Fundamental question about field extensionsLet k be an algebraically closed field. Let K⊂k be a subfield, and suppose that it has an algebraic extension K⊂L. Is it true (or at least well defined, because I'm not even sure about this) that L⊆k? One slightly different point of view is: given injections i:K→k,j:K→L, the last one being algebraic, is there an injection l:L→k such that l∘j=i (i.e. l is a homomorphism of K-algebras)?

Question

Fundamental question about field extensionsLet k be an algebraically closed field. Let K⊂k be a subfield, and suppose that it has an algebraic extension K⊂L. Is it true (or at least well defined, because I&#039;m not even sure about this) that L⊆k? One slightly different point of view is: given injections i:K→k,j:K→L, the last one being algebraic, is there an injection l:L→k such that l∘j=i (i.e. l is a homomorphism of K-algebras)?

gogrervisulget9z · Accepted Answer

You cannot prove that L⊆k, but you&#039;re right that there exists an embedding L→k that&#039;s the identity o K.Consider the set ξ of pairs (F,jF), where K⊆F⊆L (as subfields) and jF:F→k  is the identity on K.We can partially order ξ by declaring (F1,jF1)≤(F2,jF2) if (and only if) F1⊆F2 and the restriction of jF2 to F1 is jF1.Clearly ξ is not empty, because (K,i)∈ξ. Moreover any chain in ξ has an upper bound: take the union of the first components of the pair and define the homomorphism accordingly.By Zorn&#039;s lemma, there is a maximal element (F,jF) and we want to prove that F=L. Here the fact that L is algebraic over K is needed (up to now we didn&#039;t use it).If a∈LF, the element a is algebraic over K and all it takes is to prove that we can extend jF to a homomorphism j:F(a)→k. This is standard. But now (F(a), j) contradicts the maximality of (F,jF) and we&#039;re done.

Fundamental question about field extensions Let k be an algebraically closed field. Let K \su

Answered question

Answer & Explanation

New Questions in Abstract algebra