Algebraic closure

Let $𝐾$ be a field. An algebraic closure $―― 𝐾$ of $𝐾$ is an algebraically closed field such that $―― 𝐾 : 𝐾$ is an algebraic extension. #m/def/field Assuming AC,¹ an algebraic closure always exists and is unique up to isomorphism of field extensions, so one often speaks of the algebraic closure.

Proof of existence and uniqueness

The proof of existence and uniqueness requires enough lemmata to warrant a section of this Zettel.² We invoke Zorn's lemma.

Existence

Let $𝐾 = 𝐾 0$ be a field. There exists an extension $𝐾 1 : 𝐾 0$ such that every nonconstant polynomial $𝑓 (𝑥) \in 𝐾 [𝑥]$ has at least one root in $𝐾 1$ .

Proof due to Emil Artin

Let $F$ denote the set of nonconstant monic polynomials in $𝐾 0 [𝑥]$ , and let $𝐾 0 [𝑡 𝑓] 𝑓 \in F$ be the corresponding polynomial ring, potentially in infinitely many indeterminates. Consider the ideal

𝐼 = ⟨ 𝑓 (𝑡 𝑓) : 𝑓 \in F ⟩ ◃ 𝐾 0 [𝑥],

which we will show must be proper. Suppose towards contradiction that $𝐼 = ⟨ 1 ⟩$ , so

1 = 𝑛 \sum 𝑖 = 1 𝑎 𝑖 𝑓 𝑖 (𝑡 𝑓 𝑖)

for some $𝑎 𝑖 \in 𝐾 0 [𝑡 𝑓] 𝑓 \in F$ and $𝑓 𝑖 \in F$ . We can then construct an extension $𝐹 : 𝐾$ where the polynomials $(𝑓 𝑖) 𝑛 𝑖 = 1$ have roots $(𝛼 𝑖) 𝑛 𝑖 = 1$ , by iteratively Adjoining a root to a field. If we evaluate

1 = 𝑛 \sum 𝑖 = 1 𝑎 𝑖 𝑓 𝑖 (𝛼 𝑖) = 𝑛 \sum 𝑖 = 1 𝑎 𝑖 \cdot 0 = 0

which is a contradiction. Since $𝐼$ is proper, invoking Zorn it is contained in a maximal ideal $𝔪$ , giving the the field extension

𝐾 0 [𝑡 𝑓] 𝑓 \in F 𝔪 : = 𝐾 1 : 𝐾 0,

where by construction every nonconstant monic (and thus nonconstant general) polynomial $𝑓 (𝑥)$ has a root $𝜋 (𝑡 𝑓)$ .

To guarantee the existence of all roots we iterate this process ad infinitum, so not only does $𝑓 (𝑥) \in 𝐾 0 [𝑥]$ have a root $𝛼 1 \in 𝐾 1$ , but $𝑓 (𝑥) / (𝑥 - 𝛼 1) \in 𝐾 1 [𝑥]$ has a root $𝛼 2 \in 𝐾 2$ , &c. This yields a chain of extensions

𝐾 0 ↪ 𝐾 1 ↪ 𝐾 2 ↪ \dots

Let $𝐿$ be the union or limit of this chain. $𝐿$ is an algebraically closed field, and thus $(𝐿 : 𝐾) \circ$ is an algebraic closure of $𝐾$ .

Proof

For every $𝑎, 𝑏 \in 𝐿$ we have $𝑎, 𝑏 \in 𝐾 𝑖$ for some $𝑖 \in ℕ 0$ , so we can just work within whatever $𝐾 𝑖$ is necessary, since the result is independent. Thus $𝐿$ is a field. If $𝑓 (𝑥) \in 𝐿 [𝑥]$ , then $𝑓 (𝑥) \in 𝐾 𝑖 [𝑥]$ for some $𝑖 \in ℕ 0$ , and thus it has a root in $𝐾 𝑖 + 1 \leq 𝐿$ .

Uniqueness

See Embedding an algebraic extension into an algebraically closed field.

Suppose $―― 𝐾 1$ and $―― 𝐾 2$ are algebraic closures of $𝐾$ . Then there exists an isomorphism of field extensions $𝜓 \in 𝖥 𝗅 𝖽 𝐾 (―― 𝐾 1, ―― 𝐾 2)$ .