Embedding an algebraic extension into an algebraically closed field
Assume Zorn’s lemma.
If
Proof
Consider the restricted Poset of intermediate extensions
𝑍 = { ( 𝑀 , 𝑖 𝑀 ) ∣ 𝐾 : 𝑀 : 𝐹 , 𝑖 𝑀 ∈ 𝖥 𝗅 𝖽 𝐾 ( 𝑀 , 𝐿 ) } where we note all fields in
are algebraic extensions of 𝑍 . We show that every ^chain 𝐾 in 𝐶 has an upper bound. Let 𝑍 which is a field. If 𝑀 𝐶 = ⋃ ( 𝑀 , 𝑖 𝑀 ) ∈ 𝑍 𝑀 , then we define 𝛼 ∈ 𝑀 𝐶 for some 𝑖 𝑀 𝐶 ( 𝛼 ) = 𝑖 𝑀 ( 𝛼 ) ∈ 𝐿 , which is clearly independent of the choice of ( 𝑀 , 𝑖 𝑀 ) ∈ 𝐶 . Then 𝑀 is an upper bound of ( 𝑀 𝐶 , 𝑖 𝑀 𝐶 ) ∈ 𝑍 . 𝐶 By Zorn’s lemma,
has a maximal element 𝑍 . Since ( 𝐺 , 𝑖 𝐺 ) is algebraic, 𝐺 𝐻 : = 𝑖 𝐺 ( 𝐺 ) ≤ ―― 𝐾 = ( 𝐿 : 𝐾 ) ∘ ≤ 𝐿 We claim that
, whence 𝐻 = 𝐹 is the desired morphism. 𝑖 𝐹 = 𝑖 𝐻 ∈ 𝖥 𝗅 𝖽 𝐾 ( 𝐹 , 𝐿 ) 𝐿 | ( 𝐿 : 𝐾 ) ∘ ≥ 𝐻 = 𝑖 𝐺 ( 𝐺 ) ≅ 𝐺 | 𝐹 ∥ 𝐺 | 𝑀 | 𝐾 Suppose towards contradiction there exists
and consider the simple extension 𝛼 ∈ 𝐹 ∖ 𝐺 . Since 𝐺 ( 𝛼 ) : 𝐺 is algebraic over 𝛼 ∈ 𝐹 , it is algebraic over 𝐾 , thus it is the root of an irreducible 𝐺 . Abusing notation to invoke the induced homomorphism 𝑝 ( 𝑥 ) ∈ 𝐺 [ 𝑥 ] 𝑖 𝐺 : 𝐺 [ 𝑥 ] → 𝐻 [ 𝑥 ] let
, which is irreducible over ℎ ( 𝑥 ) = 𝑖 𝐺 ( 𝑔 ( 𝑥 ) ) , and has a root 𝐻 in 𝛽 — here we use that 𝐿 is algebraically closed. Now by ^P2, the isomorphism 𝐿 lifts to an isomorphism 𝑖 𝐺 : 𝐺 → 𝐻 𝑖 𝐺 ( 𝛼 ) : 𝐺 ( 𝛼 ) → 𝐻 ( 𝛽 ) ≤ 𝐿 sending
to 𝛼 , contradicting the maximality of 𝛽 . Therefore ( 𝐺 , 𝑖 𝐺 ) , and we’re done. 𝐺 = 𝐹
This lemma is used to prove uniqueness of the Algebraic closure.