Definition: Let \(f(x)\in F[x]\) be a polynomial and let \(K/F\) be an extension field. \(K\) is called a splitting field for \(f(x)\) if
Proposition: Any polynomial \(f(x)\in F[x]\) has a splitting field.
Proof: If all irreducible factors of \(f(x)\) have degree 1 then \(F\) is a splitting field. Otherwise, let \(\alpha\) be a root of an irreducible factor of \(f\) of degree greater than \(1\) and let \(F_1=F(\alpha)\). Write \(f(x)=(x-\alpha)f_1(x)\) and, by induction, let \(E\) be a splitting field for \(f_1(x)\) over \(F(\alpha)\). Then all the roots of \(f(x)\) belong to \(E\). Let \(K\) be the subfield of \(E\) generated over \(F\) by the roots of \(f(x)\). This is your splitting field.
Remark: Some books say that if \(K/F\) is the splitting field over \(F\) for a polynomial, then \(K\) is called a normal extension.
Proposition: If \(f(x)\in F[x]\) has degree \(n\) then its splitting field has degree at most \(n!\).
Proof: It can be obtained by adjoining roots successively of polynomials of degree \(n\), \(n-1\), .
Theorem: (DF Theorem 27 p. 541) Let \(\phi: F\to F'\) be a field isomorphism. Let \(f(x)\in F[x]\) and let \(f'(x)\in F'[x]\) be the polynomial obtained from \(f\) by applying \(\phi\) to its coefficients. Let \(E/F\) be a splitting field of \(f\) and let \(E'/F'\) be a splitting field of \(f'\). Then there is an isomorphism \(\sigma:E\to E'\) which makes the following diagram commutative (the vertical arrows are the inclusion maps):
\[ \begin{xy}\xymatrix{ E \ar[r]^{\sigma} & E' \\ F\ar[u]\ar[r]^{\phi} & F'\ar[u] \\ } \end{xy} \]
Corollary: Any two splitting fields for \(f(x)\) are isomorphic via an isomorphism that is the identity on \(F\).
The extension theorem can seem a little mysterious. Let’s look more closely at an application.
Let \(f(x)=x^3-2\) and let \(E/\Q\) be its splitting field (which has degree 6 over \(\Q\)). Inside this field there are three isomorphic cubic extensions: \(L_1=\Q(\sqrt[3]{2})\), \(L_2=\Q(\omega\sqrt[3]{2})\), and \(L_3=\Q(\omega^2\sqrt[3]{2})\) where \(\omega=e^{2\pi i/3}\) is a cube root of unity.
\[ \begin{xy} \xymatrix{ & E & \\ L_1\ar[ur] & L_2\ar[u] & L_3\ar[ul] \\ & \Q\ar[ur]\ar[u]\ar[ul] &\\ } \end{xy} \]
Now \(E\) is a splitting field for \(f(x)\) over each of \(L_1\), \(L_2\), and \(L_3\).
We can apply the theorem to (for example) the diagram
\[ \begin{xy}\xymatrix{ E \ar[r]^{\sigma} & E \\ \Q(\sqrt[3]{2})\ar[u]\ar[r]^{\phi} & \Q(\omega\sqrt[3]{2})\ar[u] \\ } \end{xy} \]
where \(\phi\) is the isomorphism that sends \(\sqrt[3]{2}\to\omega\sqrt[3]{2}\) and fixes \(\Q\). It follows that there is an automorphism \(\sigma\) of the splitting field that extends \(\phi\).
In general, if \(f(x)\) is an irreducible polynomial over \(F\), and \(\alpha\) and \(\beta\) are two roots of \(f(x)\) in its splitting field \(E/F\), then there is an automorphism \(E\to E\) fixing \(F\) sending \(\alpha\) to \(\beta\). In particular the automorphism group of \(E\) fixing \(F\) permutes the roots of \(f(x)\) transitively.
The proof is by induction. If all roots of \(f(x)\) belong to \(F\), then all roots of \(f'(x)\) belong to \(F'\), and \(E=F\) and \(E'=F'\) so the identity map works. Now suppose we know the result for all \(f\) of degree less than \(n\) and suppose that \(f\) is of degree \(n\). Choose an irreducible factor \(p(x)\) of \(f(x)\) of degree at least \(2\), and the corresponding factor \(p'(x)\) of \(f'(x)\). Since \(F[x]/p(x)\) is isomorphic to \(F'[x]/p'(x)\), we have an isomorphism \(\\phi': F(\alpha)\to F'(\beta)\) that restricts to \(\phi:F\to F'\).
Let \(f(x)=(x-\alpha)f_1(x)\) and \(f'(x)=(x-\beta)f'_1(x)\). Now \(E\) (resp. \(E'\)) is a splitting field for \(f_1\) (resp \(f_1'\)) and by induction we have an isomorphism \(\sigma: E\to E'\) that restricts to \(\phi':F(\alpha)\to F'(\beta)\). This \(\sigma\) also restricts to \(\phi:F\to F'\) (since \(\phi'\) does).
Proposition: Let \(K/F\) be the splitting field of a polynomial. Then if \(g(x)\in F[x]\) is any irreducible polynomial over \(F\), and \(\alpha\in K\) is a root of \(g(x)\), then all roots of \(g(x)\) belong to \(K\). (In other words, if \(K/F\) is a splitting field for some polynomial, then any polynomial in \(F[x]\) is either irreducible or splits into linear factors over \(K\).)
Proof: Suppose that \(K\) is the splitting field of \(f(x)\in F[x]\). Suppose that \(\alpha\in K\) and let \(\beta\) be another root of \(g(x)\) and consider the field \(K(\beta)\). Then \(K(\beta)\) is the splitting field of \(f(x)\) over \(F(\beta)\). (\(K\) contains all the roots of \(f(x)\), and it must contain \(\beta\) if it contains \(F(\beta)\). ) But then we have the diagram:
\[ \begin{xy} \xymatrix{ K\ar[r] & K(\beta) \\ F(\alpha)\ar[u]\ar[r] & F(\beta)\ar[u]\\ } \end{xy} \]
The extension theorem tells us that there is an isomorphism from \(K\) to \(K(\beta)\) carrying \(F(\alpha)\) to \(F(\beta)\) and fixing the field \(F\). Therefore \([K:F]=[K(\beta):F]\). But then \[[K(\beta):F]=[K(\beta):K][K:F].\] This forces \([K(\beta):K]=1\) so \(\beta\in K\).
Definition: A field \(F\) is algebraically closed if it has no nontrivial algebraic extensions; in other words, if every irreducible polynomial over \(F\) has degree \(1\).
Definition: If \(F\) is a field, then \(\overline{F}\) is an algebraic closure of \(F\) if \(\overline{F}/F\) is algebraic and every polynomial in \(F[x]\) splits completely in \(\overline{F}\).
So notice that the complex numbers are algebraically closed, but they are not an algebraic closure of \(\Q\), because they contain transcendental elements.
Lemma: If \(\overline{F}\) is an algebraic closure of \(F\), then \(\overline{F}\) is algebraically closed.
This lemma says that if every polynomial with coefficients in \(F\) has a root in \(\overline{F}\), then every polynomial with coefficients in \(\overline{F}\) has a root in \(F\).
To prove this, let \(f(x)\in\overline{F}[x]\). Let \(F_1/F\) be the extension of \(F\) generated by the coefficients of \(f\). Since \(F_1\) is generated by finitely many algebraic elements, \(F_1/F\) is finite and a root \(\alpha\) of \(f(x)\in F_1[x]\) is finite over \(F_1\). Therefore \(f\) has a root in a finite extension of \(F\), which is therefore in \(\overline{F}\).
Theorem: Given a field \(F\), there exists an algebraically closed field containing \(F\).
Proof: See Proposition 30 in DF on p. 544.
Theorem: If \(K/F\) is algebraically closed, then the collection of elements of \(K\) that are algebraic over \(F\) is an algebraic closure of \(F\).
Since \(\C\) is algebraically closed, the set of algebraic numbers inside \(\C\) is an algebraic closure of \(\Q\). The construction of \(\R\) and \(\C\) is primarily by analysis, and the proof that \(\C\) is algebraically closed is also analytic – at least, the usual proof.