March | 2019 | Arithmetic variety

A complex number is algebraic if it is the root of some polynomial $P(x)$ with rational coefficients. $\sqrt{2}$ is algebraic (e.g. the polynomial $x^2 -2$ ); $i$ is algebraic (e.g. the polynomial $x^2 + 1$ ); $\pi$ and $e$ are not. A complex number that is not algebraic is called transcendental.

Is the sum of algebraic numbers always algebraic? What about the product of algebraic numbers? For example, given that $\sqrt{2}$ and $\sqrt{3}$ are algebraic, how do we know $\alpha = \sqrt{2} + \sqrt{3}$ is also algebraic?

We can try to repeatedly square the equation $x = \sqrt{2} + \sqrt{3}$ . This gives us $x^2 = 2 + 3 + 2\sqrt{6}$ . Then isolating the radical, we have $x^2 - 5 = 2\sqrt{6}$ . Squaring again, we get $x^4 - 10x^2 + 25 = 24$ , so $\alpha$ is a root of $x^4 - 10x^2 + 1$ . This is in fact the unique monic polynomial of minimum degree that has $\alpha$ as a root (called the minimal polynomial of $\alpha$ ) which shows $\alpha$ is algebraic. But a sum like $\sqrt{2} + \sqrt{3} + \sqrt{5} + i$ would probably have a minimal polynomial of much greater degree, and it would be much more complicated to construct this polynomial and verify the number is algebraic.

It is also increasingly difficult to construct polynomials for numbers like $\sqrt{2} + \sqrt[3]{3}$ . And, apparently there also exist algebraic numbers that cannot be expressed as radicals at all. This further complicates the problem.

In fact, the sum and product of algebraic numbers is algebraic – in fact, any number which can be obtained by adding, subtracting, multiplying and dividing algebraic numbers is algebraic. This means that a number like

$\frac{\sqrt{2} + i\sqrt[3]{25} - 2 + \sqrt{3}}{15 - 3i + 4e^{5i\pi / 12}}$

is algebraic – there is some polynomial which has it as a root. But we will prove this result non-constructively; that is, we will prove that such a number must be algebraic, without providing an explicit process to actually obtain the polynomial. To establish this result, we will try to look for a deeper structure in the algebraic numbers, which is elucidated, perhaps surprisingly, using the tools of linear algebra.

Definition: Let $S$ be a set of complex numbers that contains $0$ . We say $S$ is an abelian group if for all $x, y \in S$ , $x + y \in S$ and $x - y \in S$ . In other words, $S$ is closed under addition and subtraction.

Some examples: $\mathbb{Z}$ , $\mathbb{Q}$ , $\mathbb{R}$ , $\mathbb{C}$ , the Gaussian integers $\mathbb{Z}[i]$ (i.e. the set of expressions $a+bi$ where $a$ and $b$ are integers)

Definition: An abelian group is a field if it contains $1$ and for all $x, y \in S$ , $xy \in S$ , and if $y \neq 0$ , $x/y \in S$ . In other words, $S$ is closed under multiplication and division.

We generally use the letters $k$ , $F$ , $E$ for arbitrary fields. Of the previous examples, only $\mathbb{Q}$ , $\mathbb{R}$ , $\mathbb{C}$ are fields.

Exercise: Show that if $k$ is a field, $\mathbb{Q} \subseteq k$ .

Exercise: Show that the set $\{a + b\sqrt{2} \ | \ a, b \in \mathbb{Q} \}$ is a field. We call this field $\mathbb{Q}(\sqrt{2})$ . (Hint: rationalize the denominator).

Exercise: Describe the the smallest field which contains both $\sqrt{2}$ and $\sqrt{3}$ (this is called $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ )

Generally if $k$ is a field and $x_1, \dots x_n$ some complex numbers, we will denote the smallest field that contains all of $k$ and the elements $x_1, \dots x_n$ as $k(x_1, \dots x_n)$ .

Definition: Let $k$ be a field. A $k$ -vector space is an abelian group $V$ such that if $c \in k$ , $x \in V$ , then $cx \in V$ . Intuitively, the elements of $V$ are closed under scaling by $k$ . (We also sometimes use the phrase “ $V$ is a vector space over $k$ “)

Examples: $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ and $\mathbb{Q}(\sqrt{2})$ are both $\mathbb{Q}$ -vector spaces. In fact, $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is also a $\mathbb{Q}(\sqrt{2})$ vector space.

With this language in place, we can state the main goal of this post as follows:

Theorem: If $\alpha_1, \dots \alpha_n$ are algebraic numbers and $x \in \mathbb{Q}(\alpha_1, \dots \alpha_n)$ , then $x$ is algebraic.

Note how this encompasses our previous claim that any number obtained by adding, subtracting, multiplying and dividing algebraic numbers is algebraic. For example, we can deduce the giant fraction above is algebraic by applying the theorem to $\mathbb{Q}(\sqrt{2}, i, \sqrt[3]{25}, \sqrt{3}, e^{5i/12})$ .

Definition: A field extension of a field $k$ is just a field $F$ that contains $k$ . We will write this as “ $F \supseteq k$ is a field extension”.

Exercise: If $F \supseteq k$ is a field extension, then $F$ is a $k$ -vector space

Consider a field like $\mathbb{Q}(\sqrt{2})$ . Not only is it a field, but it is also a $\mathbb{Q}$ -vector space – its elements can be scaled by rational numbers. We want to adapt concepts from ordinary linear algebra to this setting. We want to think of $\mathbb{Q}(\sqrt{2})$ as a two dimensional vector space, with basis $1$ and $\sqrt{2}$ – every element is can be uniquely written as a $\mathbb{Q}$ -linear combination of $1$ and $\sqrt{2}$ . Similarly, we would like to think of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ as a $\mathbb{Q}$ -vector space with basis $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ . Note that if we regard $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ as a $\mathbb{Q}(\sqrt{2})$ -vector space, its basis is $1, \sqrt{3}$ . This is because if an element is written as $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$ , where the coefficients are rational, we can rewrite it as $(a + b\sqrt{2})1 + (c + d\sqrt{2})\sqrt{3}$ , now with coefficients in $\mathbb{Q}(\sqrt{2})$ .

On the other hand, consider $\mathbb{Q}(\pi)$ . Since $\pi$ is not algebraic, no linear combination (with rational coefficients) of the numbers $1, \pi, \pi^2, \pi^3, \dots$ equals $0$ without all the coefficients being zero. This makes them linearly independent – and makes $\mathbb{Q}(\pi)$ an infinite-dimensional vector space.

This finite-versus-infinite dimensionality difference will lie at the crux of our argument. Here is a rough summary of the argument to prove the Theorem:

If we add an algebraic number to a field, we get a finite-dimensional vector space over that field.
By repeatedly adding algebraic numbers to $\mathbb{Q}$ , the resulting field will still be a finite-dimensional $\mathbb{Q}$ -vector space.
Any element of a field which is a finite-dimensional $\mathbb{Q}$ -vector space must be algebraic.

From here on, we will assume that the reader has some familiarity with ideas from linear algebra, such as the notions of linear combination, linear independence, and basis. We will only sketch proofs (maybe I will add details later). We would like to warn the reader that the proof sketches have many gaps and may be difficult to follow.

Definition: A field extension $F \supseteq k$ is finite if there is a finite set of elements $e_1, \dots e_n \in F$ such that every element of $F$ can be represented as $c_1e_1 + \dots c_ne_n$ , where all the $c_i \in k$ . We say the elements $e_1, \dots e_n$ generate (aka span) $F$ over $k$ . If, furthermore, every element can be represented uniquely in this form, then we say $e_1, \dots e_n$ is a $k$ –basis (or just basis) for $F$ .

Examples: $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}$ are finite extensions, while $\mathbb{Q}(\pi)/\mathbb{Q}$ is not.

Exercise: Show $\mathbb{C}/\mathbb{Q}$ is not a finite extension. (Hint: $\mathbb{C}$ is uncountable)

Proposition 1: Every finite field extension $F/k$ has a basis.

Proof sketch: This is a special case of a famous theorem in linear algebra. Assume $e_1, \dots e_n$ generate $F$ . Start with the last element. If $e_n$ can be represented as a linear combination of $e_1 \dots e_{n-1}$ , remove it from the list. If we removed $e_n$ , we now have $e_1 \dots e_{n-1}$ , and we can repeat the process with $e_{n-1}$ . Continue this process until we cannot remove any more elements. Then (it can be checked that) we obtain a set whose elements are linearly independent. Then (it can be checked that) these form a $k$ -basis for $F$ . $\square$

For the next theorem, keep in mind the example of the extensions $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}(\sqrt{2})$ .

Proposition 2: Suppose $F \supseteq k$ and $E \supseteq F$ are finite field extensions. Then $E \supseteq k$ is a finite extension.

Proof: By Proposition 1, both these field extensions have bases. Label the $F$ -basis of $E$ as $e_1, \dots e_n$ , and the $k$ -basis for $f$ as $e'_1, \dots e'_m$ . Then every element of $F$ can be uniquely represented as $c_1e_1 + \dots + c_ne_n$ , for $c_i \in F$ . Furthermore, each $c_i$ can be written uniquely as $c_i = a_{i, 1}e'_1 + \dots + a_{i, m}e'_m$ for $a_{i, j} \in k$ . Plugging these in for the $c_i$ shows that the $mn$ elements of the form $e_ie'_j$ form a $k$ -basis for $E$ . $\square$

Proposition 3: If $\alpha$ is algebraic, then $k(\alpha) \supseteq k$ is a finite extension.

Proof sketch: Suppose $x \in k(\alpha)$ . First we will show every element in $k(\alpha)$ is a polynomial in $\alpha$ with coefficients in $k$ . $k(\alpha)$ consists of all numbers that can be obtained by repeatedly adding, subtracting, multiplying and dividing $\alpha$ and elements of $k$ . By performing some algebraic manipulations and combining fractions, we can see that every element can be written in the form $p(\alpha)/q(\alpha)$ , where $p$ and $q$ are polynomials with coefficients in $k$ .

Now I claim that for any polynomial $q$ where $q(\alpha) \neq 0$ , there exists a polynomial $s$ such that $s(\alpha) = 1/q(\alpha)$ . Let $m$ be the minimal polynomial of $\alpha$ . Since $q(\alpha) \neq 0$ and $m(\alpha) = 0$ , $q$ and $m$ are relatively prime as polynomials. So there exist polynomials $r$ and $s$ such that $rq+sm = 1$ . Plugging in $\alpha$ into the polynomials gives us $r(\alpha)q(\alpha) + s(\alpha)m(\alpha) = 1$ . Since $m(\alpha) = 0$ , $s(\alpha) = 1/q(\alpha)$ , as desired.

Thus any element in $k(\alpha)$ , written as $p(\alpha)/q(\alpha)$ , can be written as $p(\alpha)s(\alpha)$ for an appropriate polynomial $s$ ; in other words, every element of $k(\alpha)$ is a polynomial of $\alpha$ with coefficients in $k$ . Now, suppose the minimal polynomial of $\alpha$ is $a_nx^n + \dots a_1x + a_0$ . Then $\alpha^n = \frac{1}{a_n}(-a_0 - a_1\alpha \dots - a_{n-1}\alpha^{n-1})$ . So $\alpha^n$ (and all higher powers of $\alpha$ ) can be written as rational linear combinations of $1, \alpha, \dots \alpha^{n-1}$ . Then it can be verified that every element in $k(\alpha)$ can be written uniquely as a $k$ -linear combination of $1, \alpha, \dots \alpha^{n-1}$ . This establishes that $k(\alpha)/k$ is a finite extension; in particular, it has the basis $1, \alpha, \dots \alpha^{n-1}.$ $\square$

Proposition 4: Suppose $k \supseteq \mathbb{Q}$ is a finite extension. Then any $x \in k$ is algebraic.

Proof sketch: Consider the elements $1, x, x^2, \dots…$ . They cannot all be linearly independent. This is because in a finite-dimensional vector space of dimension $n$ any set with at least $n+1$ elements must be linearly dependent. So there must be some linear combination of them which equals zero. But this simply means that for some constants $a_i$ and positive integer $n$ , $a_nx^n + \dots a_1x + a_0 = 0$ . Letting the $a_i$ be the coefficients of a polynomial $P$ , $P(x) = 0$ , so $x$ is algebraic. $\square$

We have developed enough theory now to prove the result.

Proof of main Theorem: Let $\alpha_1, \alpha_2, \dots \alpha_n$ be some collection of algebraic numbers. Then by Proposition 3, $\mathbb{Q}(\alpha_1, \dots \alpha_n) \supseteq \mathbb{Q}(\alpha_1, \dots \alpha_{n-1}), \dots \mathbb{Q}(\alpha_1, \alpha_2) \supseteq \mathbb{Q}(\alpha_1), \mathbb{Q}(\alpha_1) \supseteq \mathbb{Q}$ are all finite field extensions. Applying Proposition 2 repeatedly gives us $\mathbb{Q}(\alpha_1, \dots \alpha_n) \supseteq \mathbb{Q}$ is a finite extension. Then by Proposition 4, any element of $\mathbb{Q}(\alpha_1, \dots \alpha_n)$ is algebraic.

This theorem essentially says that given a collection of algebraic numbers, by repeatedly adding, subtracting, multiplying and dividing them, we can only obtain algebraic numbers. But what about radicals? For example, we have now established that $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is algebraic. Is $\sqrt{\sqrt{2} + \sqrt{3} + \sqrt{5}}$ algebraic as well?

In fact, we can generalize our result to the following: any number obtained by repeatedly adding, subtracting, multiplying, dividing algebraic numbers, as well as taking $m$ -th roots (for positive integers $m$ ) will be algebraic.

This is not too hard now that we have the main theorem. It suffices to show that if $x$ is algebraic, $\sqrt[m]{x}$ is algebraic. If $x$ is algebraic, then there exist constants $a_0, \dots a_n$ such that $a_nx^n + \dots a_1x + a_0 = 0$ . This can be rewritten as $a_n(\sqrt[m]{x})^{mn} + \dots a_1(\sqrt[m]{x})^m + a_0 = 0$ . Thus $\sqrt[m]{x}$ is algebraic.

Acknowledgements: Thanks to Anton Cao and Nagaganesh Jaladanki for reviewing this article.

Arithmetic variety

Monthly Archives: March 2019

On algebraic numbers