Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came up with the following proof of the above theorem which is the key to the Galois's theory of algebraic equations. The usual proof uses Lagrange resolvent or Hilbert 90 which uses a similar trick. This proof is conceptual. Is this proof well-known?

Proposition

Let $K$ be a field. Let $n$ be a positive integer which is not divisible by the characteristic of $K$. Suppose $K$ contains an $n$-th primitive root of unity. Let $L$ be a cyclic extension of degree n over a field $K$. Then there is an element $y$ of $L$ such that $L = K(y)$ and $y^n$ is an element of $K$.

Proof: Let $\sigma$ be a generator of the Galois group of $L/K$. It is well-known that $1,\sigma,\dotsc,\sigma^{n-1}$ are linearly independent over $K$. Hence, since $\sigma^n = 1$, $X^n - 1$ is the minimal polynomial of $\sigma$ over $K$. Let $f(x)$ be the characteristic polynomial of $\sigma$. By the Cayley-Hamilton theorem, $f(\sigma) = 0$. Hence $f(X)$ is divisible by $X^n - 1$. Since $f(X)$ is monic and the degree of $f(X)$ is $n$, $f(X) = X^n - 1$. Let $\zeta$ be an $n$-th primitive root of unity. Since $f(\zeta) = 0$, $\zeta$ is an eigenvalue of $\sigma$. Hence there is an element $y \neq 0$ of $L$ such that $\sigma(y) = \zeta y$. Since $\sigma(y^n) = (\zeta y)^n = y^n$, $y^n$ is an element of $K$ by the fundamental theorem of Galois theory. $\sigma(y^i) = (\zeta y)^i = (\zeta^i)y^i,i = 0,1,\dotsc,n - 1$. Hence $1,y,\dotsc,y^{n-1}$ are eigenvectors of $1,\zeta,\dotsc,\zeta^{n-1}$ respectively. Hence $1,y,\dotsc,y^{n-1}$ are linearly independent over K. Hence $L = K(y)$.

share|improve this question
add comment

1 Answer

up vote 6 down vote accepted

First, a minor comment: this result is normally referred to as "Kummer theory" rather than "Lagrange's theorem" (at least in the number-theoretic circles that I am part of).

Regarding your argument: this proof (or, more precisely, a variation of it which replaces the explicit char. poly. computation with an appeal to the representation theory of cyclic groups) is the one I usually give when I have to prove Kummer theory in a class. However, I haven't seen it written down (I came up with it myself at some point). I presume that it is well-known to some, and not to others; in any case, I agree that it is a nice, and pleasingly conceptual, proof.

share|improve this answer
    
As for the name "Kummer theory", the theorem was proved by Galois using Lagrange's technique. I guess Kummer learned it from Galois's theory. In any case, I think the priority for the theorem was on Galois or Lagrange, not on Kummer. –  Makoto Kato Apr 15 '12 at 2:01
    
@Makoto: Dear Makoto, Thank you for the additional historical background. My comment was only intended to indicate the standard name for the result, in case you want to pursue it further in the literature. (And maybe the emphasis of the phrase "Kummer theory" is slightly different, in that it's supposed to also include the converse result, that if $K$ contains the $n$th roots of $1$, then any extension of the form $K^(a^{1/n})$, for $a \in K$, is cyclic. Presumably this converse statement was also known before Kummer, though?) Regards, –  Matt E Apr 15 '12 at 11:55
    
Thanks. Yes, Galois knew the converse. –  Makoto Kato Apr 16 '12 at 14:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.