Embedding an algebraic extension into an algebraically closed field

Assume Zorn's lemma. If $𝐿 : 𝐾$ is a field extension with $𝐿$ algebraically closed and $𝐹 : 𝐾$ is any algebraic extension, then there exists a morphism of field extensions

𝑖 𝐹 \in 𝖥 𝗅 𝖽 𝐾 (𝐹, 𝐿)

Proof

Consider the restricted Poset of intermediate extensions

𝑍 = {(𝑀, 𝑖 𝑀) ∣ 𝐾 : 𝑀 : 𝐹, 𝑖 𝑀 \in 𝖥 𝗅 𝖽 𝐾 (𝑀, 𝐿)}

where we note all fields in $𝑍$ are algebraic extensions of $𝐾$ . We show that every chain $𝐶$ in $𝑍$ has an upper bound. Let $𝑀 𝐶 = ⋃ (𝑀, 𝑖 𝑀) \in 𝑍 𝑀$ which is a field. If $𝛼 \in 𝑀 𝐶$ , then we define $𝑖 𝑀 𝐶 (𝛼) = 𝑖 𝑀 (𝛼) \in 𝐿$ for some $(𝑀, 𝑖 𝑀) \in 𝐶$ , which is clearly independent of the choice of $𝑀$ . Then $(𝑀 𝐶, 𝑖 𝑀 𝐶) \in 𝑍$ is an upper bound of $𝐶$ .

By Zorn's lemma, $𝑍$ has a maximal element $(𝐺, 𝑖 𝐺)$ . Since $𝐺$ is algebraic,

𝐻 : = 𝑖 𝐺 (𝐺) \leq ―― 𝐾 = (𝐿 : 𝐾) \circ \leq 𝐿

We claim that $𝐻 = 𝐹$ , whence $𝑖 𝐹 = 𝑖 𝐻 \in 𝖥 𝗅 𝖽 𝐾 (𝐹, 𝐿)$ is the desired morphism.

𝐿 | (𝐿 : 𝐾) \circ \geq 𝐻 = 𝑖 𝐺 (𝐺) ≅ 𝐺 | 𝐹 ∥ 𝐺 | 𝑀 | 𝐾

Suppose towards contradiction there exists $𝛼 \in 𝐹 ∖ 𝐺$ and consider the simple extension $𝐺 (𝛼) : 𝐺$ . Since $𝛼 \in 𝐹$ is algebraic over $𝐾$ , it is algebraic over $𝐺$ , thus it is the root of an irreducible $𝑝 (𝑥) \in 𝐺 [𝑥]$ . Abusing notation to invoke the induced homomorphism

𝑖 𝐺 : 𝐺 [𝑥] \to 𝐻 [𝑥]

let $ℎ (𝑥) = 𝑖 𝐺 (𝑔 (𝑥))$ , which is irreducible over $𝐻$ , and has a root $𝛽$ in $𝐿$ — here we use that $𝐿$ is algebraically closed. Now by ^P2, the isomorphism $𝑖 𝐺 : 𝐺 \to 𝐻$ lifts to an isomorphism

𝑖 𝐺 (𝛼) : 𝐺 (𝛼) \to 𝐻 (𝛽) \leq 𝐿

sending $𝛼$ to $𝛽$ , contradicting the maximality of $(𝐺, 𝑖 𝐺)$ . Therefore $𝐺 = 𝐹$ , and we're done.

This lemma is used to prove uniqueness of the Algebraic closure.

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