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Section21.1Extension Fields

A field \(E\) is an extension field of a field \(F\) if \(F\) is a subfield of \(E\). The field \(F\) is called the base field. We write \(F \subset E\).

Example21.1

For example, let \begin{equation*}F = {\mathbb Q}( \sqrt{2}\,) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \}\end{equation*} and let \(E = {\mathbb Q }( \sqrt{2} + \sqrt{3}\,)\) be the smallest field containing both \({\mathbb Q}\) and \(\sqrt{2} + \sqrt{3}\). Both \(E\) and \(F\) are extension fields of the rational numbers. We claim that \(E\) is an extension field of \(F\). To see this, we need only show that \(\sqrt{2}\) is in \(E\). Since \(\sqrt{2} + \sqrt{3}\) is in \(E\), \(1 / (\sqrt{2} + \sqrt{3}\,) = \sqrt{3} - \sqrt{2}\) must also be in \(E\). Taking linear combinations of \(\sqrt{2} + \sqrt{3}\) and \(\sqrt{3} - \sqrt{2}\), we find that \(\sqrt{2}\) and \(\sqrt{3}\) must both be in \(E\).

Example21.2

Let \(p(x) = x^2 + x + 1 \in {\mathbb Z}_2[x]\). Since neither 0 nor 1 is a root of this polynomial, we know that \(p(x)\) is irreducible over \({\mathbb Z}_2\). We will construct a field extension of \({\mathbb Z}_2\) containing an element \(\alpha\) such that \(p(\alpha) = 0\). By Theorem 17.22, the ideal \(\langle p(x) \rangle\) generated by \(p(x)\) is maximal; hence, \({\mathbb Z}_2[x] / \langle p(x) \rangle\) is a field. Let \(f(x) + \langle p(x) \rangle\) be an arbitrary element of \({\mathbb Z}_2[x] / \langle p(x) \rangle\). By the division algorithm, \begin{equation*}f(x) = (x^2 + x + 1) q(x) + r(x),\end{equation*} where the degree of \(r(x)\) is less than the degree of \(x^2 + x + 1\). Therefore, \begin{equation*}f(x) + \langle x^2 + x + 1 \rangle = r(x) + \langle x^2 + x + 1 \rangle.\end{equation*} The only possibilities for \(r(x)\) are then \(0\), \(1\), \(x\), and \(1 + x\). Consequently, \(E = {\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle\) is a field with four elements and must be a field extension of \({\mathbb Z}_2\), containing a zero \(\alpha\) of \(p(x)\). The field \({\mathbb Z}_2( \alpha)\) consists of elements \begin{align*} 0 + 0 \alpha & = 0\\ 1 + 0 \alpha & = 1\\ 0 + 1 \alpha & = \alpha\\ 1 + 1 \alpha & = 1 + \alpha. \end{align*} Notice that \({\alpha}^2 + {\alpha} + 1 = 0\); hence, if we compute \((1 + \alpha)^2\), \begin{equation*}(1 + \alpha)(1 + \alpha)= 1 + \alpha + \alpha + (\alpha)^2 = \alpha.\end{equation*} Other calculations are accomplished in a similar manner. We summarize these computations in the following tables, which tell us how to add and multiply elements in \(E\).

\begin{equation*}\begin{array}{c|cccc} + & 0 & 1 & \alpha & 1 + \alpha \\ \hline 0 & 0 & 1 & \alpha & 1 + \alpha \\ 1 & 1 & 0 & 1 + \alpha & \alpha \\ \alpha & \alpha & 1 + \alpha & 0 & 1 \\ 1 + \alpha & 1 + \alpha & \alpha & 1 & 0 \end{array}\end{equation*}

Table21.3Addition Table for \({\mathbb Z}_2(\alpha)\)

\begin{equation*}\begin{array}{c|cccc} \cdot & 0 & 1 & \alpha & 1 + \alpha \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & 1 + \alpha \\ \alpha & 0 & \alpha & 1 + \alpha & 1 \\ 1 + \alpha & 0 & 1 + \alpha & 1 & \alpha \end{array}\end{equation*}

Table21.4Multiplication Table for \({\mathbb Z}_2(\alpha)\)

The following theorem, due to Kronecker, is so important and so basic to our understanding of fields that it is often known as the Fundamental Theorem of Field Theory.

Proof
Example21.6

Let \(p(x) = x^5 + x^4 + 1 \in {\mathbb Z}_2[x]\). Then \(p(x)\) has irreducible factors \(x^2 + x + 1\) and \(x^3 + x + 1\). For a field extension \(E\) of \({\mathbb Z}_2\) such that \(p(x)\) has a root in \(E\), we can let \(E\) be either \({\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle\) or \({\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle\). We will leave it as an exercise to show that \({\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle\) is a field with \(2^3=8\) elements.

SubsectionAlgebraic Elements

An element \(\alpha\) in an extension field \(E\) over \(F\) is algebraic over \(F\) if \(f(\alpha)=0\) for some nonzero polynomial \(f(x) \in F[x]\). An element in \(E\) that is not algebraic over \(F\) is transcendental over \(F\). An extension field \(E\) of a field \(F\) is an algebraic extension of \(F\) if every element in \(E\) is algebraic over \(F\). If \(E\) is a field extension of \(F\) and \(\alpha_1, \ldots, \alpha_n\) are contained in \(E\), we denote the smallest field containing \(F\) and \(\alpha_1, \ldots, \alpha_n\) by \(F( \alpha_1, \ldots, \alpha_n)\). If \(E = F( \alpha )\) for some \(\alpha \in E\), then \(E\) is a simple extension of \(F\).

Example21.7

Both \(\sqrt{2}\) and \(i\) are algebraic over \({\mathbb Q}\) since they are zeros of the polynomials \(x^2 -2\) and \(x^2 + 1\), respectively. Clearly \(\pi\) and \(e\) are algebraic over the real numbers; however, it is a nontrivial fact that they are transcendental over \({\mathbb Q}\). Numbers in \({\mathbb R}\) that are algebraic over \({\mathbb Q}\) are in fact quite rare. Almost all real numbers are transcendental over \({\mathbb Q}\). 6  (In many cases we do not know whether or not a particular number is transcendental; for example, it is still not known whether \(\pi + e\) is transcendental or algebraic.)

A complex number that is algebraic over \({\mathbb Q}\) is an algebraic number. A transcendental number is an element of \({\mathbb C}\) that is transcendental over \({\mathbb Q}\).

Example21.8

We will show that \(\sqrt{2 + \sqrt{3} }\) is algebraic over \({\mathbb Q}\). If \(\alpha = \sqrt{2 + \sqrt{3} }\), then \(\alpha^2 = 2 + \sqrt{3}\). Hence, \(\alpha^2 - 2 = \sqrt{3}\) and \(( \alpha^2 - 2)^2 = 3\). Since \(\alpha^4 - 4 \alpha^2 + 1 = 0\), it must be true that \(\alpha\) is a zero of the polynomial \(x^4 - 4 x^2 + 1 \in {\mathbb Q}[x]\).

It is very easy to give an example of an extension field \(E\) over a field \(F\), where \(E\) contains an element transcendental over \(F\). The following theorem characterizes transcendental extensions.

Proof

We have a more interesting situation in the case of algebraic extensions.

Proof

Let \(E\) be an extension field of \(F\) and \(\alpha \in E\) be algebraic over \(F\). The unique monic polynomial \(p(x)\) of the last theorem is called the minimal polynomial for \(\alpha\) over \(F\). The degree of \(p(x)\) is the degree of \(\alpha\) over \(F\).

Example21.11

Let \(f(x) = x^2 - 2\) and \(g(x) = x^4 - 4 x^2 + 1\). These polynomials are the minimal polynomials of \(\sqrt{2}\) and \(\sqrt{2 + \sqrt{3} }\), respectively.

Proof
Example21.14

Since \(x^2 + 1\) is irreducible over \({\mathbb R}\), \(\langle x^2 + 1 \rangle\) is a maximal ideal in \({\mathbb R}[x]\). So \(E = {\mathbb R}[x]/\langle x^2 + 1 \rangle\) is a field extension of \({\mathbb R}\) that contains a root of \(x^2 + 1\). Let \(\alpha = x + \langle x^2 + 1 \rangle\). We can identify \(E\) with the complex numbers. By Proposition 21.12, \(E\) is isomorphic to \({\mathbb R}( \alpha ) = \{ a + b \alpha : a, b \in {\mathbb R} \}\). We know that \(\alpha^2 = -1\) in \(E\), since \begin{align*} \alpha^2 + 1 & = (x + \langle x^2 + 1 \rangle)^2 + (1 + \langle x^2 + 1 \rangle)\\ & = (x^2 + 1) + \langle x^2 + 1 \rangle\\ & = 0. \end{align*} Hence, we have an isomorphism of \({\mathbb R}( \alpha )\) with \({\mathbb C}\) defined by the map that takes \(a + b \alpha\) to \(a + bi\).

Let \(E\) be a field extension of a field \(F\). If we regard \(E\) as a vector space over \(F\), then we can bring the machinery of linear algebra to bear on the problems that we will encounter in our study of fields. The elements in the field \(E\) are vectors; the elements in the field \(F\) are scalars. We can think of addition in \(E\) as adding vectors. When we multiply an element in \(E\) by an element of \(F\), we are multiplying a vector by a scalar. This view of field extensions is especially fruitful if a field extension \(E\) of \(F\) is a finite dimensional vector space over \(F\), and Theorem 21.13 states that \(E = F(\alpha )\) is finite dimensional vector space over \(F\) with basis \(\{ 1, \alpha, {\alpha}^2, \ldots, {\alpha}^{n - 1} \}\).

If an extension field \(E\) of a field \(F\) is a finite dimensional vector space over \(F\) of dimension \(n\), then we say that \(E\) is a finite extension of degree \(n\) over \(F\). We write \begin{equation*}[E:F]= n.\end{equation*} to indicate the dimension of \(E\) over \(F\).

Proof
Remark21.16

Theorem 21.15 says that every finite extension of a field \(F\) is an algebraic extension. The converse is false, however. We will leave it as an exercise to show that the set of all elements in \({\mathbb R}\) that are algebraic over \({\mathbb Q}\) forms an infinite field extension of \({\mathbb Q}\).

The next theorem is a counting theorem, similar to Lagrange's Theorem in group theory. Theorem 21.17 will prove to be an extremely useful tool in our investigation of finite field extensions.

Proof

The following corollary is easily proved using mathematical induction.

Example21.20

Let us determine an extension field of \({\mathbb Q}\) containing \(\sqrt{3} + \sqrt{5}\). It is easy to determine that the minimal polynomial of \(\sqrt{3} + \sqrt{5}\) is \(x^4 - 16 x^2 + 4\). It follows that \begin{equation*}[{\mathbb Q}( \sqrt{3} + \sqrt{5}\, ) : {\mathbb Q} ] = 4.\end{equation*} We know that \(\{ 1, \sqrt{3}\, \}\) is a basis for \({\mathbb Q}( \sqrt{3}\, )\) over \({\mathbb Q}\). Hence, \(\sqrt{3} + \sqrt{5}\) cannot be in \({\mathbb Q}( \sqrt{3}\, )\). It follows that \(\sqrt{5}\) cannot be in \({\mathbb Q}( \sqrt{3}\, )\) either. Therefore, \(\{ 1, \sqrt{5}\, \}\) is a basis for \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) = ( {\mathbb Q}(\sqrt{3}\, ))( \sqrt{5}\, )\) over \({\mathbb Q}( \sqrt{3}\, )\) and \(\{ 1, \sqrt{3}, \sqrt{5}, \sqrt{3} \sqrt{5} = \sqrt{15}\, \}\) is a basis for \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) = {\mathbb Q}( \sqrt{3} + \sqrt{5}\, )\) over \({\mathbb Q}\). This example shows that it is possible that some extension \(F( \alpha_1, \ldots, \alpha_n )\) is actually a simple extension of \(F\) even though \(n \gt 1\).

Example21.21

Let us compute a basis for \({\mathbb Q}( \sqrt[3]{5}, \sqrt{5} \, i )\), where \(\sqrt{5}\) is the positive square root of 5 and \(\sqrt[3]{5}\) is the real cube root of 5. We know that \(\sqrt{5} \, i \notin {\mathbb Q}(\sqrt[3]{5}\, )\), so \begin{equation*}[ {\mathbb Q}(\sqrt[3]{5}, \sqrt{5}\, i) : {\mathbb Q}(\sqrt[3]{5}\, )] = 2.\end{equation*} It is easy to determine that \(\{ 1, \sqrt{5}i\, \}\) is a basis for \({\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )\) over \({\mathbb Q}( \sqrt[3]{5}\, )\). We also know that \(\{ 1, \sqrt[3]{5}, (\sqrt[3]{5}\, )^2 \}\) is a basis for \({\mathbb Q}(\sqrt[3]{5}\, )\) over \({\mathbb Q}\). Hence, a basis for \({\mathbb Q}(\sqrt[3]{5}, \sqrt{5}\, i )\) over \({\mathbb Q}\) is \begin{equation*}\{ 1, \sqrt{5}\, i, \sqrt[3]{5}, (\sqrt[3]{5}\, )^2, (\sqrt[6]{5}\, )^5 i, (\sqrt[6]{5}\, )^7 i = 5 \sqrt[6]{5}\, i \text{ or } \sqrt[6]{5}\, i \}.\end{equation*} Notice that \(\sqrt[6]{5}\, i\) is a zero of \(x^6 + 5\). We can show that this polynomial is irreducible over \({\mathbb Q}\) using Eisenstein's Criterion, where we let \(p = 5\). Consequently, \begin{equation*}{\mathbb Q} \subset {\mathbb Q}( \sqrt[6]{5}\, i) \subset {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i ).\end{equation*} But it must be the case that \({\mathbb Q}( \sqrt[6]{5}\, i) = {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )\), since the degree of both of these extensions is 6.

Proof

SubsectionAlgebraic Closure

Given a field \(F\), the question arises as to whether or not we can find a field \(E\) such that every polynomial \(p(x)\) has a root in \(E\). This leads us to the following theorem.

Proof

Let \(E\) be a field extension of a field \(F\). We define the algebraic closure of a field \(F\) in \(E\) to be the field consisting of all elements in \(E\) that are algebraic over \(F\). A field \(F\) is algebraically closed if every nonconstant polynomial in \(F[x]\) has a root in \(F\).

Proof

It is a nontrivial fact that every field has a unique algebraic closure. The proof is not extremely difficult, but requires some rather sophisticated set theory. We refer the reader to [3], [4], or [8] for a proof of this result.

We now state the Fundamental Theorem of Algebra, first proven by Gauss at the age of 22 in his doctoral thesis. This theorem states that every polynomial with coefficients in the complex numbers has a root in the complex numbers. The proof of this theorem will be given in Chapter 23.