Linearly Independent Automorphisms

From Galois theory. I've rewritten the presentation in [1, §124, p. 106].

The Theorem

“If $\phi_1, \phi_2, \ldots, \phi_n$ are distinct automorphisms of $E$, then the set $\{\phi_1, \phi_2, \ldots, \phi_n\}$ is linearly independent over $E$.”


Definition of “Linearly Dependent” and “Linearly Independent”

From the start of the previously cited §124 of [1] comes an important definition:

“A finite set of automorphisms of a field $E$ is called linearly dependent over $E$ if there are elements of $E$, $c_1, c_2, \ldots, c_n \in E$, not all zero, such that

\begin{align} c_1(\phi_1\alpha) + c_2(\phi_2\alpha) + \ldots + c_n(\phi_n\alpha) = 0 \end{align}

for all elements $\alpha$. Otherwise $\{\phi_1, \phi_2, \ldots, \phi_n\}$ is called linearly independent.”

Other Equations

To make the presentation less cluttered I'll present the other equations I'll use here:

\begin{align} c_1(\phi_1\alpha) + c_2(\phi_2\alpha) + \ldots + c_r(\phi_r\alpha) = 0 \end{align}
\begin{align} c_1(\phi_1\beta)(\phi_1\alpha) + c_2(\phi_2\beta)(\phi_2\alpha) + \ldots + c_r(\phi_r\beta)(\phi_r\alpha) = 0 \end{align}
\begin{align} c_1(\phi_r\beta)(\phi_1\alpha) + c_2(\phi_r\beta)(\phi_2\alpha) + \ldots + c_r(\phi_r\beta)(\phi_r\alpha) = 0 \end{align}
\begin{align} c'_1(\phi_1\beta)(\phi_1\alpha) + c'_2(\phi_2\beta)(\phi_2\alpha) + \ldots + c'_{r-1}(\phi_{r-1}\beta)(\phi_{r-1}\alpha) = 0 \end{align}

Reverse Belief-based Reasoning

“If $\phi_1, \phi_2, \ldots, \phi_n$ are distinct automorphisms of $E$, then the set $\{\phi_1, \phi_2, \ldots, \phi_n\}$ is linearly independent over $E$.”

If you believe…

The claim that $\{\phi_1, \phi_2, \ldots, \phi_n\}$ is linearly dependent over $E$ will get you a contradiction.

If you believe…

Equation (1) implies equation (2), which has the smallest number $r\leq n$ of non-zero coefficients, and further that equation (2) implies an equation (5) with even fewer non-zero coefficients, which makes for a contradiction.

If you believe…

To get from equations (1) to (2) you can remove all the zero coefficients and relabel the remaining coefficients.

To get from equations (2) to (5) you can subtract equation (4) from (3). The last term of (3) and (4) are equal, so (5) has $r-1$ non-zero coefficients.

If you believe…

The definition of a linearly dependent set of automorphisms says equation (1) is allowed, and not all coefficients $c$ are zero. You can't remove any more terms from (2) because there are no more zero coefficients.

Meanwhile, you can replace $\alpha$ in equation (2) with $\beta\alpha$ to produce equation (3).

Also, you can multiply (2) by $\phi_r\beta$ to produce equation (4).

If you believe…

After you replace $\alpha$ in equation (2) with $\beta\alpha$ you have terms of the form $\phi(\beta\alpha)$. By the definition of an automorphism, $\phi(\beta\alpha)$ = $(\phi\beta)(\phi\alpha)$.


1. Clark, A. Elements of Abstract Algebra. New York: Dover, 1984 [orig. 1971].


Please note that I am not a mathematician and so the presentation of proofs that I make may be deeply flawed. I'm using this writing process to figure out what I'm reading. Please consult more authoritative sources as well.

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