Linear Independence in Number Theory

In this article, we discuss some questions I had regarding linear independence as it shows up in subjects like transcendental number theory.

We consider the following conjecture of mine:

Conjecture. Let $E$ be a field extension of $F$ , and let $a_{1},\ldots,a_{n} \in E$ . Then, the $a_{i}$ are linearly independent over $F$ if and only if $c_{1}a_{1} + \ldots + c_{n}a_{n} \notin F$ for all $c_{1},\ldots,c_{n} \in F$ .

The reason why there is some doubt to this is the lack of a “constant term” from $F$ . For example, it is strongly believed that $\pi$ and $e$ are linearly independent over the rationals, and this seems to be recognized as a stronger statement than $\pi + e$ being irrational (which incidentally is also unknown.) But translated literally, linear independence means that $r\pi + se = 0 \rightarrow r = s = 0$ for all rational $r,s$ . How does this translate into $\pi + e$ being irrational without a “constant term,” like $r\pi + se + t = 0 \rightarrow r = s = t = 0$ ?

If true, this conjecture would reformulate linear independence in an arguably easier-to-interpret way, especially for people who haven’t studied abstract algebra but know about rationality/irrationality.

Attempt. First, assume that $c_{1}a_{1} + \ldots + c_{n}a_{n} \notin F$ for all $c_{1},\ldots,c_{n} \in F$ ; we show the $a_{i}$ are linearly independent over $F$ . Assume they were linearly dependent, then there would exist a linear relation over them with coefficients in $F$ equaling 0. But $0 \in F$ , contradiction.

Next, assume the $a_{i}$ are linearly independent; we show that $c_{1}a_{1} + \ldots + c_{n}a_{n} \notin F$ for all $c_{i} \in F$ . Assume there was $c_{i} \in F$ such that $c_{1}a_{1} + \ldots + c_{n}a_{n} \in F$ . Let $f = c_{1}a_{1} + \ldots + c_{n}a_{n}$ . If $f = 0$ then the $a_{i}$ are linearly dependent, contradiction. Otherwise,

$\displaystyle \frac{c_{1}}{f}a_{1} + \ldots + \frac{c_{n}}{f}a_{n} = 1.$

so $c_{1}a_{1} + \ldots + c_{n}a_{n} + ( - f)1 = 0$ .

Actually, we can present a counterexample. Take $a = \pi$ and $b = 1 - 2\pi$ . We have that $2a + b = 1$ is rational. However, $a,b$ are linearly independent over $\mathbb{Q}$ : indeed, if we have $c,d\mathbb{\in Q}$ with

$\displaystyle ca + db = c\pi + d(1 - 2\pi) = 0,$

then

$\displaystyle (c - 2d)\pi + d = 0,$

and since $\pi$ is irrational we must have $d = 0$ , so then $c\pi = 0 \rightarrow c = 0$ . Thus, the conjecture above is not correct.

Instead, the proper formulation just involves adding the element 1 to the set! We have our result:

Theorem. Let $E$ be a field extension of $F$ , and let $a_{1},\ldots,a_{n} \in E$ . Then, $a_{1},\ldots,a_{n},1$ are linearly independent over $F$ if and only if $c_{1}a_{1} + \ldots + c_{n}a_{n} \notin F$ for all $c_{1},\ldots,c_{n} \in F$ .

Hence, all the results in transcendental number theory involving linear (or algebraic) independence can be applied to determine irrationality (or transcendentality) by considering the number 1 (or some rational number) as part of the “input” set.

Corollary. To say that $r\pi + se$ is irrational for all $r,s\in\mathbb{Q}$ is the same as saying that $\pi$ , $e$ , and 1 are linearly independent over $\mathbb{Q}$ .

Saying that $r\pi + se$ is irrational for all rational $r,s$ is the same as saying that for $m,n$ integers, $m\pi + ne$ can never be an integer: if $r\pi + se$ were rational for some choice of $r,s$ then that yields a choice of $m,n$ where $m\pi + ne$ is integral, and conversely if $m\pi + ne$ were integral then that yields a choice of $r,s$ where $r\pi + se$ is rational.

In fact, in general, if we are willing to extend the notion of linear independence beyond fields (using the formulation given above), linear independence over $\mathbb{Q}$ is the same as linear independence over $\mathbb{Z}$ . Even more generally, given an integral domain $D$ and its field of fractions $F$ (and within a universe of a field extension of $F$ ), linear independence over $F$ is the same as linear independence over $D$ : if we have $a_{1},\ldots,a_{n}$ in the universe, then given a choice of $f_{1},\ldots,f_{n} \in F$ such that $f_{1}a_{1} + \ldots + f_{n}a_{n} = 0$ we can yield a choice of $d_{1},\ldots,d_{n} \in D$ such that $d_{1}a_{1} + \ldots + d_{n}a_{n} = 0$ , and conversely.

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