Embedding problem for two algebraically closed fields

Give two algebraically closed fields with same characteristic. Can we always embed one of them to the other?

I will appreciate any references.

• @DietrichBurde Yes, but the algebraic numbers can be embedded into $\mathbb C$... – Ege Erdil Apr 19 '17 at 13:50
• @DietrichBurde Yes, you are given two algebraically closed fields, and you are asked to embed one of them to another. You're not asked to provide embeddings in both directions, just one direction. – Ege Erdil Apr 19 '17 at 13:51
• @DietrichBurde You're right that the question as it is stated is a bit confusing, but you may not even have an embedding of sets if you interpret it the way you did, let alone an embedding of fields, so I think my interpretation is what the OP meant. – Ege Erdil Apr 19 '17 at 13:54
• @DietrichBurde FWIW I agree with the others. This may be somewhat idiomatic mathematical English, but I don't recall ever having seen a context where "one" used this way would not come with a tacitly implied freedom of choice. A shorthand for "one of them" – Jyrki Lahtonen Apr 19 '17 at 14:00
• Use of "one" here is not unlike German "einer", for example (one of first Google hits) *Damit ist bewiesen, dass A zu einer der Zahlen a, a + 1, a + 2, ..., a + m − 1 kongruent ist". Choice is there (probably the correct parallel is between "one of" and "einer der". Anyway, I agree that non-trivial set theory must be used to answer the question. You probably need the axiom of choice to get the existence of transcendence bases needed in Starfall's argument. But, I believe in Zorn's lemma :-) – Jyrki Lahtonen Apr 19 '17 at 14:09

Yes. If $L$ and $M$ are algebraically closed fields of the same characteristic, then let $K$ be their common prime field; and pick transcendence bases $S_L, S_M$ respectively. Without loss of generality, assume $|S_L| \geq |S_M|$. Then, there is an injective map of sets $S_M \to S_L$, which extends to an embedding $K(S_M) \to K(S_L) \subset L$ of fields. $M/K(S_M)$ is an algebraic extension and $L$ is algebraically closed, thus the embedding $K(S_M) \to L$ extends to an embedding $M \to L$.