Example21.1

For example, let

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,

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

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

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

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 $E\text{.}$

Theorem21.5

Let $F$ be a field and let $p(x)$ be a nonconstant polynomial in $F[x]\text{.}$ Then there exists an extension field $E$ of $F$ and an element $\alpha \in E$ such that $p(\alpha) = 0\text{.}$

Proof

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

and

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

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

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.

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 ${\mathbb Q}\text{.}$⁶The probability that a real number chosen at random from the interval $[0, 1]$ will be transcendental over the rational numbers is one. (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.)

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 $x^4 - 4 x^2 + 1 \in {\mathbb Q}[x]\text{.}$

Theorem21.9

Let $E$ be an extension field of $F$ and $\alpha \in E\text{.}$ Then $\alpha$ is transcendental over $F$ if and only if $F( \alpha )$ is isomorphic to $F(x)\text{,}$ the field of fractions of $F[x]\text{.}$

Proof

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.

Theorem21.10

Let $E$ be an extension field of a field $F$ and $\alpha \in E$ with $\alpha$ algebraic over $F\text{.}$ Then there is a unique irreducible monic polynomial $p(x) \in F[x]$ of smallest degree such that $p( \alpha ) = 0\text{.}$ If $f(x)$ is another polynomial in $F[x]$ such that $f(\alpha) = 0\text{,}$ then $p(x)$ divides $f(x)\text{.}$

Proof

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.

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.

Proposition21.12

Let $E$ be a field extension of $F$ and $\alpha \in E$ be algebraic over $F\text{.}$ Then $F( \alpha ) \cong F[x] / \langle p(x) \rangle\text{,}$ where $p(x)$ is the minimal polynomial of $\alpha$ over $F\text{.}$

Proof

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 $\alpha\text{.}$

Theorem21.13

Let $E = F( \alpha )$ be a simple extension of $F\text{,}$ where $\alpha \in E$ is algebraic over $F\text{.}$ Suppose that the degree of $\alpha$ over $F$ is $n\text{.}$ Then every element $\beta \in E$ can be expressed uniquely in the form

for $b_i \in F\text{.}$

Proof

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

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

Similarly,

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

To show uniqueness, suppose that

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

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,

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

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

Theorem21.15

Every finite extension field $E$ of a field $F$ is an algebraic extension.

Proof

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

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

Therefore,

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

Theorem21.17

If $E$ is a finite extension of $F$ and $K$ is a finite extension of $E\text{,}$ then $K$ is a finite extension of $F$ and

Proof

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

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

implies that

Let

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

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

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.

Corollary21.18

If $F_i$ is a field for $i = 1, \dots, k$ and $F_{i+1}$ is a finite extension of $F_i\text{,}$ then $F_k$ is a finite extension of $F_1$ and

Corollary21.19

Let $E$ be an extension field of $F\text{.}$ If $\alpha \in E$ is algebraic over $F$ with minimal polynomial $p(x)$ and $\beta \in F( \alpha )$ with minimal polynomial $q(x)\text{,}$ then $\deg q(x)$ divides $\deg p(x)\text{.}$

Proof

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

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

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 ${\mathbb Q}( \sqrt[3]{5}, \sqrt{5} \, i )\text{,}$ where $\sqrt{5}$ is the positive square root of 5 and $\sqrt[3]{5}$ is the real cube root of 5. We know that $\sqrt{5} \, i \notin {\mathbb Q}(\sqrt[3]{5}\, )\text{,}$ so

It is easy to determine that $\{ 1, \sqrt{5}i\, \}$ is a basis for ${\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )$ over ${\mathbb Q}( \sqrt[3]{5}\, )\text{.}$ We also know that $\{ 1, \sqrt[3]{5}, (\sqrt[3]{5}\, )^2 \}$ is a basis for ${\mathbb Q}(\sqrt[3]{5}\, )$ over ${\mathbb Q}\text{.}$ Hence, a basis for ${\mathbb Q}(\sqrt[3]{5}, \sqrt{5}\, i )$ over ${\mathbb Q}$ is

Notice that $\sqrt[6]{5}\, i$ 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,

But it must be the case that ${\mathbb Q}( \sqrt[6]{5}\, i) = {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )\text{,}$ since the degree of both of these extensions is 6.

Theorem21.22

Let $E$ be a field extension of $F\text{.}$ Then the following statements are equivalent.

1. $E$ is a finite extension of $F\text{.}$

2. There exists a finite number of algebraic elements $\alpha_1, \ldots, \alpha_n \in E$ such that $E = F(\alpha_1, \ldots, \alpha_n)\text{.}$

3. There exists a sequence of fields

   \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, \end{equation*}

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

Proof

(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

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

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

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

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

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

Theorem21.23

Let $E$ be an extension field of $F\text{.}$ The set of elements in $E$ that are algebraic over $F$ form a field.

Proof

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.

Corollary21.24

The set of all algebraic numbers forms a field; that is, the set of all complex numbers that are algebraic over ${\mathbb Q}$ makes up a field.

Theorem21.25

A field $F$ is algebraically closed if and only if every nonconstant polynomial in $F[x]$ factors into linear factors over $F[x]\text{.}$

Proof

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

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.

Corollary21.26

An algebraically closed field $F$ has no proper algebraic extension $E\text{.}$

Proof

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 $F = E\text{.}$

Theorem21.27

Every field $F$ has a unique algebraic closure.

Theorem21.28Fundamental Theorem of Algebra

The field of complex numbers is algebraically closed.