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## Section21.1Extension Fields

A field $E$ is an extension field of a field $F$ if $F$ is a subfield of $E\text{.}$ The field $F$ is called the base field. We write $F \subset E\text{.}$

###### 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 $\sqrt{2} + \sqrt{3}\text{.}$ Both $E$ and $F$ are extension fields of the rational numbers. We claim that $E$ is an extension field of $F\text{.}$ To see this, we need only show that $\sqrt{2}$ is in $E\text{.}$ Since $\sqrt{2} + \sqrt{3}$ is in $E\text{,}$ $1 / (\sqrt{2} + \sqrt{3}\,) = \sqrt{3} - \sqrt{2}$ must also be in $E\text{.}$ Taking linear combinations of $\sqrt{2} + \sqrt{3}$ and $\sqrt{3} - \sqrt{2}\text{,}$ we find that $\sqrt{2}$ and $\sqrt{3}$ must both be in $E\text{.}$

###### Example21.2.

Let $p(x) = x^2 + x + 1 \in {\mathbb Z}_2[x]\text{.}$ Since neither 0 nor 1 is a root of this polynomial, we know that $p(x)$ is irreducible over ${\mathbb Z}_2\text{.}$ We will construct a field extension of ${\mathbb Z}_2$ containing an element $\alpha$ such that $p(\alpha) = 0\text{.}$ 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 ${\mathbb Z}_2[x] / \langle p(x) \rangle\text{.}$ By the division algorithm,

\begin{equation*} f(x) = (x^2 + x + 1) q(x) + r(x)\text{,} \end{equation*}

where the degree of $r(x)$ is less than the degree of $x^2 + x + 1\text{.}$ Therefore,

\begin{equation*} f(x) + \langle x^2 + x + 1 \rangle = r(x) + \langle x^2 + x + 1 \rangle\text{.} \end{equation*}

The only possibilities for $r(x)$ are then $0\text{,}$ $1\text{,}$ $x\text{,}$ and $1 + x\text{.}$ 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 ${\mathbb Z}_2\text{,}$ containing a zero $\alpha$ of $p(x)\text{.}$ 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\text{.} \end{align*}

Notice that ${\alpha}^2 + {\alpha} + 1 = 0\text{;}$ hence, if we compute $(1 + \alpha)^2\text{,}$

\begin{equation*} (1 + \alpha)(1 + \alpha)= 1 + \alpha + \alpha + (\alpha)^2 = \alpha\text{.} \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 $E\text{.}$

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.

To prove this theorem, we will employ the method that we used to construct Example 21.2. Clearly, we can assume that $p(x)$ is an irreducible polynomial. We wish to find an extension field $E$ of $F$ containing an element $\alpha$ such that $p(\alpha) = 0\text{.}$ The ideal $\langle p(x) \rangle$ generated by $p(x)$ is a maximal ideal in $F[x]$ by Theorem 17.22; hence, $F[x]/\langle p(x) \rangle$ is a field. We claim that $E = F[x]/\langle p(x) \rangle$ is the desired field.

We first show that $E$ is a field extension of $F\text{.}$ We can define a homomorphism of commutative rings by the map $\psi:F \rightarrow F[x]/\langle p(x) \rangle\text{,}$ where $\psi(a) = a + \langle p(x)\rangle$ for $a \in F\text{.}$ It is easy to check that $\psi$ is indeed a ring homomorphism. Observe that

\begin{equation*} \psi( a ) + \psi( b ) = (a + \langle p(x) \rangle) + (b + \langle p(x) \rangle) = (a + b) + \langle p(x) \rangle = \psi( a + b ) \end{equation*}

and

\begin{equation*} \psi( a ) \psi( b ) = (a + \langle p(x) \rangle) (b + \langle p(x) \rangle) = ab + \langle p(x) \rangle = \psi( ab )\text{.} \end{equation*}

To prove that $\psi$ is one-to-one, assume that

\begin{equation*} a + \langle p(x) \rangle = \psi(a) = \psi(b) = b + \langle p(x) \rangle\text{.} \end{equation*}

Then $a - b$ is a multiple of $p(x)\text{,}$ since it lives in the ideal $\langle p(x) \rangle\text{.}$ Since $p(x)$ is a nonconstant polynomial, the only possibility is that $a - b = 0\text{.}$ Consequently, $a = b$ and $\psi$ is injective. Since $\psi$ is one-to-one, we can identify $F$ with the subfield $\{ a + \langle p(x) \rangle : a \in F \}$ of $E$ and view $E$ as an extension field of $F\text{.}$

It remains for us to prove that $p(x)$ has a zero $\alpha \in E\text{.}$ Set $\alpha = x + \langle p(x) \rangle\text{.}$ Then $\alpha$ is in $E\text{.}$ If $p(x) = a_0 + a_1 x + \cdots + a_n x^n\text{,}$ then

\begin{align*} p( \alpha ) & = a_0 + a_1( x + \langle p(x) \rangle) + \cdots + a_n ( x + \langle p(x) \rangle)^n\\ & = a_0 + ( a_1 x + \langle p(x) \rangle) + \cdots + (a_n x^n + \langle p(x) \rangle)\\ & = a_0 + a_1 x + \cdots + a_n x^n + \langle p(x) \rangle\\ & = 0 + \langle p(x) \rangle\text{.} \end{align*}

Therefore, we have found an element $\alpha \in E = F[x]/\langle p(x) \rangle$ such that $\alpha$ is a zero of $p(x)\text{.}$

###### Example21.6.

Let $p(x) = x^5 + x^4 + 1 \in {\mathbb Z}_2[x]\text{.}$ Then $p(x)$ has irreducible factors $x^2 + x + 1$ and $x^3 + x + 1\text{.}$ For a field extension $E$ of ${\mathbb Z}_2$ such that $p(x)$ has a root in $E\text{,}$ we can let $E$ be either ${\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle$ or ${\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle\text{.}$ 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 $f(x) \in F[x]\text{.}$ An element in $E$ that is not algebraic over $F$ is transcendental over $F\text{.}$ An extension field $E$ of a field $F$ is an algebraic extension of $F$ if every element in $E$ is algebraic over $F\text{.}$ If $E$ is a field extension of $F$ and $\alpha_1, \ldots, \alpha_n$ are contained in $E\text{,}$ we denote the smallest field containing $F$ and $\alpha_1, \ldots, \alpha_n$ by $F( \alpha_1, \ldots, \alpha_n)\text{.}$ If $E = F( \alpha )$ for some $\alpha \in E\text{,}$ then $E$ is a simple extension of $F\text{.}$

###### Example21.7.

Both $\sqrt{2}$ and $i$ are algebraic over ${\mathbb Q}$ since they are zeros of the polynomials $x^2 -2$ and $x^2 + 1\text{,}$ respectively. Clearly $\pi$ and $e$ are algebraic over the real numbers; however, it is a nontrivial fact that they are transcendental over ${\mathbb Q}\text{.}$ Numbers in ${\mathbb R}$ that are algebraic over ${\mathbb Q}$ are in fact quite rare. Almost all real numbers are transcendental over ${\mathbb Q}\text{.}$ 7 (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.)

The probability that a real number chosen at random from the interval $[0, 1]$ will be transcendental over the rational numbers is one.

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 ${\mathbb Q}\text{.}$

###### Example21.8.

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

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

Let $\phi_{\alpha} : F[x] \rightarrow E$ be the evaluation homomorphism for $\alpha\text{.}$ Then $\alpha$ is transcendental over $F$ if and only if $\phi_{\alpha} (p(x)) = p(\alpha) \neq 0$ for all nonconstant polynomials $p(x) \in F[x]\text{.}$ This is true if and only if $\ker \phi_{\alpha} = \{ 0 \}\text{;}$ that is, it is true exactly when $\phi_{\alpha}$ is one-to-one. Hence, $E$ must contain a copy of $F[x]\text{.}$ The smallest field containing $F[x]$ is the field of fractions $F(x)\text{.}$ By Theorem 18.4, $E$ must contain a copy of this field.

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

Let $\phi_{\alpha} : F[x] \rightarrow E$ be the evaluation homomorphism. The kernel of $\phi_{\alpha}$ is a principal ideal generated by some $p(x) \in F[x]$ with $\deg p(x) \geq 1\text{.}$ We know that such a polynomial exists, since $F[x]$ is a principal ideal domain and $\alpha$ is algebraic. The ideal $\langle p(x) \rangle$ consists exactly of those elements of $F[x]$ having $\alpha$ as a zero. If $f( \alpha ) = 0$ and $f(x)$ is not the zero polynomial, then $f(x) \in \langle p(x) \rangle$ and $p(x)$ divides $f(x)\text{.}$ So $p(x)$ is a polynomial of minimal degree having $\alpha$ as a zero. Any other polynomial of the same degree having $\alpha$ as a zero must have the form $\beta p( x)$ for some $\beta \in F\text{.}$

Suppose now that $p(x) = r(x) s(x)$ is a factorization of $p(x)$ into polynomials of lower degree. Since $p( \alpha ) = 0\text{,}$ $r( \alpha ) s( \alpha ) = 0\text{;}$ consequently, either $r( \alpha )=0$ or $s( \alpha ) = 0\text{,}$ which contradicts the fact that $p$ is of minimal degree. Therefore, $p(x)$ must be irreducible.

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

###### Example21.11.

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

Let $\phi_{\alpha} : F[x] \rightarrow E$ be the evaluation homomorphism. The kernel of this map is $\langle p(x) \rangle\text{,}$ where $p(x)$ is the minimal polynomial of $\alpha\text{.}$ By the First Isomorphism Theorem for rings, the image of $\phi_{\alpha}$ in $E$ is isomorphic to $F( \alpha )$ since it contains both $F$ and $\alpha\text{.}$

Since $\phi_{\alpha} ( F[x] ) \cong F( \alpha )\text{,}$ every element in $E = F( \alpha )$ must be of the form $\phi_{\alpha} ( f(x) ) = f( \alpha )\text{,}$ where $f(\alpha)$ is a polynomial in $\alpha$ with coefficients in $F\text{.}$ Let

\begin{equation*} p(x) = x^n + a_{n - 1} x^{n - 1} + \cdots + a_0 \end{equation*}

be the minimal polynomial of $\alpha\text{.}$ Then $p( \alpha ) = 0\text{;}$ hence,

\begin{equation*} {\alpha}^n = - a_{n - 1} {\alpha}^{n - 1} - \cdots - a_0\text{.} \end{equation*}

Similarly,

\begin{align*} {\alpha}^{n + 1} & = {\alpha} {\alpha}^n\\ & = - a_{n - 1} {\alpha}^n - a_{n - 2} {\alpha}^{n - 1} - \cdots - a_0 {\alpha}\\ & = - a_{n - 1}( - a_{n - 1} {\alpha}^{n - 1} - \cdots - a_0) - a_{n - 2} {\alpha}^{n - 1} - \cdots - a_0 {\alpha}\text{.} \end{align*}

Continuing in this manner, we can express every monomial ${\alpha}^m\text{,}$ $m \geq n\text{,}$ as a linear combination of powers of ${\alpha}$ that are less than $n\text{.}$ Hence, any $\beta \in F( \alpha )$ can be written as

\begin{equation*} \beta = b_0 + b_1 \alpha + \cdots + b_{n - 1} \alpha^{n - 1}\text{.} \end{equation*}

To show uniqueness, suppose that

\begin{equation*} \beta = b_0 + b_1 \alpha + \cdots + b_{n-1} \alpha^{n-1} = c_0 + c_1 \alpha + \cdots + c_{n - 1} \alpha^{n - 1} \end{equation*}

for $b_i$ and $c_i$ in $F\text{.}$ Then

\begin{equation*} g(x) = (b_0 - c_0) + (b_1 - c_1) x + \cdots + (b_{n - 1} - c_{n - 1})x^{n - 1} \end{equation*}

is in $F[x]$ and $g( \alpha ) = 0\text{.}$ Since the degree of $g(x)$ is less than the degree of $p( x )\text{,}$ the irreducible polynomial of $\alpha\text{,}$ $g(x)$ must be the zero polynomial. Consequently,

\begin{equation*} b_0 - c_0 = b_1 - c_1 = \cdots = b_{n - 1} - c_{n - 1} = 0\text{,} \end{equation*}

or $b_i = c_i$ for $i = 0, 1, \ldots, n-1\text{.}$ Therefore, we have shown uniqueness.

###### Example21.14.

Since $x^2 + 1$ is irreducible over ${\mathbb R}\text{,}$ $\langle x^2 + 1 \rangle$ is a maximal ideal in ${\mathbb R}[x]\text{.}$ So $E = {\mathbb R}[x]/\langle x^2 + 1 \rangle$ is a field extension of ${\mathbb R}$ that contains a root of $x^2 + 1\text{.}$ Let $\alpha = x + \langle x^2 + 1 \rangle\text{.}$ We can identify $E$ with the complex numbers. By Proposition 21.12, $E$ is isomorphic to ${\mathbb R}( \alpha ) = \{ a + b \alpha : a, b \in {\mathbb R} \}\text{.}$ We know that $\alpha^2 = -1$ in $E\text{,}$ 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\text{.} \end{align*}

Hence, we have an isomorphism of ${\mathbb R}( \alpha )$ with ${\mathbb C}$ defined by the map that takes $a + b \alpha$ to $a + bi\text{.}$

Let $E$ be a field extension of a field $F\text{.}$ If we regard $E$ as a vector space over $F\text{,}$ 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 $F\text{,}$ 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 $F\text{,}$ and Theorem 21.13 states that $E = F(\alpha )$ is finite dimensional vector space over $F$ with basis $\{ 1, \alpha, {\alpha}^2, \ldots, {\alpha}^{n - 1} \}\text{.}$

If an extension field $E$ of a field $F$ is a finite dimensional vector space over $F$ of dimension $n\text{,}$ then we say that $E$ is a finite extension of degree $n$ over $F$. We write

\begin{equation*} [E:F]= n\text{.} \end{equation*}

to indicate the dimension of $E$ over $F\text{.}$

Let $\alpha \in E\text{.}$ Since $[E:F] = n\text{,}$ the elements

\begin{equation*} 1, \alpha, \ldots, {\alpha}^n \end{equation*}

cannot be linearly independent. Hence, there exist $a_i \in F\text{,}$ not all zero, such that

\begin{equation*} a_n {\alpha}^n + a_{n - 1} {\alpha}^{n - 1} + \cdots + a_1 \alpha + a_0 = 0\text{.} \end{equation*}

Therefore,

\begin{equation*} p(x) = a_n x^n + \cdots + a_0 \in F[x] \end{equation*}

is a nonzero polynomial with $p( \alpha ) = 0\text{.}$

###### 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 ${\mathbb Q}\text{.}$

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.

Let $\{ \alpha_1, \ldots, \alpha_n \}$ be a basis for $E$ as a vector space over $F$ and $\{ \beta_1, \ldots, \beta_m \}$ be a basis for $K$ as a vector space over $E\text{.}$ We claim that $\{ \alpha_i \beta_j \}$ is a basis for $K$ over $F\text{.}$ We will first show that these vectors span $K\text{.}$ Let $u \in K\text{.}$ Then $u = \sum_{j = 1}^{m} b_j \beta_j$ and $b_j = \sum_{i = 1}^{n} a_{ij} \alpha_i\text{,}$ where $b_j \in E$ and $a_{ij} \in F\text{.}$ Then

\begin{equation*} u = \sum_{j = 1}^{m} \left( \sum_{i = 1}^{n} a_{ij} \alpha_i \right) \beta_j = \sum_{i,j} a_{ij} ( \alpha_i \beta_j )\text{.} \end{equation*}

So the $mn$ vectors $\alpha_i \beta_j$ must span $K$ over $F\text{.}$

We must show that $\{ \alpha_i \beta_j \}$ are linearly independent. Recall that a set of vectors $v_1, v_2, \ldots, v_n$ in a vector space $V$ are linearly independent if

\begin{equation*} c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0 \end{equation*}

implies that

\begin{equation*} c_1 = c_2 = \cdots = c_n = 0\text{.} \end{equation*}

Let

\begin{equation*} u = \sum_{i,j} c_{ij} ( \alpha_i \beta_j ) = 0 \end{equation*}

for $c_{ij} \in F\text{.}$ We need to prove that all of the $c_{ij}$'s are zero. We can rewrite $u$ as

\begin{equation*} \sum_{j = 1}^{m} \left( \sum_{i = 1}^{n} c_{ij} \alpha_i \right) \beta_j = 0\text{,} \end{equation*}

where $\sum_i c_{ij} \alpha_i \in E\text{.}$ Since the $\beta_j$'s are linearly independent over $E\text{,}$ it must be the case that

\begin{equation*} \sum_{i = 1}^n c_{ij} \alpha_i = 0 \end{equation*}

for all $j\text{.}$ However, the $\alpha_j$ are also linearly independent over $F\text{.}$ Therefore, $c_{ij} = 0$ for all $i$ and $j\text{,}$ which completes the proof.

The following corollary is easily proved using mathematical induction.

We know that $\deg p(x) = [F( \alpha ) : F ]$ and $\deg q(x) = [F( \beta ) : F ]\text{.}$ Since $F \subset F( \beta ) \subset F( \alpha )\text{,}$

\begin{equation*} [F( \alpha ) : F ]= [ F( \alpha ) : F( \beta ) ] [ F( \beta ) : F ]\text{.} \end{equation*}
###### Example21.20.

Let us determine an extension field of ${\mathbb Q}$ containing $\sqrt{3} + \sqrt{5}\text{.}$ It is easy to determine that the minimal polynomial of $\sqrt{3} + \sqrt{5}$ is $x^4 - 16 x^2 + 4\text{.}$ It follows that

\begin{equation*} [{\mathbb Q}( \sqrt{3} + \sqrt{5}\, ) : {\mathbb Q} ] = 4\text{.} \end{equation*}

We know that $\{ 1, \sqrt{3}\, \}$ is a basis for ${\mathbb Q}( \sqrt{3}\, )$ over ${\mathbb Q}\text{.}$ Hence, $\sqrt{3} + \sqrt{5}$ cannot be in ${\mathbb Q}( \sqrt{3}\, )\text{.}$ 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 ${\mathbb Q}\text{.}$ 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 $n \gt 1\text{.}$

###### Example21.21.

Let us compute a basis for ${\mathbb Q}( \sqrt{5}, \sqrt{5} \, i )\text{,}$ where $\sqrt{5}$ is the positive square root of $5$ and $\sqrt{5}$ is the real cube root of $5\text{.}$ We know that $\sqrt{5} \, i \notin {\mathbb Q}(\sqrt{5}\, )\text{,}$ so

\begin{equation*} [ {\mathbb Q}(\sqrt{5}, \sqrt{5}\, i) : {\mathbb Q}(\sqrt{5}\, )] = 2\text{.} \end{equation*}

It is easy to determine that $\{ 1, \sqrt{5}i\, \}$ is a basis for ${\mathbb Q}( \sqrt{5}, \sqrt{5}\, i )$ over ${\mathbb Q}( \sqrt{5}\, )\text{.}$ We also know that $\{ 1, \sqrt{5}, (\sqrt{5}\, )^2 \}$ is a basis for ${\mathbb Q}(\sqrt{5}\, )$ over ${\mathbb Q}\text{.}$ Hence, a basis for ${\mathbb Q}(\sqrt{5}, \sqrt{5}\, i )$ over ${\mathbb Q}$ is

\begin{equation*} \{ 1, \sqrt{5}\, i, \sqrt{5}, (\sqrt{5}\, )^2, (\sqrt{5}\, )^5 i, (\sqrt{5}\, )^7 i = 5 \sqrt{5}\, i \text{ or } \sqrt{5}\, i \}\text{.} \end{equation*}

Notice that $\sqrt{5}\, i$ is a zero of $x^6 + 5\text{.}$ We can show that this polynomial is irreducible over ${\mathbb Q}$ using Eisenstein's Criterion, where we let $p = 5\text{.}$ Consequently,

\begin{equation*} {\mathbb Q} \subset {\mathbb Q}( \sqrt{5}\, i) \subset {\mathbb Q}( \sqrt{5}, \sqrt{5}\, i )\text{.} \end{equation*}

But it must be the case that ${\mathbb Q}( \sqrt{5}\, i) = {\mathbb Q}( \sqrt{5}, \sqrt{5}\, i )\text{,}$ since the degree of both of these extensions is $6\text{.}$

(1) $\Rightarrow$ (2). Let $E$ be a finite algebraic extension of $F\text{.}$ Then $E$ is a finite dimensional vector space over $F$ and there exists a basis consisting of elements $\alpha_1, \ldots, \alpha_n$ in $E$ such that $E = F(\alpha_1, \ldots, \alpha_n)\text{.}$ Each $\alpha_i$ is algebraic over $F$ by Theorem 21.15.

(2) $\Rightarrow$ (3). Suppose that $E = F(\alpha_1, \ldots, \alpha_n)\text{,}$ where every $\alpha_i$ is algebraic over $F\text{.}$ Then

\begin{equation*} E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n - 1} ) \supset \cdots \supset F( \alpha_1 ) \supset F\text{,} \end{equation*}

where each field $F(\alpha_1, \ldots, \alpha_i)$ is algebraic over $F(\alpha_1, \ldots, \alpha_{i - 1})\text{.}$

(3) $\Rightarrow$ (1). Let

\begin{equation*} E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n - 1} ) \supset \cdots \supset F( \alpha_1 ) \supset F\text{,} \end{equation*}

where each field $F(\alpha_1, \ldots, \alpha_i)$ is algebraic over $F(\alpha_1, \ldots, \alpha_{i - 1})\text{.}$ Since

\begin{equation*} F(\alpha_1, \ldots, \alpha_i) = F(\alpha_1, \ldots, \alpha_{i - 1} )(\alpha_i) \end{equation*}

is simple extension and $\alpha_i$ is algebraic over $F(\alpha_1, \ldots, \alpha_{i - 1})\text{,}$ it follows that

\begin{equation*} [ F(\alpha_1, \ldots, \alpha_i) : F(\alpha_1, \ldots, \alpha_{i - 1} )] \end{equation*}

is finite for each $i\text{.}$ Therefore, $[E : F]$ is finite.

### SubsectionAlgebraic Closure

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

Let $\alpha, \beta \in E$ be algebraic over $F\text{.}$ Then $F( \alpha, \beta )$ is a finite extension of $F\text{.}$ Since every element of $F( \alpha, \beta )$ is algebraic over $F\text{,}$ $\alpha \pm \beta\text{,}$ $\alpha \beta\text{,}$ and $\alpha / \beta$ ($\beta \neq 0$) are all algebraic over $F\text{.}$ Consequently, the set of elements in $E$ that are algebraic over $F$ form a field.

Let $E$ be a field extension of a field $F\text{.}$ 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 $F\text{.}$ A field $F$ is algebraically closed if every nonconstant polynomial in $F[x]$ has a root in $F\text{.}$

Let $F$ be an algebraically closed field. If $p(x) \in F[x]$ is a nonconstant polynomial, then $p(x)$ has a zero in $F\text{,}$ say $\alpha\text{.}$ Therefore, $x-\alpha$ must be a factor of $p(x)$ and so $p(x) = (x - \alpha) q_1(x)\text{,}$ where $\deg q_1(x) = \deg p(x) - 1\text{.}$ Continue this process with $q_1(x)$ to find a factorization

\begin{equation*} p(x) = (x - \alpha)(x - \beta)q_2(x)\text{,} \end{equation*}

where $\deg q_2(x) = \deg p(x) -2\text{.}$ The process must eventually stop since the degree of $p(x)$ is finite.

Conversely, suppose that every nonconstant polynomial $p(x)$ in $F[x]$ factors into linear factors. Let $ax - b$ be such a factor. Then $p( b/a ) = 0\text{.}$ Consequently, $F$ is algebraically closed.

Let $E$ be an algebraic extension of $F\text{;}$ then $F \subset E\text{.}$ For $\alpha \in E\text{,}$ the minimal polynomial of $\alpha$ is $x - \alpha\text{.}$ Therefore, $\alpha \in F$ and $F = E\text{.}$

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 , , or  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.