# Can two fields which are extensions of one another be non isomorphic

Let $$\mathbb L$$ and $$\mathbb K$$ be two fields. We say that $$\mathbb L$$ is an extension of $$\mathbb K$$ if there exists a ring homomorphism $$\varphi$$ from $$\mathbb K$$ to $$\mathbb L$$. In that case, $$\varphi$$ is necessarily one-to-one and it is customary to identify $$x\in\mathbb K$$ with its image $$\varphi(x)\in\mathbb L$$, so that we can think of it as an inclusion $$\mathbb K\subset\mathbb L$$.

I am not really clear with the caveats of such an identification. In particular there is a question I could not find an answer to: if $$\mathbb K\subset\mathbb L$$ and $$\mathbb L\subset\mathbb K$$ for the real inclusion, we obviously have $$\mathbb K=\mathbb L$$. But what about with the aforementioned identifications? That is, if $$\mathbb L$$ is an extension of $$\mathbb K$$, and $$\mathbb K$$ is an extension of $$\mathbb L$$, do we necessarily have that $$\mathbb K$$ and $$\mathbb L$$ are isomorphic fields?

What I have tried: Suppose that $$\mathbb L$$ and $$\mathbb K$$ are extensions of one another. Then there exist ring homomorphisms $$\varphi:\mathbb K\to\mathbb L$$ and $$\psi:\mathbb L\to\mathbb K$$. So $$\varphi\circ\psi$$ is a ring homomorphism from $$\mathbb L$$ to itself. To conclude it would suffice to prove that its image is $$\mathbb L$$, or at least isomorphic to $$\mathbb L$$. Why would that be true? I do not know. Is there a counter-example? I do not know.

• Yes, there are counter-examples. Commented Oct 4, 2023 at 9:12
• Minor quibble: the image of $\varphi \circ \psi$ is always isomorphic to $L$. To conclude you would have needed to show that it is exactly $L$ (which is false). Commented Oct 4, 2023 at 10:08
• Hmm I got confused, I don't know why I wrote this, and I don't understand why you say that.
– Will
Commented Oct 4, 2023 at 10:13
• How do you see that the image of $\varphi\circ\psi$ is isomorphic to $\mathbb L$?
– Will
Commented Oct 4, 2023 at 14:08
• Any homomorphism between fields is injective, so it's always an isomorphism onto its image. Commented Oct 4, 2023 at 17:05

That is, if $$\mathbb L$$ is an extension of $$\mathbb K$$, and $$\mathbb K$$ is an extension of $$\mathbb L$$, do we necessarily have that $$\mathbb K$$ and $$\mathbb L$$ are isomorphic fields?

No. For example, $$\mathbb{C}(x)$$ is clearly an extension of $$\mathbb{C}$$. Much less obviously, $$\mathbb{C}$$ is also an extension of $$\mathbb{C}(x)$$, because for general field-theoretic reasons the algebraic closure $$\overline{\mathbb{C}(x)}$$ must be abstractly isomorphic to $$\mathbb{C}$$ (assuming the axiom of choice). And $$\mathbb{C}$$ and $$\mathbb{C}(x)$$ are not isomorphic because the latter is not algebraically closed, e.g. the polynomial $$t^2 - x$$ doesn't have a root.

This implies that $$\mathbb{C}$$ embeds nontrivially into itself, although this embedding can't really be written down explicitly and is very poorly behaved, e.g. it is nowhere measurable.

In the positive direction a field which is finite-dimensional over its prime subfield clearly can't embed nontrivially into itself, so you are safe if $$L$$ and $$K$$ are either finite fields or number fields.

• There are more explicit counterexamples that don't depend on the axiom of choice but it's a little trickier to show that the corresponding fields aren't isomorphic, or at least I don't know any really easy examples. Commented Oct 4, 2023 at 10:09
• @Martin: don't you still need (some) choice to write down the monomorphism $K(x_0) \to K$? Commented Oct 4, 2023 at 17:06
Here is another counterexample for fields: If $$K=\overline{\mathbb{Q}(x_1,x_2,...)}$$, then there are monomorphisms $$K(x_0) \to K \to K(x_0)$$, but no isomorphism since $$K(x_0)$$ is not algebraically closed.