The following are equivalent for a field \(L\text{:}\)
\(L\) is algebraically closed.
Every non-constant polynomial with coefficients in \(L\) splits completely into (not necessarily distinct) linear factors.
There are no non-trivial algebraic extensions of \(L\text{:}\) If \(L \subseteq E\) is an algebraic field extension then \(L = E\text{.}\)
Proof.
\(\Rightarrow\) (2): Given a non-constant \(f(x) \in L[x]\text{,}\) by assumption \(f\) has a root \(a \in L\) and thus \(f(x) = (x - a)g(x)\) with \(g(x) \in L[x]\text{.}\) But then \(g\) also has a root and so it too factors, and so on.
\(\Rightarrow\) (3): Say \(L \subseteq E\) is algebraic. Pick \(\a \in E\text{.}\) Then \(\a\) is a root of some \(f(x) \in F[x]\text{.}\) But since \(f\) factors completely, \(\a \in L\text{.}\)
\(\Rightarrow\) (1). Pick a non-constant \(f(x) \in L[x]\text{.}\) By [provisional cross-reference: cite] , there there is finite extension \(L'\) of \(L\) in which \(f\) does have a root. By assumption \(L = L'\) and so this root must be in \(L\text{.}\)
Definition16.62.Algebraic Closure.
Given a field \(F\text{,}\) a field \(\ov{F}\) is called an algebraic closure of \(F\) if \(\ov{F}\) is an algebraic field extension of \(F\) and \(\ov{F}\) is algebraically closed.
Example16.63.
\(\C\) is an algebraic closure of \(\R\text{.}\) This follows from the fact that \(\R \subseteq \C\) is a finite extension, hence algebraic, and the Fundamental Theorem of Algebra, which we will not prove.
Proposition16.64.Collection of Algebraic Elements is Algebraically Closed.
Let
\begin{equation*}
\ov{\Q} := \{ z \in \C \mid z\text{ is algebraic over }\Q\}.
\end{equation*}
Then \(\ov{\Q}\) is an algebraic closure of \(\Q\text{.}\)
More generally, if \(F \subseteq L\) is a field extension and \(L\) is algebraically closed, then the collection \(\ov{F}\) of elements of \(L\) that are algebraic over \(F\) is an algebraic closure of \(F\text{.}\)
Proof.
It is far from clear that \(\ov{F}\) is a subfield of \(L\text{,}\) and so we first prove that: Given \(\a, \beta \in \ov{F}\text{,}\) we have that \([F(\a): F]\) and \([F(\a, \beta): F(\a)]\) are finite and hence so is \([F(\a, \beta): F]\text{.}\) Thus, every element of \(F(\a, \beta)\) is algebraic over \(F\text{;}\) that is, \(F(\a, \beta) \subseteq \ov{F}\text{.}\) Since \(F(\a, \beta)\) is a field, it follows that \(\ov{F}\) contains \(\a \pm \beta\text{,}\)\(\a \cdot \beta\) and \(\a^{-1}\) if \(\a \ne 0\text{.}\) This proves that \(\ov{F}\) is indeed a subfield of \(L\text{.}\)
It is clear from the definition that \(F \subseteq \ov{F}\) is an algebraic field extension.
Given a non-constant \(g(x) \in \ov{F}[x]\text{,}\) let \(\a\) be one of its roots in \(L\) (which exists since we assume \(L\) is algebraically closed). Then \(\ov{F} \subseteq \ov{F}(\a)\) is an algebraic extension and hence so is \(F \subseteq \ov{F}(\a)\) by Proposition . This proves \(\a \in \ov{F}\) and hence that \(\ov{F}\) is algebraically closed.
Theorem16.65.Existence and Uniqueness of Algebraic Closures.
For any field \(F\text{,}\) there exists an algebraic closure of \(F\text{.}\) If \(L\) and \(L'\) are two algebraic closures of the same field \(F\text{,}\) then there exists a field isomorphism \(\phi: L \xra{\cong} L'\) such that \(\phi|_F = \id_F\) (i.e., \(\phi(a) = a\) for all \(a \in F\)).
Proof.
Fake Proof of Existence:
Let \(\cS\) be the collection of all algebraic field extensions of \(F\text{.}\) Make \(\cS\) into a poset by declaring \(L \leq L'\) iff \(L \subseteq L'\text{.}\) We prove \(\cS\) has a maximal element.
Let \(\cT\) be any totally ordered subset of \(\cS\text{.}\) If \(\cT\) is empty, then \(F \in \cS\) is an upper bound for \(\cT\text{.}\) If \(\cT\) is non-empty, set \(K := \cup_{E \in \cT} E\text{.}\) Using that \(\cT\) is totally ordered, it is not hard to see that \(K\) is indeed a field. It clearly contains \(F\) as a subfield and every element of it is algebraic over \(F\text{.}\) So \(K \in \cS\) and it is an upper bound for \(\cT\text{.}\) By Zorn’s Lemma, \(\cS\) has a maximal member \(L\text{.}\)
By construction \(L\) is algebraic over \(F\text{.}\) If \(L\) were not algebraically closed, then there would be a non-trivial algebraic extension \(L \subsetneq E\) of it, by [provisional cross-reference: cite] . But then \(F \subseteq E\) is algebraic by Proposition [provisional cross-reference: cite], and this contradicts the maximality of \(L\text{.}\)
Why is this only a fake proof? It’s because \(\cS\text{,}\) as we’ve defined it, is not a set but rather it is something bigger than that. Zorn’s Lemma only applies to posets. How annoying!
Theorem16.66.Relatively Prime Degrees and Irreducibility.
Assume that \(F \sse K\) is a finite extension of fields of degree \(n = [K : F ]\text{.}\) If \(f \in F [x]\) is irreducible of degree \(d\) and \(\gcd(d, n) = 1\) then \(f\) remains irreducible when regarded as an element of the ring \(K[x]\text{.}\)
Proof.
Let \(F \subseteq K\) be a finite extension of fields of degree \(n = [K : F ]\text{.}\)
Suppose that \(f \in F[x]\) is irreducible of degree \(d\) and \(\gcd(d, n) = 1\text{.}\)
First, note that if \(d=1\) then \(f(x)\) will remain irreducible in \(K[x]\)[provisional cross-reference: empty]. Suppose then that \(d>1\text{.}\) There exists an algebraically closed extension \(\widetilde F\) such that \(f\) has a root \(r\)[provisional cross-reference: empty]. Consider \(K(r)\text{.}\) As \(r\) is algebraic in \(K(r)\) we know there exists some unique irreducible minimum polynomial \(g\) of degree \(q\text{,}\) and thus that \([K(r):K]=q\)[provisional cross-reference: empty]. Using the The Degree Formula we see that
However, \([K(r):F]=[K(r):F(r)][F(r):F]\) and so \(qn=md\) for some \(m\in\Z\text{,}\) so \(d|qn\text{.}\) As \(\gcd(n,d)=1\) we must have \(d|q\text{.}\) But \(q\) was defined to be the degree of \(g\text{,}\) which divides \(f\text{.}\) As \(d|q\) and \(q\leq d\text{,}\) we see that \(d=q\text{,}\) so \(f=kg\) for some \(k\in F\text{.}\) As irreducible polynomials multiplied by a constant are still irreducible, we see that \(f\) is indeed irreducible in \(K[x]\text{.}\)
The statement in part (a) would become false if the assumption that \(\gcd(n, d) = 1\) were omitted.
Example16.67.
Let \(p(x)\) be a non-constant irreducible polynomial of degree \(d\) in \(F[x]\text{.}\) Let \(K=F[x]/p(x)\text{.}\) Because \(p(x)\) is irreducible and \(F[x]\) is a PID, \((p(x))\) is a maximal ideal. Thus \(K = F[x]/(p(x))\) is a field [provisional cross-reference: empty], \([K:F]=d\text{,}\) and \(x+p(x)\) is a root of \(p(x)\) in \(K\)[provisional cross-reference: empty]. Hence \(p(x)\) is no longer irreducible by Theorem 2.2
Qual Watch.
Proving Theorem 16.66 was [cross-reference to target(s) "jan-2023-7" missing or not unique] on the [cross-reference to target(s) "jan-2023" missing or not unique] qualifying exam and [cross-reference to target(s) "may-2017-7" missing or not unique] on the [cross-reference to target(s) "may-2017" missing or not unique] qualifying exam.
