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 $E$.

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.

##### 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.

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.

##### 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 $F$.

##### 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

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.
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 $F$.