Definition 9.10.1. A field $F$ is said to be *algebraically closed* if every algebraic extension $E/F$ is trivial, i.e., $E = F$.

## 9.10 Algebraic closure

The “fundamental theorem of algebra” states that $\mathbf{C}$ is algebraically closed. A beautiful proof of this result uses Liouville's theorem in complex analysis, we shall give another proof (see Lemma 9.23.1).

This may not be the definition in every text. Here is the lemma comparing it with the other one.

Lemma 9.10.2. Let $F$ be a field. The following are equivalent

$F$ is algebraically closed,

every irreducible polynomial over $F$ is linear,

every nonconstant polynomial over $F$ has a root,

every nonconstant polynomial over $F$ is a product of linear factors.

**Proof.**
If $F$ is algebraically closed, then every irreducible polynomial is linear. Namely, if there exists an irreducible polynomial of degree $> 1$, then this generates a nontrivial finite (hence algebraic) field extension, see Example 9.7.6. Thus (1) implies (2). If every irreducible polynomial is linear, then every irreducible polynomial has a root, whence every nonconstant polynomial has a root. Thus (2) implies (3).

Assume every nonconstant polynomial has a root. Let $P \in F[x]$ be nonconstant. If $P(\alpha ) = 0$ with $\alpha \in F$, then we see that $P = (x - \alpha )Q$ for some $Q \in F[x]$ (by division with remainder). Thus we can argue by induction on the degree that any nonconstant polynomial can be written as a product $c \prod (x - \alpha _ i)$.

Finally, suppose that every nonconstant polynomial over $F$ is a product of linear factors. Let $E/F$ be an algebraic extension. Then all the simple subextensions $F(\alpha )/F$ of $E$ are necessarily trivial (because the only irreducible polynomials are linear by assumption). Thus $E = F$. We see that (4) implies (1) and we are done. $\square$

Now we want to define a “universal” algebraic extension of a field. Actually, we should be careful: the algebraic closure is *not* a universal object. That is, the algebraic closure is not unique up to *unique* isomorphism: it is only unique up to isomorphism. But still, it will be very handy, if not functorial.

Definition 9.10.3. Let $F$ be a field. We say $F$ is *algebraically closed* if every algebraic extension $E/F$ is trivial, i.e., $E = F$. An *algebraic closure* of $F$ is a field $\overline{F}$ containing $F$ such that:

$\overline{F}$ is algebraic over $F$.

$\overline{F}$ is algebraically closed.

If $F$ is algebraically closed, then $F$ is its own algebraic closure. We now prove the basic existence result.

Theorem 9.10.4. Every field has an algebraic closure.

The proof will mostly be a red herring to the rest of the chapter. However, we will want to know that it is *possible* to embed a field inside an algebraically closed field, and we will often assume it done.

**Proof.**
Let $F$ be a field. By Lemma 9.8.9 the cardinality of an algebraic extension of $F$ is bounded by $\max (\aleph _0, |F|)$. Choose a set $S$ containing $F$ with $|S| > \max (\aleph _0, |F|)$. Let's consider triples $(E, \sigma _ E, \mu _ E)$ where

$E$ is a set with $F \subset E \subset S$, and

$\sigma _ E : E \times E \to E$ and $\mu _ E : E \times E \to E$ are maps of sets such that $(E, \sigma _ E, \mu _ E)$ defines the structure of a field extension of $F$ (in particular $\sigma _ E(a, b) = a +_ F b$ for $a, b \in F$ and similarly for $\mu _ E$), and

$F \subset E$ is an algebraic field extension.

The collection of all triples $(E, \sigma _ E, \mu _ E)$ forms a set $I$. For $i \in I$ we will denote $E_ i = (E_ i, \sigma _ i, \mu _ i)$ the corresponding field extension to $F$. We define a partial ordering on $I$ by declaring $i \leq i'$ if and only if $E_ i \subset E_{i'}$ (this makes sense as $E_ i$ and $E_{i'}$ are subsets of the same set $S$) and we have $\sigma _ i = \sigma _{i'}|_{E_ i \times E_ i}$ and $\mu _ i = \mu _{i'}|_{E_ i \times E_ i}$, in other words, $E_{i'}$ is a field extension of $E_ i$.

Let $T \subset I$ be a totally ordered subset. Then it is clear that $E_ T = \bigcup _{i \in T} E_ i$ with induced maps $\sigma _ T = \bigcup \sigma _ i$ and $\mu _ T = \bigcup \mu _ i$ is another element of $I$. In other words every totally order subset of $I$ has a upper bound in $I$. By Zorn's lemma there exists a maximal element $(E, \sigma _ E, \mu _ E)$ in $I$. We claim that $E$ is an algebraic closure. Since by definition of $I$ the extension $E/F$ is algebraic, it suffices to show that $E$ is algebraically closed.

To see this we argue by contradiction. Namely, suppose that $E$ is not algebraically closed. Then there exists an irreducible polynomial $P$ over $E$ of degree $> 1$, see Lemma 9.10.2. By Lemma 9.8.5 we obtain a nontrivial finite extension $E' = E[x]/(P)$. Observe that $E'/F$ is algebraic by Lemma 9.8.8. Thus the cardinality of $E'$ is $\leq \max (\aleph _0, |F|)$. By elementary set theory we can extend the given injection $E \subset S$ to an injection $E' \to S$. In other words, we may think of $E'$ as an element of our set $I$ contradicting the maximality of $E$. This contradiction completes the proof. $\square$

Lemma 9.10.5. Let $F$ be a field. Let $\overline{F}$ be an algebraic closure of $F$. Let $M/F$ be an algebraic extension. Then there is a morphism of $F$-extensions $M \to \overline{F}$.

**Proof.**
Consider the set $I$ of pairs $(E, \varphi )$ where $F \subset E \subset M$ is a subextension and $\varphi : E \to \overline{F}$ is a morphism of $F$-extensions. We partially order the set $I$ by declaring $(E, \varphi ) \leq (E', \varphi ')$ if and only if $E \subset E'$ and $\varphi '|_ E = \varphi $. If $T = \{ (E_ t, \varphi _ t)\} \subset I$ is a totally ordered subset, then $\bigcup \varphi _ t : \bigcup E_ t \to \overline{F}$ is an element of $I$. Thus every totally ordered subset of $I$ has an upper bound. By Zorn's lemma there exists a maximal element $(E, \varphi )$ in $I$. We claim that $E = M$, which will finish the proof. If not, then pick $\alpha \in M$, $\alpha \not\in E$. The $\alpha $ is algebraic over $E$, see Lemma 9.8.4. Let $P$ be the minimal polynomial of $\alpha $ over $E$. Let $P^\varphi $ be the image of $P$ by $\varphi $ in $\overline{F}[x]$. Since $\overline{F}$ is algebraically closed there is a root $\beta $ of $P^\varphi $ in $\overline{F}$. Then we can extend $\varphi $ to $\varphi ' : E(\alpha ) = E[x]/(P) \to \overline{F}$ by mapping $x$ to $\beta $. This contradicts the maximality of $(E, \varphi )$ as desired.
$\square$

Lemma 9.10.6. Any two algebraic closures of a field are isomorphic.

**Proof.**
Let $F$ be a field. If $M$ and $\overline{F}$ are algebraic closures of $F$, then there exists a morphism of $F$-extensions $\varphi : M \to \overline{F}$ by Lemma 9.10.5. Now the image $\varphi (M)$ is algebraically closed. On the other hand, the extension $\varphi (M) \subset \overline{F}$ is algebraic by Lemma 9.8.4. Thus $\varphi (M) = \overline{F}$.
$\square$

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