Algebraic Closure

Definition / Theorem (Algebraic closure inside a field).
Let $E/F$ be a field extension, and suppose $\alpha, \beta \in E$ are algebraic over $F$. Then so are:

  • $\alpha \pm \beta$
  • $\alpha \beta$
  • $\alpha / \beta$ (provided $\beta \ne 0$)

As a corollary, the set:

\[\overline{F}_E := \{ \alpha \in E : \alpha \text{ is algebraic over } F \}\]

is a subfield of $E$ containing $F$. It is called the algebraic closure of $F$ in $E$, and is the largest algebraic subextension of $F$ inside $E$.

Proof.
It suffices to observe that if $\beta$ is algebraic over $F$, then it is also algebraic over $F[\alpha]$.
Since $F[\alpha, \beta]$ is a finite extension of $F[\alpha]$, and $F[\alpha]/F$ is also finite, the Tower Law gives that

\[[F[\alpha, \beta] : F] < \infty\]

so $F[\alpha, \beta]/F$ is a finite (and hence algebraic) extension. Therefore, any element formed from $\alpha$ and $\beta$ using field operations (except division by 0) is algebraic over $F$.

Definition / Theorem (Algebraically closed field).
A field $F$ is called algebraically closed if any of the following equivalent conditions hold:

  1. Every non-constant polynomial $p(x) \in F[x]$ has a root in $F$.

  2. Every non-constant polynomial $p(x) \in F[x]$ splits completely as

    \[p(x) = c \prod_{i=1}^n (x - \alpha_i),\quad \alpha_i \in F\]
  3. If $E/F$ is an algebraic extension, then $E = F$.

Proof (Sketch).

  • $1 \Rightarrow 2$: Use induction on the degree of $p(x)$. If $p(x)$ has a root $\alpha \in F$, factor it as $(x - \alpha)q(x)$ and apply the inductive hypothesis to $q(x)$.

  • $2 \Rightarrow 3$: If $E/F$ is algebraic and $\alpha \in E$, then $\alpha$ is a root of its minimal polynomial $m_{\alpha, F}(x) \in F[x]$, which must split in $F$ by assumption. Hence, $\alpha \in F$, so $E = F$.

  • $3 \Rightarrow 1$: Let $p(x) \in F[x]$ be a non-constant polynomial. If $\alpha$ is a root of $p(x)$ in some extension, then $F(\alpha)/F$ is algebraic. By assumption, $F(\alpha) = F$, so $\alpha \in F$.

Definition.
Let $F$ be a field. An algebraic closure of $F$ is a field extension $\overline{F}$ such that:

  1. $\overline{F}/F$ is an algebraic extension.
  2. $\overline{F}$ is algebraically closed.

Theorem (Existence of Algebraic Closure).
Let $F$ be a field. Then there exists an algebraically closed field $\overline{F}$ such that $\overline{F}/F$ is algebraic.

Proof (Sketch).
We construct the algebraic closure using Zorn’s Lemma.

Let $E/F$ be any algebraic extension (possibly $F$ itself). Define:

\[\mathcal{G} := \{ L \mid E \subseteq L,\ L/F \text{ is algebraic} \}\]

partially ordered by inclusion.

Let $\mathcal{C} \subset \mathcal{G}$ be a chain, and define:

\[\widetilde{L} := \bigcup_{L \in \mathcal{C}} L\]

Since all the fields in the chain are algebraic over $F$ and nested by inclusion, their union $\widetilde{L}$ is also a field and an algebraic extension of $F$ (proof omitted). Thus, $\widetilde{L} \in \mathcal{G}$, so every chain has an upper bound.

By Zorn’s Lemma, $\mathcal{G}$ has a maximal element $\overline{F}$.

We claim that $\overline{F}$ is algebraically closed. Suppose not: then there exists a non-constant polynomial $p(x) \in \overline{F}[x]$ with no root in $\overline{F}$.

Let $\alpha$ be a root to $p(x)$. Then $\alpha$ is algebraic over $\overline{F}$, and hence algebraic over $F$. The field $F(\alpha)$ is an algebraic extension of $F$ containing $\overline{F}$ properly, contradicting maximality. Hence, $\overline{F}$ is algebraically closed.

Remark: Why Algebraic Closures Matter

The key benefit of knowing that an algebraic closure exists is this:

There is always a field where every polynomial in $F[x]$ splits completely.

In other words, every root of any polynomial over $F$ can be found somewhere inside a single field — the algebraic closure $\overline{F}$. This allows us to treat roots as concrete objects and study how they interact — rather than worry about whether they exist.

This simplifies many arguments in field theory. For example:

  • We can always speak of ‘the roots’ of a polynomial, even if they don’t live in $F$.

  • We can define a splitting field as a subfield of $\overline{F}$ generated by the roots.

  • We can study automorphisms of $\overline{F}$ that permute these roots — which is the starting point of Galois theory.

Even though the algebraic closure is huge and hard to describe explicitly, its mere existence is what makes the whole theory work.

Theorem (Uniqueness of Algebraic Closure)
Let $F$ be a field. Then all its algebraic closure are closed upto isomorphism.

We will prove this theorem later when discussing the extension theorems.