# A proof of the Lagrange's theorem on cyclic extention fields

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 $L = K(y)$.

-

@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