## Section8.6Extension Fields as Vector Spaces

### Subsection8.6.1Back to Extension Fields

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 8.18 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 $F\text{.}$

###### Proof.

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 $p( \alpha ) = 0\text{.}$

###### Remark8.36.

Theorem 8.35 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 8.37 will prove to be an extremely useful tool in our investigation of finite field extensions.

###### 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

\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.

###### Proof.

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

\begin{equation*} [F( \alpha ) : F ]= [ F( \alpha ) : F( \beta ) ] [ F( \beta ) : F ]\text{.} \end{equation*}
###### Example8.40.

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 $n \gt 1\text{.}$

###### Example8.41.

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 $5\text{.}$ We know that $\sqrt{5} \, i \notin {\mathbb Q}(\sqrt[3]{5}\, )\text{,}$ so

\begin{equation*} [ {\mathbb Q}(\sqrt[3]{5}, \sqrt{5}\, i) : {\mathbb Q}(\sqrt[3]{5}\, )] = 2\text{.} \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 ${\mathbb Q}( \sqrt[3]{5}\, )\text{.}$ 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

\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 \}\text{.} \end{equation*}

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,

\begin{equation*} {\mathbb Q} \subset {\mathbb Q}( \sqrt[6]{5}\, i) \subset {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )\text{.} \end{equation*}

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 $6\text{.}$

###### 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 8.35.

(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.

### Worksheet8.6.2Activity: Dimension and Degree

Recall that a basis for a vector space is a linearly independent spanning set, and that the dimension of a vector space is the size of a (any) basis for the space.

If $K$ is an extension field of $F\text{,}$ we can view $K$ as a vector space over the field of scalars $F\text{.}$ In this case, we say the degree of $K$ over $F\text{,}$ written $[K:F]$ is the dimension of this vector space.

###### 1.

Find a basis for $\Q(\sqrt{7})$ over $\Q\text{.}$ What is $[\Q(\sqrt{7}):\Q]\text{?}$

Solution

One possible basis is $\{1, \sqrt{7}\}\text{,}$ making $[\Q(\sqrt{7}):\Q] = 2$

###### 2.

Find a basis for $\Q(\sqrt[3]{5})$ over $\Q\text{.}$ What is $[\Q(\sqrt[3]{5}):\Q]\text{?}$

Solution

One possible basis is $\{1, \sqrt[3]{5}, \sqrt[3]{5}^2\}\text{.}$ Thus $[\Q(\sqrt[3]{5}):\Q] = 3\text{.}$

###### 3.

Suppose $\alpha$ is a root of $p(x) = x^5 -6x^4 + 9x^2 + 3\text{.}$ Find a basis for $\Q(\alpha)$ over $\Q\text{.}$ What is $[\Q(\alpha):\Q]\text{.}$

Solution

The polynomial $p(x)$ is irreducible by Eisenstein's criterion. Therefore, it is the minimal polynomial for any of its roots, including $\alpha\text{.}$ We then know that $\{1, \alpha, \alpha^2, \alpha^3, \alpha^4\}$ will be a basis for $\Q(\alpha)\text{,}$ making $[\Q(\alpha):\Q] = 5\text{.}$

###### 4.

What is the general rule here? Some things to think about: If you claim that you can always find a basis in some systematic way, how do you know it is really a basis? How do you know the basis is linearly independent? How do you know it spans?

Solution

We claim that $[\Q(c):\Q]$ is equal to the degree of the minimal polynomial for $c$ (so the degree of the extension is the degree of the polynomial). In fact, we know that $\{1, c, c^2, \ldots, c^{n-1}\}$ will always be a basis (where $n$ is the degree).

Why is this? We know that $\Q(c)$ must contain all powers of $c\text{.}$ However, $c^m$ for $m\ge n$ can always be expressed as a linear combination of $c^i$ for $i \lt n\text{,}$ using the minimal polynomial for $c\text{.}$ This proves that $\{1, c, c^2, \ldots, c^{n-1}\}$ spans $\Q(c)\text{.}$ To see why this set is linearly independent, note that if it were not, then there would be a linear combination of powers of $c$ that was zero. But that would give a polynomial equal to zero at $c\text{,}$ making $c$ a root of a polynomial of degree less than $n\text{.}$

###### 5.

The polynomial $q(x) = x^5 -7x^3 - 5x^2 + 35$ has $\sqrt{7}$ and $\sqrt[3]{5}$ as roots. Does this mean $[\Q(\sqrt{7}):\Q] = [\Q(\sqrt[3]{5}):\Q] = 5\text{?}$ Why not?

Solution

Clearly this cannot be true, since we already said that these degrees were $2$ and $3$ respectively. They key thing that made our argument work above was that the polynomial was a minimal polynomial.

To further see that this is silly, what would the basis for $\Q(\sqrt{7})$ be if it had size 7? Would it be $\{1, \sqrt{7}, \sqrt{7}^2,\ldots, \sqrt{7}^4\}\text{?}$ That is clearly linearly dependent, since $\sqrt{7}^2$ is a multiple of 1.

We now have a fairly good idea how to work with $\Q(\alpha)\text{.}$ What if we consider $\Q(\sqrt[3]{5}, \sqrt{7})\text{,}$ the smallest field containing $\Q\text{,}$ $\sqrt{7}\text{,}$ and also $\sqrt[3]{5}\text{?}$

###### 6.

We can think of this as an extension of an extension. Take $\Q(\sqrt[3]{5})$ as our base field. Adjoin to that $\sqrt{7}$ to get $\Q(\sqrt[3]{5}, \sqrt{7})\text{.}$ What is $[\Q(\sqrt[3]{5}, \sqrt{7}):\Q(\sqrt[3]{5})]\text{?}$ Use the general rule we discovered above and also find a basis

Solution

Over $\Q(\sqrt[3]{5})\text{,}$ the minimal polynomial for $\sqrt{7}$ is still $x^2 - 7\text{.}$ For what else could it be? The only way that there could be a smaller degree polynomial would be for there to be a degree 1 polynomial, but in that case, we would have $\sqrt{7} \in \Q(\sqrt[3]{5})\text{.}$ It is not obvious, but this is not the case.

Given that, we know that the $[\Q(\sqrt[3]{5}, \sqrt{7}):\Q(\sqrt[3]{5})] = 2\text{.}$

A basis will be $\{1, \sqrt{7}\}\text{.}$

###### 7.

Using the basis above and the basis for $\Q(\sqrt[3]{5})$ over $\Q\text{,}$ find a basis for $\Q(\sqrt[3]{5}, \sqrt{7})$ over $\Q\text{.}$

Solution

A basis for $\Q(\sqrt[3]{5})$ over $\Q$ is $\{1, \sqrt[3]{5}, \sqrt[3]{5}^2\}\text{.}$ So every element of $\Q(\sqrt[3]{5})$ can be written in the form $a+b\sqrt[3]{5} + c \sqrt[3]{5}^2\text{.}$ These are the coefficients we can multiply by basis elements $\{1, \sqrt{7}\}$ for $\Q(\sqrt[3]{5},\sqrt{7})$ over $\Q(\sqrt[3]{5})\text{.}$ So a general element in $\Q(\sqrt[3]{5}, \sqrt{7})$ has the form

\begin{equation*} a+b\sqrt[3]{5} + c\sqrt[3]{5}^2 + a'\sqrt{7} + b'\sqrt[3]{5}\sqrt{7} + c'\sqrt[3]{5}^2\sqrt{7}. \end{equation*}

A basis is thus

\begin{equation*} \{1, \sqrt[3]{5}, \sqrt[3]{5}^2, \sqrt{7}, \sqrt[3]{5}\sqrt{7}, \sqrt[3]{5}^2\sqrt{7} \}. \end{equation*}
###### 8.

What is $[\Q(\sqrt[3]{5}, \sqrt{7}):\Q]\text{?}$ What is the general rule for degrees of extensions of extensions?

Solution

From looking at the basis above, we see that the degree over $\Q$ is 6, which is of course, $2 \cdot 3\text{.}$ It certainly appears this will always work. In other words, if $K$ is an extension of $F\text{,}$ which is an extension of $E\text{,}$ we have

\begin{equation*} [K:E] = [K:F]\cdot [F:E]. \end{equation*}
###### 9.

What if we started with $\Q(\sqrt{7})$ and then adjoined $\sqrt[3]{5}\text{?}$ Repeat the analysis you did above to make sure we get the same results about degree and basis.

Solution

We do indeed get the same thing. The minimal polynomial for $\sqrt[3]{5}$ is $x^3 - 5\text{,}$ even over the extension field $\Q(\sqrt{7})\text{.}$ Thus $[\Q(\sqrt{7},\sqrt[3]{5}):\Q(\sqrt{7})] = 3\text{.}$ A basis is $\{1, \sqrt[3]{5}, \sqrt[3]{5}^2\}\text{.}$

If we combine this with the basis for $\Q(\sqrt{7})$ over $\Q\text{,}$ we get a basis for $\Q(\sqrt[3]{5}, \sqrt{7})$ over $\Q\text{:}$ $\{1, \sqrt{7}, \sqrt[3]{5}, \sqrt{7}\sqrt[3]{5}, \sqrt[3]{5}^2, \sqrt{7}\sqrt[3]{5}^2\}\text{,}$ containing 6 elements, as expected.

###### 10.

What is $[\Q(\sqrt{2}, \sqrt[4]{2}):\Q]\text{?}$

Solution

At first, this might look like a counterexample to the “tower rule” we gave above. The minimum polynomial for $\sqrt[4]{2}$ over $\Q$ is $x^4 - 2\text{,}$ which has degree 4. The minimum polynomial for $\sqrt{2}$ is $x^2 - 2\text{.}$ However, $[\Q(\sqrt{2}, \sqrt[4]{2}):\Q] = 4\text{,}$ not 8. This can be seen easily by noticing that $\Q(\sqrt{2}, \sqrt[4]{2}) = \Q(\sqrt[4]{2})\text{,}$ as this already contains $\sqrt{2}\text{.}$

The point is that the tower rule is correct: you do multiply the degrees of each extension. But these degrees are not always the degree of the minimal polynomials over the base field. Each time you go up one floor in the tower, you must again ask what the minimum polynomial for the next element you adjoin is.

### Subsection8.6.3Algebraic 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.

###### 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.

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 $F\text{.}$

###### 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

\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.

###### 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 $F = E\text{.}$

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 9.

### Exercises8.6.4Collected Homework

All the questions below will consider the number $a = \sqrt{3+\sqrt[3]{2}}\text{.}$

###### 1.

Verify that $a = \sqrt{3 + \sqrt[3]{2}}$ is algebraic. Then explain why you don't yet know whether the polynomial you found was the minimum polynomial.

###### 2.

Write down a basis for $\Q(\sqrt[3]{2})$ and say what $[\Q(\sqrt[3]{2}):\Q]$ is, and why this makes sense. Then find $[\Q(a):\Q(\sqrt[3]{2})]$ and a basis for $\Q(a)$ over $\Q(\sqrt[3]{2})\text{.}$

###### 3.

Use the previous question to find a basis for $\Q(a)$ over $\Q$ and then say what $[\Q(a):\Q]$ must be. Then give a different basis for $\Q(a)$ over $\Q\text{.}$

###### 4.

Using all of your work above, find (or explain why you have already found) the minimal polynomial for $a$ over $\Q$ and prove it is irreducible (using the previous question).