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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 16.5 Splitting Fields

Subsection Split Up and Look For Clues

“The best way to carve is not to split.”
―Laozi

Definition 16.68. Splitting Field.

For a field \(F\) and non-constant polynomial \(f(x) \in F[x]\text{,}\) a splitting field of \(f(x)\) over \(F\) is a field extension \(F \subseteq L\) such that
  1. \(f(x)\) splits completely into linear factors in \(L[x]\text{;}\) that is, \(f(x) = c\prod_{i=1}^n (x - \a_i)\) for some \(c, \a_1, \dots, \a_n \in L\text{,}\) and
  2. \(L = F(\a_1, \dots, \a_n)\text{;}\) that is, \(L\) is the smallest subfield of \(L\) that contains \(F\) and all the roots of \(f(x)\text{.}\)

Example 16.69. Splitting Fields.

  • As a silly example, if \(f(x)\) already splits into linear factors over \(F[x]\text{,}\) then \(F\) itself is the splitting field of \(f(x)\) over \(F\text{.}\)
  • The splitting field of \(x^2 + 1\) over \(\R\) is \(\C\text{.}\)
  • If \(q(x)\) is any irreducible quadratic polynomial in \(\R[x]\text{,}\) then the splitting field of \(q(x)\) is \(\C\text{.}\)
  • In general, given \(f(x) \in \Q[x]\text{,}\) a splitting field of \(f(x)\) is given by \(\Q(\a_1, \dots, \a_n)\) where \(\a_1, \dots, \a_n \in \C\) are all of the roots of \(f(x)\) in \(\C\text{.}\)

Remark 16.70.

More generally, we may speak of the splitting field of a list of polynomials in \(F[x]\text{:}\) it is a field extension over which each polynomial factors completely and which is generated by all the roots of all the polynomials.
Note that given a finite list \(f_1(x), \dots, f_k(x) \in F[x]\text{,}\) a splitting field for this list is identical to a splitting field for their product \(f_1(x)\cdots f_k(x)\text{.}\)

Proof.

For (1), let \(\ov{F}\) be an algebraic closure of \(F\text{,}\) which exists by the previous Theorem. Let \(\a_1, \dots, \a_m\) be the roots of \(f(x)\) in \(\ov{F}\text{,}\) and set \(L = F(\a_1, \dots, \a_m)\text{.}\)
It is clear \(L\) is a splitting field.
To prove (2), we proceed by induction on the degree of \(f(x)\text{.}\) If \(f\) is linear, then the only splitting field of \(f\) over \(F\) is \(F\) itself and so the result is clear in this case. Say \(\a_1, \dots, \a_m\) and \(\a'_1, \dots, \a'_m\) are the roots of \(f(x)\) in \(L\) and \(L'\text{,}\) respectively, and say they are ordered so that, \(\a_m\) and \(\a'_m\) are roots of the same irreducible factor of \(f(x)\) in \(F[x]\text{.}\)
By Corollary there is an isomorphism \(\phi: F(\a_n) \xra{\cong} F(\a_n')\) that fixes \(F\text{.}\) Note that \(f(x)\) factors as \((x-\a_n) g(x)\) in \(F(\a_n)[x]\) and that \(L\) is the splitting field of \(g(x)\) over \(F(\a_n)\text{,}\) and similarly \(f(x)\) factors as \((x-\a'_n) g'(x)\) in \(F(\a'_n)[x]\) and that \(L\) is the splitting field of \(g'(x)\) over \(F(\a'_n)\text{.}\) If we blur our eyes slightly and pretend \(\phi\) is the identity map, we can apply the inductive hypothesis, since \(\deg(g) < \deg(f)\text{,}\) to conclude that there is an isomorphism \(\s\) as in the statement. I leave a more rigorous argument to your imaginations.
To prove (3), we also proceed by induction on the degree of \(f\text{,}\) using the same notation as in the proof of (2).
Since \(\a_n\) is a root of \(f(x)\text{,}\) we have \(m_{F,\a_n} \mid f(x)\) and hence
\begin{equation*} [F(\a_n): F] = \deg(m_{F,\a_n}) \leq n. \end{equation*}
In \(F(\a_n)\) we have \(f(x) = (x-\a_n) g(x)\) with \(g(x) \in F(\a_n)[x]\) and, as before, \(L\) is the splitting field of \(g(x)\) over \(F(\a_n)\text{,}\) so that by induction \([L: F(\a_n)] \leq (n-1)!\text{.}\) By the The Degree Formula
\begin{equation*} [L: F] = [L: F(\a_n)][F(\a_n): F] \leq (n-1)! \cdot n = n!. \end{equation*}

Remark 16.72.

Recall from before that we proved there exists a field extension \(F \subseteq E\) in which \(f(x)\) has at least one root. So \(f(x) = (x-\a_1) g(x)\) for some \(\a_1 \in E\) and \(g(x) \in E[x]\text{.}\) We can then find a field extension of \(E\) in which \(g\) has at least one root \(\a_2\text{,}\) and so on. In this way we build a field extension \(F \subseteq K\) such that \(g\) factors in \(K\) as \(c\prod_{i=1}^n (x - \a_i)\) for some \(c, \a_1, \dots, \a_n \in K\text{.}\) Finally, \(L = F(\a_1, \dots, \a_n)\) is a splitting field of \(f\text{.}\)

Example 16.73.

The splitting field \(L\) of \(x^3 -2\) over \(\Q\) is \(L = \Q(\sqrt[3]{2}, \zeta_3 \sqrt[3]{2}, \zeta_3^2 \sqrt[3]{2})\text{,}\) where \(\zeta_3 = e^{\frac{2 \pi i}{3}}\text{.}\) It is not hard to see that \(L = \Q(\sqrt[3]{2}, \zeta_3)\text{.}\) We have \([\Q(\sqrt[3]{2}): \Q] = 3\) since \(x^3 - 2\) is irreducible in \(\Q[x]\) (by Eisenstein’s Criterion). Since \(\Q(\sqrt[3]{2}) \subseteq \R\) and thus \(\Q(\sqrt[3]{2}) \ne L\text{,}\) the The Degree Formula gives that \([L: \Q] \geq 6\text{.}\) By the Theorem, \([L: \Q] \leq 6\) and hence \([L: \Q] = 6\text{.}\) (We could also have proven this without appealing to the Theorem.)

Example 16.74.

The splitting field of \(f(x) = x^4 - 5x^2 + 6\) is
\begin{equation*} \Q(\sqrt{2}, -\sqrt{2}, \sqrt{3}, -\sqrt{3}) = \Q(\sqrt{2}, \sqrt{3}) = \Q(\sqrt{2} + \sqrt{3}). \end{equation*}
This holds since \(f(x) = (x^2-3)(x^2-2)\text{.}\) It is not too hard to see that the degree of this splitting field over \(\Q\) is \(4\text{,}\) far smaller than the \(4!\) upper bound given by the Theorem.

Example 16.75.

Let \(f(x) = x^n -1 \in \Q[x]\text{.}\) Then \(f(x)\) splits completely in \(\C[x]\) and its roots are the \(n\) \(n\)-th roots of \(1\text{.}\) One of these is \(\zeta_n := e^{2 \pi i/n}\text{.}\) Notice that every other \(n\)-th root of \(1\) is a power of this one. We thus see that \(\Q(\zeta_n)\) is the splitting field of \(x^n-1\) over \(\Q\text{.}\) This field is called the {} of \(n-th\) roots of 1 over \(\Q\text{.}\) This is a somewhat special example: upon joining one of the roots of \(f\) we got all the others for free. This happens in other examples too, but is certainly {} a general principle.
In particular, we see that the degree of \(\Q \subseteq \Q(\zeta_n)\) is at most \(n\text{,}\) far less than the bound of \(n!\) given by the Proposition above. In fact, it is at most \(n-1\) since \(f\) factors as \((x-1)(x^{n-1} + \cdots + x + 1)\text{,}\) and hence the minimum polynomial of \(\zeta_n\) is a divisor of \(x^{n-1} + \cdots + x + 1\text{.}\)
When \(n = p\) is prime, then \(x^{p-1} + \cdots + x + 1\) is irreducible, as we proved before, and hence it must equal the minimum polynomial of \(\zeta_p\text{.}\) So, in this case, the degree of \(\Q \subseteq \Q(\zeta_p)\) is exactly \(p-1\text{,}\) but it can be smaller than \(n-1\) in general; for example, when \(n = 4\text{,}\) \(\zeta_4= i\) and \([\Q(i): \Q] = 2\text{.}\) Note that \(x^3 + x^2 + x + 1\) factors as \((x^2+1)(x+1)\) and of course \(m_{i, \Q}(x) = x^2 + 1\text{.}\)
The irreducible polynomial \(m_{\zeta_n, \Q}(x)\) is known as the {}.

Proof.

We basically already proved this, but since it is of large importance, let’s do so again:
Since \(\a, \beta\) are roots of the same irreducible polynomial, by Corollary there is an isomorphism \(\tau: F(\a) \to F(\beta)\) such that \(\tau|_F = \id_F\) and \(\tau(\a) = \beta\text{.}\) We have two field maps, \(F(\a) \into L\) (actual inclusion) and the composition of \(F(\a) \xra{\tau} F(\beta)\into L\text{,}\) and they realize \(L\) as the splitting field of \(f(x)\) over \(F(\a)\) in two different ways. Since splitting fields are unique, an isomorphism such as \(\s\) exists.

Example 16.77.

Let \(L\) be the splitting field of \(x^3 - 2\) over \(\Q\text{;}\) so \(L = \Q(\sqrt[3]{2}, e^{2 \pi i/3}\sqrt[3]{2}, e^{4 \pi i/3}\sqrt[3]{2})\text{.}\)
Corollary 16.76 gives that there is a field automorphism \(\s\) of \(L\) such that \(\s(e^{2 \pi i/3}\sqrt[3]{2}) = e^{4 \pi i/3}\sqrt[3]{2}\text{.}\) Complex conjugation gives such an isomorphism.
It also gives there is a field automorphism \(\tau\) of \(L\) such that \(\tau(\sqrt[3]{2}) = e^{2 \pi i/3}\sqrt[3]{2}\text{.}\) This is less obvious.
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