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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).

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

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

  1. $F$ is algebraically closed,

  2. every irreducible polynomial over $F$ is linear,

  3. every nonconstant polynomial over $F$ has a root,

  4. 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. An algebraic closure of $F$ is a field $\overline{F}$ containing $F$ such that:

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

  2. $\overline{F}$ is algebraically closed.

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

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

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

  2. $\sigma _ E : E \times E \to E$ 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

  3. $E/F$ 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 $\varphi (M) = \overline{F}$. $\square$

Comments (2)

Comment #7845 by Stéphane on

Being algebraically closed is defined (identically) in both 9.10.1 and 9.10.3.

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