0

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.

math maniac.
  • 1,993

1 Answers1

5

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.

Arturo Magidin
  • 398,050
  • @Arturo Magidin how do I show isomorphism between F $(x^2)$ and F $(x)$? I guess $x \mapsto x^2$ with identity on F will work well. Am I right? – math maniac. Nov 25 '19 at 05:45
  • @mathmaniac. They are both purely transcendental extensions of rank 1. – Arturo Magidin Nov 25 '19 at 05:50
  • Sorry @Arturo Magidin I haven't studied transcendental extension yet. So it will be better for me to digest if you kindly demonstrate an isomorphism explicitly between the two fields. – math maniac. Nov 25 '19 at 05:52
  • @mathmaniac. Yes, that isomorphism works. The map $x^2\to x$ induces a homomorphism $\mathbf{F}[x^2]\to\mathbf{F}(x)$, and since you are mapping into a field, it factors through the field of fractions. giving the desired isomorphism. – Arturo Magidin Nov 25 '19 at 05:53
  • Yeah now I got it. Thank you very much @Arturo Magidin sir for your kind help. If you allow me can I ask you a question? – math maniac. Nov 25 '19 at 05:55
  • @mathmaniac. You already did; but you can ask another... – Arturo Magidin Nov 25 '19 at 05:58
  • Ok. Thanks. I don't want to create annoyance to anyone all the time to fulfill my requirement. – math maniac. Nov 25 '19 at 06:02