The Algebraic Closure

Previously, we explored extending fields using irreducible polynomials. One classic result arises from the complex numbers, where

$\mathbb C \cong \mathbb R[x]/\langle x^2 + 1\rangle.$

The fundamental theorem of algebra tells us that any polynomial $f(x) \in \mathbb C[x]$ has a root $\alpha \in \mathbb C$ . In this case, we say that $\mathbb C$ is algebraically closed. The word “closed” broadly translates to “properties are more or less preserved”.

This closure property turns out to work for any field. Let $\mathbb K \supseteq \mathbb F$ .

Definition 1. An element $\alpha \in \mathbb K \supseteq \mathbb F$ is algebraic over $\mathbb F$ if there exists a polynomial $f \in \mathbb F[x]$ such that $f(\alpha) = 0$ . We say that $\mathbb K$ is algebraic over $\mathbb F$ if every element in $\mathbb K$ is algebraic over $\mathbb F$ .

We remark that being “algebraic” is a property of a field and its elements, not the polynomial. That is, given the element $\alpha$ , there exists a sufficiently nice polynomial $f$ such that $f(\alpha) = 0$ .

Lemma 1. The set $\mathbb K_{\mathbb F} := \{ u \in \mathbb K : u\ \text{is algebraic over}\ \mathbb F\}$ forms a sub-field of $\mathbb K$ .

Proof. Fix $u,v \in \mathbb K_{\mathbb F}$ . Recall that $\mathbb F(u, v) \supseteq \mathbb F$ is the smallest field that contains $\mathbb F$ as a subfield. Since the extension is finite, it is algebraic, so that $\mathbb F(u, v) \subseteq \mathbb K_{\mathbb F}$ . In particular, $u-v \in \mathbb K_{\mathbb F}$ and $u \cdot v^{-1} \in \mathbb K_{\mathbb F}$ , so that $\mathbb K_{\mathbb F}$ is closed under the field operations inherited from $\mathbb K$ . Therefore, $\mathbb K_{\mathbb F}$ forms a field.

In particular, $\mathbb K$ is algebraic over $\mathbb F$ if and only if $\mathbb K = \mathbb K_{\mathbb F}$ .

To attain algebraic closure, we ask the reverse question: given the polynomial $f$ , is there a sufficiently nice element $\alpha$ such that $f(\alpha) = 0$ ? Notice that in general, $\mathbb F[x] \supsetneq \mathbb F$ .

Definition 2. We call a field $\mathbb F$ algebraically closed if every polynomial $f \in \mathbb F[x]$ has at least one root $\alpha \in \mathbb F$ such that $f(\alpha) = 0$ .

Lemma 2. If $\mathbb F$ is algebraically closed, then any non-constant polynomial $f \in \mathbb F$ can be decomposed into a product of linear factors (i.e. polynomials with degree $1$ ).

Proof. We prove this result holds by induction: For any $n$ , if $f \in \mathbb F[x]$ has degree $n$ , then it can be written as a product of linear factors. The result is obvious for $n = 1$ . Suppose the result holds for $\deg(f) = k$ . Fix $f \in \mathbb F[x]$ with $\deg(f) = k +1$ . Since $\mathbb F$ is algebraically closed, there exists $\alpha \in \mathbb F$ such that $f(\alpha) = 0$ . By the factor theorem, there exists a polynomial $\tilde{f}$ with $\deg(\tilde{f}) = k$ such that $f(x) = (x-\alpha) \cdot \tilde{f}(x)$ . By the induction hypothesis, $\tilde f$ can be written as a product of linear factors, so that this holds for $f$ as well, as required.

We claim that for any field $\mathbb F$ , there exists an essentially unique field extension $\mathbb F' \supseteq \mathbb F$ that is algebraically closed.

Lemma 3. There exists a field extension $\mathbb K \supseteq \mathbb F$ such that every non-constant polynomial $f \in \mathbb F[x]$ has at least one root $\alpha \in \mathbb K$ . We remark the following caveats:

$\mathbb K$ may not be algebraically closed. That is, there may exist a non-constant polynomial $g \in \mathbb K[x]$ with no roots in $\mathbb K$ .
$\mathbb K$ may not be algebraic over $\mathbb F$ . That is, there may exist an element $u \in \mathbb K$ that is not algebraic in $\mathbb F$ .

Proof. For any non-constant $f \in \mathbb F[x]$ , let $x_f$ denote a variable. Then given non-constant polynomials $f,g$ , the finitely many nonzero constants $c_{i,j} \in \mathbb F$ , generate a polynomial $\psi_{f,g}$ defined by

$\displaystyle \psi_{f,g}(x_f, x_g) = \sum_{i=0}^\infty \sum_{j=0}^\infty c_{i,j} \cdot x_f^i \cdot x_g^j \in \mathbb F[x_f, x_g],$

where $c_{i,j} \in \mathbb F$ . This observation can be generalised. For non-constant polynomials $\mathbf f = (f_1,\dots, f_n)$ whose corresponding variables are denoted $\mathbf x = (x_{f_1},\dots, x_{f_n})$ , the finitely many nonzero constants $c_{\mathbf v} \in \mathbb F$ , where $\mathbf v = (v_1,\dots, v_n) \in (\mathbb N^+)^n$ , generate a polynomial $\psi_{\mathbf f}$ defined by

$\displaystyle \psi_{\mathbf f}(\mathbf x) = \sum_{\mathbf v \in (\mathbb N^+)^n} c_{\mathbf v} \cdot x_{f_1}^{v_1} \cdot \cdots \cdot x_{f_n}^{v_n} \in \mathbb F[\mathbf x].$

Define $\mathbf x := \{x_f : f \in \mathbb F[x]^*\}$ and the ring

$\mathbb F[\mathbf x] := \{ \psi_{\mathbf f} : \mathbf f \subseteq \mathbb F[x]^*\}.$

By considering the polynomial $f(x) = x$ , $\mathbb F[x] \subseteq \mathbb F[\mathbf x]$ . Now given a non-constant polynomial $f = \sum_{i=0}^\infty c_i \mu_i \in \mathbb F[x]$ and its corresponding variable $x_f$ ,

$\displaystyle f(x_f) = \sum_{i=0}^\infty c_i \cdot \mu_i(x_f) = \sum_{i=0}^\infty c_i \cdot x_f^i \in \mathbb F[\mathbf x],$

where the right-hand side means that only finitely many $c_i$ is non-zero. Define the ideal

$I := \langle \{ f(x_f) : f \in \mathbb F[x]^* \} \rangle \subseteq \mathbb F[\mathbf x].$

We claim that $I \neq \mathbb F[\mathbf x]$ . Suppose instead that $I = \mathbb F[\mathbf x]$ . Since $1 \in I$ , there exist finitely many $w_i \in \mathbb F[\mathbf x]$ such that

$\displaystyle 1 = \sum_{i=1}^\infty w_i \cdot f_i(x_{f_i}).$

Define $A_i := \{ \alpha_{i,1},\dots,\alpha_{i,\deg(f_i)} \}$ , where $f_i(\alpha_{i,j}) = 0$ . Define

$\displaystyle A := \bigcup_{i=1}^\infty \{\alpha \in \mathbb F : f_i(\alpha) = 0\},$

and $\mathbb K := \mathbb F(A)$ , and the ring homomorphism $\phi : \mathbb F[\mathbf x] \to \mathbb K$ by $\phi|_{\mathbb F} = \mathrm{id}|_{\mathbb F}$ and

$\phi(x_{f_i}) \mapsto \alpha_{i,1}.$

Then

$\displaystyle 1 = \phi(1) = \sum_{i=1}^r \phi(w_i) \cdot f(\phi(x_{f_i})) = \sum_{i=1}^r \phi(w_i) \cdot f(\alpha_{i,1}) = 0,$

a contradiction. Now define

$\mathcal S := \{J \subseteq \mathbb F[\mathbf x] : J \supseteq I\ \text{as proper ideals}\} \ni I.$

Fix any ascending chain $\mathcal C = \{J_t\}$ in $\mathcal S$ . Since the set $K := \bigcup_{t} J_t$ is also an ideal that contains $I$ , it is an upper bound in $\mathcal C$ . By Zorn’s lemma, $\mathcal S$ has a maximal element $M$ .

To show that $M$ is a maximal ideal, we observe that for any ideal $M'$ such that $M \subseteq M' \subseteq \mathbb F[\mathbf x]$ , if $M' \neq \mathbb F[\mathbf x]$ , then $M'$ being a proper ideal of $\mathbb F[x]$ implies that $M' \in \mathcal S$ , so that $M \subseteq M'$ implies $M = M'$ , as required.

Finally, since $M \subseteq \mathbb F[\mathbf x]$ as a maximal ideal, $\mathbb K :=\mathbb F[\mathbf x]/M$ forms a field that contains $\mathbb F$ via the natural inclusion map $f \mapsto f + M$ , so that

$f(x_f + M) = f(x_f) + M = I +M = M = 0.$

Hence any polynomial $f \in \mathbb F[x]$ has a root $x_f + M \in \mathbb K$ .

Theorem 1. For any field $\mathbb F$ , there exists an algebraic extension $\mathbb F' \supseteq \mathbb F$ that is algebraically closed.

Proof. Use Lemma 3 to construct field extensions

$\mathbb F \subseteq \mathbb K_1 \subseteq \mathbb K_2 \subseteq \cdots$

and define the field extension $\mathbb K := \bigcup_{i=1}^\infty \mathbb K_i \supseteq \mathbb F$ . By Lemma 1, the set $\mathbb F' := \mathbb K_{\mathbb F}$ forms a sub-field of $\mathbb K$ . In fact, by definition, $\mathbb F' \supseteq \mathbb F$ is algebraic.

It suffices to show that $\mathbb F'$ is algebraically closed. To do that, fix $f = \sum_{i=0}^n c_i \cdot \mu_i \in \mathbb F'[x]^*$ . By definition, $c_i \in \mathbb K_{m_i}$ for some $n_i$ . Define $m := m_1 + \dots + m_n$ . Then $c_i \in \mathbb K_m$ for each $i$ . In particular, $f$ has a root in $\mathbb K_{m+1} \subseteq \mathbb F'$ .

Furthermore, this field extends uniquely with respect to the base field in the following sense.

Theorem 2. Let $\mathbb F_1, \mathbb F_2$ be field extensions of $\mathbb F$ that are algebraically closed. Then there exists a field isomorphism $\phi : \mathbb F_1 \to \mathbb F_2$ such that $\phi|_{\mathbb F} = \mathrm{id}_{\mathbb F}$ .

Proof. Define the collection of $\mathbb F$ –homomorphisms by

$\mathrm{Hom}_{\mathbb F}(\mathbb F_1, \mathbb F_2) := \{\phi \in \mathrm{Hom}(\mathbb F_1, \mathbb F_2) : \phi|_{\mathbb F} = \mathrm{id}_{\mathbb F}\}.$

Consider the field homomorphism $\mathrm{id}|_{\mathbb F} : \mathbb F \to \mathbb F_2$ . Define the collection $\mathcal P$ of pairs

$\mathcal P := \{ (\mathbb K, \phi_{\mathbb K}) : \mathbb F_1 \supseteq \mathbb K \supseteq \mathbb F, \phi \in \mathrm{Hom}_{\mathbb F}(\mathbb F_1, \mathbb F_2) \}$

with the partial ordering $(\mathbb K_1, \phi_{\mathbb K_1}) \leq (\mathbb K_2, \phi_{\mathbb K_2})$ if and only if $\mathbb K_1 \subseteq \mathbb K_2$ and $\phi_{\mathbb K_2} |_{\mathbb K_1} = \phi_{\mathbb K_1}$ . Fix any chain

$\mathcal C = \{ (\mathbb K_{\alpha}, \phi_{\mathbb K_\alpha}) : \alpha \in I \} \subseteq \mathcal P.$

Define $\mathbb K := \bigcup_{\alpha} \mathbb K_\alpha$ and $\phi_{\mathbb K}$ by $\phi_{\mathbb K}(x) = \phi|_{\mathbb K_\alpha}(x) \cdot \mathbb I_{\mathbb K_\alpha}(x)$ . It is a simple exercise to check that $(\mathbb K, \phi_{\mathbb K}) \in \mathcal P$ is an upper-bound for $\mathcal C$ . By Zorn’s lemma, $\mathcal P$ contains a maximal element $(\mathbb K_{\max}, \phi_{\mathbb K_{\max}})$ .

It is clear that $\mathbb K_{\max} \subseteq \mathbb F_1$ . If $\mathbb K_{\max} \neq \mathbb F_1$ , then there exists some $u \in \mathbb F_1 \backslash \mathbb K_{\max}$ so that $(\mathbb K_{\max}, \phi_{\mathbb K_{\max}})< \mathbb (\mathbb F_1(u), \phi_{\mathbb F_1(u)})$ , contradicting the maximality of $\mathbb K_{\max}$ . Therefore, $\mathbb K_{\max} = \mathbb F_1$ , and the field homomorphism $\phi := \phi_1 : \mathbb F_1 \to \mathbb F_2$ satisfies $\phi|_{\mathbb F} = \mathrm{id}_{\mathbb F}$ . As a field homomorphism, $\phi$ must be injective. Therefore, it suffices to check that $\phi$ is surjective, that is, $\phi(\mathbb F_1) = \mathbb F_2$ .

Now $\phi(\mathbb F_1) \supseteq \mathbb F$ is algebraic, since for any $u \in \mathbb F_1$ ,

$f_{\mathbb F_1}^u(\phi(u)) = \phi(f_{\mathbb F_1}^u(u)) = \phi(0) = 0,$

so that $\phi(u)$ is algebraic over $\mathbb F$ . Since $\phi(\mathbb F_1) \subseteq \mathbb F_2$ , it suffices to check that the reverse inclusion holds.

We claim that $\phi(\mathbb F_1)$ is not just algebraic, but furthermore, algebraically closed. Fix $f = \sum_i \alpha_i x^i \in \phi(\mathbb F_1)[x]$ . Since $\alpha_i \in \phi(\mathbb F_1)$ , there exists $\beta_i \in \mathbb F_1$ such that $\alpha_i = \phi(\beta_i)$ . Since $\mathbb F_1$ is algebraically closed, the polynomial $g := \sum_i \beta_i x^i \in \mathbb F_1[x]$ has a root $\beta \in \mathbb F_1$ . Since $\phi$ is a field homomorphism,

$\begin{aligned} f(\phi(\beta)) &= \sum_i \alpha_i (\phi(\beta))^i \\ &= \sum_i \phi(\beta_i) (\phi(\beta))^i \\ &= \phi \left(\sum_i \beta_i \cdot \beta^i \right) \\ &= \phi(g(\phi(\beta))) \\ &= \phi(g(\beta)) \\ &= \phi(0) = 0. \end{aligned}$

Therefore, $\phi(\mathbb F_1)$ is algebraic closed.

For any $u \in \mathbb F_2$ , $\phi(\mathbb F_1)(u) \supseteq \phi(\mathbb F_1)$ is a finite extension with minimal polynomial $f_{\mathbb F_1}^u \in \phi(\mathbb F_1)[x]$ . Since $\phi(\mathbb F_1)$ is algebraically closed, by induction on Lemma 2, all of the roots of $f_{\mathbb F_1}^u$ belong to $\phi(\mathbb F_1)$ . In particular, $u \in \phi(\mathbb F_1)$ , so that $\phi(\mathbb F_1) = \mathbb F_2)$ .

Therefore, $\phi(\mathbb F_1) = \mathbb F_2$ , and $\phi$ is a field isomorphism, as required.

Corollary 1. Any field $\mathbb F$ has an effectively unique field extension $\overline{\mathbb F} \supseteq \mathbb F$ . We call the field extension $\overline{\mathbb F} \supseteq \mathbb F$ the algebraic closure of $\mathbb F$ , which is unique up to the existence of field isomorphisms according to Theorem 2.

Remark 1. Any algebraic closure of $\mathbb R$ must be isomorphic to $\mathbb C$ . Hence, we could have started with $\mathbb C = \overline{\mathbb R}$ and defined $i$ as a solution to $x^2 + 1 \in \mathbb R[x] \subseteq \mathbb C[x]$ , so that $i = \sqrt{-1}$ in a formal sense, then subsequently derive its algebraically convenient properties. In fact, this is how mathematicians formally “construct” a bona fide $\sqrt{-1}$ from what seems to be thin air.

We remark that Theorem 2 yields a special case of a more general result.

Theorem 3. Let $\phi : \mathbb F_1 \to \mathbb F_2$ be a field isomorphism. Suppose $\mathbb F_i' \supseteq \mathbb F_i$ as algebraic closures according to Theorem 1. Then there exists a field automorphism $\Phi : \mathbb F_1' \to \mathbb F_2'$ such that $\Phi|_{\mathbb F_1} = \phi$ .

Proof Sketch. Adapt the proof of Theorem 2.

Remark 2. In the special case $\mathbb F_1 = \mathbb F_2 = \mathbb F$ , we call $\phi$ a field automorphism on $\mathbb F$ . Then $\Phi : \mathbb F_1' \to \mathbb F_2'$ is a field isomorphism such that $\Phi|_{\mathbb F} = \phi$ . In this case, we say that $\Phi$ is an $\mathbb F$ -automorphism.

Next time, we revisit automorphisms in detail—they will play crucial roles in connecting fields and groups, and in particular, the star of abstract algebra: Galois groups.

—Joel Kindiak, 3 Mar 26, 1226H

KindiakMath

The Algebraic Closure

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The Algebraic Closure

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