let $X=\{x_n\mid n\in\mathbb{N}\}$ be countably many variables, indexed by $\mathbb{N}$, and $Y=\{x_n\mid n\in\mathbb{Z}\}$ be countably many variables, indexed by $\mathbb{Z}$.

Then $K=\mathbb{Q}(X)$ is isomorphic to $L=\mathbb{Q}(Y)$; and $K\subseteq L$, but $K\neq L$.

As Lord Shark also notes, if $x$ is an indeterminate, then $\mathbf{F}(x^2)$ is isomorphic to $\mathbf{F}(x)$ (for any field $\mathbf{F}$), and $\mathbf{F}(x^2)\subseteq \mathbf{F}(x)$, but they are not equal.

Note that the example originally given as applying to rings is incorrect; $2\mathbb{Z}$ is not a ring with identity, so it is not isomorphic to $\mathbb{Z}$. But you can compare the polynomial ring $\mathbb{Z}[X]$ with $\mathbb{Z}[Y]$ ($X$ and $Y$ as above) if you don’t want fields in your example.