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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 15.3 Algebraic Extensions

Subsection 15.3.1 Algebraic Elements

“People aren’t trees, so it is false when they speak of roots.”
―Tom Robbins

Definition 15.43. Algebraic Element.

For a field extension \(F \subseteq L\) and \(\a \in L\text{,}\) we say \(\a\) is algebraic over \(F\) if \(f(\a) = 0\) for some non-zero polynomial \(f(x) \in F[x]\text{.}\) Otherwise, \(\a\) is transcendental over \(F\text{.}\)

Example 15.44. Algebraic Elements.

\(i \in \C\) is algebraic over \(\R\text{.}\) Indeed, every element of \(\C\) is algebraic over \(\R\text{.}\) (E.g., \(z = a + bi\) is a root of \(x^2 -2ax +a^2 + b^2\text{.}\)) \(e^{2 \pi i/n} \in \C\) ia algebraic over \(\Q\text{.}\) So is \(\sqrt{m}\) for any \(m \in \Z\text{.}\)

Example 15.45. Transcendental Elements.

The numbers \(\pi\) and \(e\) of \(\R\) are transcendental over \(\Q\text{;}\) these are deep facts.

Proof.

All parts use the following construction:
Let \(\phi: F[x] \to L\) be the evaluation homomorphism, given by sending \(f(x)\) to \(f(\a)\) [provisional cross-reference: empty]. Note that \(\phi\) satisfies \(\phi|_F = \id_F\) and \(\phi(x) = \alpha\text{.}\) We have \(\im(\phi) = F[\a]\) by definition of the latter. The First Isomorphism Theorem for Rings thus gives that \(\phi\) induces a ring isomorphism
\begin{equation*} \ov{\phi}: \frac{F[x]}{I} \xra{\cong} F[\a] \end{equation*}
given by \(\ov{\phi}(f(x)+(p(x)))={\phi}(f(x))=f(\a)\text{.}\) In particular, since \(F[\a]\) is a subring of \(L\text{,}\) it is an integral domain, [provisional cross-reference: empty] and hence \(I\) must be a prime ideal (possibly the \(0\) ideal). [provisional cross-reference: empty]
Let us now prove the various parts:
  1. holds because \(I\) is the kernel of the ring map \(\phi\text{.}\)
  2. is by definition.
For (3) assume \(\a\) is algebraic. Then \(I \ne 0\) and hence it has a unique monic generator, which we write \(m_{\a, F}(x)\text{.}\) Since \(I\) is prime, \(m_{\a, F}(x)\) is irreducible [provisional cross-reference: empty]; this proves (3a). Also, this shows that \(\frac{F[x]}{(m_{\a,F}(x))}\) is a field [provisional cross-reference: empty] and hence so is \(F[\a]\text{.}\) Since \(F[\a] \subseteq F(\a)\text{,}\) \(F[\a]\) contains \(F\) and \(\a\text{,}\) and \(F[\a]\) is a field, we must have \(F[\a] = F(\a)\text{.}\) This proves (3b) and (3c).
For (4), if \(\a\) is algebraic over \(F\text{,}\) then \([F(\a):F] = \deg(m_{\a, F}(x)) < \infty\) by (3b) and Proposition [provisional cross-reference: empty]. For the converse, if \([F(\a): F] < \infty\text{,}\) then the infinite list \(1, \a, \a^2, \dots\) of elements of \(F(\a)\) cannot be \(F\)-linearly independent [provisional cross-reference: empty]. So, \(a_0 + a_1 \a + \cdots a_n \a^n = 0\) for some \(n\) and some \(a_0, \dots, a_n \in F\) not all of which are \(0\text{.}\) This shows \(\a\) is the root of a non-zero polynomial.
For (5), if \(\a\) is transcendental, then \(I = 0\) and so \(\phi: F[x] \into L\) is injective. Since \(L\) is a field, \(F[x]\) is a domain [provisional cross-reference: empty], and \(\phi\) is injective [provisional cross-reference: empty], it can be extended to the field of fractions \(F(x)\) of \(F[x]\text{,}\) i.e. there is a ring homomorphism
\begin{equation*} \tilde{\phi}: F(x) \to L \end{equation*}
given by \(\tilde{\phi} \left(\frac{f(x)}{g(x)}\right) = \frac{f(\a)}{g(\a)}\) for all \(f(x), g(x) \in F[x]\) with \(g(x) \ne 0\text{.}\) The image of \(\tilde{\phi}\) is precisely \(F(\a)\text{.}\) The map \(\tilde{\phi}\) is injective since it is a ring homomorphism whose source is a field. [provisional cross-reference: empty]

Definition 15.47. Minimum Polynomial.

Let \(F \subseteq L\) be a field extension and \(\alpha \in L\text{,}\) and consider the ideal
\begin{equation*} I=\{f(x) \in F[x] \mid f(\alpha)=0\} \end{equation*}
from Theorem 15.46. The unique monic generator \(m_{\alpha, F}(x)\) for \(I\) is called the minimal polynomial of \(\alpha\) over \(F\text{.}\)

Remark 15.48.

Note that the minimum polynomial of \(\a\) over \(F\) (if it exists) divides every polynomial in \(F[x]\) that has \(\a\) as a root. In particular, it is the unique monic, irreducible polynomial having \(\a\) as a root. It can also be characterized as the monic polynomial in \(F[x]\) of least degree having \(\a\) as a root.

Example 15.49.

The minimal polynomial of \(i\) over \(\mathbb{R}\) is \(m_{i, \mathbb{R}}(x)=x^{2}+1\text{.}\)
Solution.

Example 15.50.

For any prime integer \(p\text{,}\) set \(\mu_p = e^{2 \pi i/p}\text{,}\) a so-called “primitive \(p\)-th root of unity". Let us find \(m_{\mu_p, \Q}(x)\text{.}\) Note that \(\mu_p\) is a root of \(x^p -1\) which factors as \((x-1) g(x)\) where \(g(x) = x^{p-1} + \cdots + x + 1\text{.}\) As we showed before [provisional cross-reference: empty], \(g(x)\) is irreducible in \(\Q[x]\text{.}\) Since \(\mu_p\) is not a root of \(x-1\text{,}\) it must be a root of \(g(x)\text{,}\) and since \(g(x)\) is irreducible, it must be \(m_{\mu_p, \Q}(x)\) (see the Remark).[provisional cross-reference: empty]

Proof.

Let \(A \subseteq K\) be a basis for \(K\) as an \(L\)-vector space and let \(B \subseteq L\) be a basis for \(L\) as an \(F\)-vector space. Let \(AB\) denote the subset \(\{ab \mid a \in A, b \in B\}\) of \(K\text{.}\) The Theorem follows from the following two facts:
  1. \(AB\) is a basis of \(K\) as an \(F\)-vector space and
  2. the function \(A \times B \to AB, (a,b) \mapsto ab\) is bijective (so that the cardinality of \(AB\) is \(|A| \cdot |B|\)).
Concerning (a), for \(\a \in K\text{,}\) we have \(\a = \sum_i l_i a_i\) for some \(a_1, \dots, a_m \in A\) and \(l_1, \dots, l_m \in L\text{.}\) For each \(i\text{,}\) \(l_i\) is an \(F\)-linear combination of a finite set of elements of \(B\text{.}\) Combining these gives that \(\a\) is in the \(F\)-span of \(AB\text{.}\)
To prove linear independence, it suffices to prove that if \(a_1, \dots, a_m\) and \(b_1, \dots, b_n\) be distinct elements of \(A\) and \(B\) respectively, then the set \(\{a_ib_j\}\) is linearly independent. Suppose \(\sum_{i,j} f_{i,j} a_i b_j = 0\) for some \(f_{i,j} \in F\text{.}\) Since the \(b_j\)’s are \(L\)-linearly independent and
\begin{equation*} \sum_{i,j} f_{i,j} a_i b_j = \sum_j (\sum_i f_{i,j} a_i) b_j \end{equation*}
and \(f_{i,j} a_i \in L\text{,}\) we get that, for each \(j\text{,}\) \(\sum_i f_{i,j} a_i = 0\text{.}\) Using now that the \(a_i\)’s are \(F\)-linearly independent, we have that for all \(j\) and all \(i\text{,}\) \(f_{i,j} = 0\text{.}\) This proves \(\{a_i b_j \mid i = 1, \dots, m, j = 1, \dots, n\}\) is linearly independent over \(F\text{,}\) and hence \(AB\) is linearly independent over \(F\text{.}\)
Concerning (b), if \(ab = a'b'\) for some \(a,a' \in A, b, b' \in B\text{,}\) then \(ab - a'b' = 0\text{,}\) and since the \(b\)’s are \(L\)-linearly independent, we must have \(b = b'\) and hence \(a = a'\text{.}\)

Example 15.52.

Say \(F \subseteq L\) is a field extension of prime degree \(p\text{.}\) Given \(w \in L \setminus F\text{,}\) by the The Degree Formula we have \([L:F(w)][F(w):F] = [L:F] = p\text{.}\) Since \(w \notin F\text{,}\) \(F \subsetneq F(w)\) and so \([F(w): F] > 1\text{.}\) It follows that \([L: F(w)] = 1\text{,}\) whence \(L = F(w)\text{.}\) As a (very simple) example of this, since \([\C: \R]\) is prime, \(\R(w) = \C\) for any complex number \(w\) that is not real.

Example 15.53.

Let \(F\) be the result of adjoining to \(\Q\) all of the roots in \(\C\) of \(x^2 - 5\text{.}\) That is, \(F = \Q(\a_1, \a_2, \a_3, \a_4)\) where \(\a_1 = \sqrt[4]{5}\text{,}\) \(\a_2 = i \sqrt[4]{5}\text{,}\) \(\a_3 = -\sqrt[4]{5}\text{,}\) and \(\a_4 = - i \sqrt[4]{5}\text{.}\) As we shall see later, \(F\) is an example of a “splitting field". Let’s find \([F: \Q]\text{.}\)
First, let us note that we can also describe \(F\) as \(F = \Q(\sqrt[4]{5}, i)\text{.}\) This holds since each of \(\a_1, \dots, \a_4\) belongs to \(\Q(\sqrt[4]{5}, i)\) and hence \(F \subseteq \Q(\sqrt[4]{5}, i)\text{.}\) The opposite containment holds because \(\sqrt[4]{5}, i \in F\text{,}\) with the latter being true because \(i = \a_2/\a_1\text{.}\)
Set \(E = \Q(i)\text{.}\) Then \([E: \Q] = 2\text{.}\) Since \(F = E(\sqrt[4]{5})\) and \(\sqrt[4]{5}\) is a root of \(x^4-5\text{,}\) we have \([F : E]\) is at most \(4\) and it will be exactly \(4\) if and only if \(x^4 -5\) is irreducible in \(\Q(i)[x]\text{.}\) This is unclear.
So instead let’s try a different approach. Let \(L = \Q(\sqrt[4]{5})\text{.}\) Then since \(x^4 - 5\) is irreducible in \(\Q[x]\) by Eisenstein’s Criterion, we have \([L: \Q] = 4\text{.}\) Since \(F = L(i)\) and \(i\) is a root of \(x^2 +1\text{,}\) we have \([F:L] \leq 2\text{.}\) But \(F \ne L\) since \(F \subseteq \R\text{.}\) Note that \([F:L] = 1\) if and only if \(F = L\text{.}\) Thus \([F:L] = 2\text{.}\)
By the The Degree Formula we conclude that
\begin{equation*} [F: \Q] = [F:L][L:\Q] = 2 \cdot 4 = 8. \end{equation*}
Note that, in hindsight, it must have been true that \(x^4 - 5\) is irreducible in \(\Q(i)[x]\text{,}\) since otherwise the The Degree Formula would give that \([F: \Q] < 8\text{.}\)

Subsection 15.3.2 Algebraic Field Extensions

“If you forget your roots, you’ve lost sight of everything.”
―Walter Payton

Definition 15.54. Algebraic Extension.

A field extension \(F \subseteq L\) is called algebraic if every element \(\a \in L\) is algebraic over \(F\) (i.e., if every element of \(L\) is the root of a non-zero polynomial with coefficients in \(F\)).

Proof.

Pick \(\a \in L\text{.}\) Since \(F(\a)\) is a \(F\)-vector subspace of \(L\) and \(\dim_F(L) = [L: F] < \infty\) we have \([F(\a): F] < \infty\text{.}\) (One could also appeal to the The Degree Formula for this.) So by (4) of Theorem Theorem 15.46, \(\a\) is algebraic of \(F\text{.}\)

Example 15.56. Infinite Algebraic Extension.

Let \(E_n = \Q(\sqrt[n]{2})\) and set \(F = \bigcup_n E_n\text{.}\) Then \(F\) is a subfield of \(\C\text{:}\) To see this, note first that \(E_n \subseteq E_m\) provided \(n \mid m\text{.}\) Given \(a, b \in F\text{,}\) we have \(a \in E_n\) and \(b \in E_m\) for some \(n\) and \(m\) and hence \(a,b\) are both in \(E_{mn}\text{.}\) Since \(E_{mn}\) is a field, we have \(a + b\text{,}\) \(a-b\text{,}\) \(a \cdot b\) and (provided \(a \ne 0\)) \(\frac{1}{a}\) all belong to \(E_{mn}\) and hence to \(F\text{.}\) This proves \(F\) is field extension of \(\Q\text{.}\) It is algebraic over \(\Q\) since each \(E_n\) is. But it is not a finite extension of \(\Q\text{,}\) since \([E_n: \Q] = n\) (since \(x^n - 2\) is irreducible in \(\Q[x]\) by Eisenstein’s Criterion) and hence \([F: \Q] \geq n\) for all \(n\text{.}\) [provisional cross-reference: empty]

Proof.

Let \(F \subseteq K \subseteq L\) be extensions of fields, not necessarily finite.
Suppose that \(K/F\) and \(L/K\) are algebraic extensions. Let \(\a\in L\text{.}\) Then \(\a\) is the root of the polynomial
\begin{equation*} f(x)=a_nx^n+\dots+a_0, \end{equation*}
with \(a_{i}\in K\text{.}\) Notice that \(f\) is a polynomial in \(F(a_n,\dots,a_0)[x]\text{,}\) making \(\a\) is algebraic over this as well. Consider the chain of field extensions
\begin{equation*} F \subseteq F(a_0) \subseteq F(a_0,a_1) \subseteq \cdots \subseteq F(a_0, a_1, \dots, a_{n-1} )\subseteq F(a_0, \dots, a_{n}, \a) \end{equation*}
Since \(a_i\) is algebraic over \(F\) for all \(i\) and \(\a\) is algebraic over \(F(a_0, a_1, \dots, a_{n})\text{,}\) by Theorem CITEX each step in this chain has finite degree. By the The Degree Formula, \([F(a_0, \dots, a_{n}, \a): F]\) is finite and thus so is \([F(\a):F]\text{.}\) By the Theorem CITEX again, \(\a\) is algebraic over \(F\text{.}\)
Next suppose that \(L/F\) is algebraic. Let \(\a\in K\text{.}\) Then \(\a\in L\text{,}\) and so it is algebraic over \(F\text{.}\) Now let \(\b\in L\text{.}\) Then \(\b\) is the root of a polynomial in \(F\text{,}\) which is also in \(K\text{,}\) so \(L/K\) is algebraic as well.
\(K=\Q(\sqrt{2})\) is Galois over \(F=\Q\text{,}\) and \(L=\Q(\sqrt{2},\sqrt[4]{2})\) is Galois over \(K\text{,}\) but \(L\) is not Galois over \(F,\) as the splitting field of \(x^4-2\) has degree \(8\text{.}\)

Remark 15.58.

The converse of this proposition is also true: Given field extensions \(F \subseteq L \subseteq E\text{,}\) if \(F \subseteq E\) is algebraic then so are \(L \subseteq E\) and \(F \subseteq L\text{.}\)
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