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 normal basis theorem of a cyclic extension field. Is this proof well-known?

Proposition Let $L$ be a finite cyclic extension of a field $K$. Let $n$ be the degree of $L$ over $K$. Let $\sigma$ be the generator of its Galois group. Then there is an element $y$ of $L$ such that $y, \sigma(y),\ldots,\sigma^{n-1}(y)$ is a basis of $L/K$.

Proof: It is well-known that $1, \sigma,\ldots,\sigma^{n-1}$ are linearly independent over $K$. Hence, since $σ^n = 1$, $X^n - 1$ is the minimal polynomial of $σ$ over $K$. Let $f(x)$ be the characteristic polynomial of $σ$. By the Cayley-Hamilton theorem, $f(σ) = 0$. Hence $f(X)$ is divisible by $X^n - 1$. Since $f(X)$ is monic and the degree of $f(X)$ is $n$, $f(X) = X^n - 1$. By the proposition I posted in this site under the title "Cyclic modules over a polynomial ring", $L$ is a cyclic $K[X]$-module. Let $y$ be a generator of $L$ as a $K[X]$-module. Then $y, σ(y),\ldots,σ^{n-1}(y)$ is a basis of $L/K$.

share|improve this question
1  
The proof looks OK to me. But I think the normal basis theorem states that any finite Galois extension has a normal basis, not just cyclic ones, so maybe this special case is just not considered interesting enough for a separate proof (although it does the job for finite extensions of finite fields)? –  Marc van Leeuwen Apr 14 '12 at 17:31
    
Extending Marc's comment: In the case of finite fields this is proven in more or less exactly this way in the tome by Lidl & Niederreiter - a de facto bible of finite fields. I vaguely recall that I have reproduced that proof somewhere in this forum, but I couldn't find it. –  Jyrki Lahtonen Apr 14 '12 at 18:03
1  
The same argument is in Jacobson's Basic Algebra I, so the proof is not new. –  KCd Apr 14 '12 at 18:13
    
Though some people already alluded, let me clear the things just in case. The usual proof of normal basis theorem is divided into two cases. Namely infinite and finte base field cases. The methods of two cases are totally different. Of course, my proof applies to finte base fields. There is a proof that applies to both cases, but it's less well-known. –  Makoto Kato Apr 15 '12 at 0:13
    
@MakotoKato Could you tell me what's the proof that applies to both cases? I was asked to prove the normal basis theorem as a homework, but I really don't have any idea... Thanks a lot. –  lee Aug 5 at 2:31

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.