Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F/k$ be an extension of fields with $|k|=3^6$ and $|F|=3^{60}$, how many elements $\alpha\in F$ are there with $k(\alpha)=F$?

share|cite|improve this question
up vote 2 down vote accepted

The elements of $\mathbb F_{q^s}$ which generate $\mathbb F_{q^s}$ are the elements that are not in any strict subfield of $\mathbb F_{q^s}$. These subfields are all the $\mathbb F_{q^r}$, with $r$ dividing $s$.

You can compute the cardinal of $$\bigcup_{r | s, r < s} \mathbb F_{q^r}$$ using inclusion-exclusion principle and the fact that $\mathbb F_{q^b} \cap \mathbb F_{q^a} = \mathbb F_{q^{\gcd(a, b)}}$.

For your particular example, we have $q = 3^6$ and $s = 10$. The strict divisors of $s$ are 1, 5, and 2. Thus the number you are looking for is $$ q^{10} - (q^5 + q^2 + q^1) + 2q^1 = 42391158275215997623161807840$$

share|cite|improve this answer
You're absolutely right, and I'm ashamed of that... Unfortunately, I cannot delete an accepted answer, so I'll fix it. – Lierre May 4 '12 at 19:39
No reason to delete, when fixing the answer works! Give my regards to Pascale Charpin, if you're working in her group. – Jyrki Lahtonen May 4 '12 at 20:25
If I were in the team Secret, I would know finite fields better ;), but my building is definitely close ! – Lierre May 4 '12 at 20:28

For any prime power $q$ there are only two intermediate fields between $K=F_q$ and $L=F_{q^{10}}$, namely $M_1=F_{q^2}$ and $M_2=F_{q^5}$. This follows either from the general theory of finite fields or from Galois theory. The Galois group $Gal(L/K)$ is cyclic of order 10, and hence its only non-trivial subgroups are cyclic groups of orders five and two respectively, and then $M_1$ and $M_2$ are the corresponding fixed fields.

The element $x\in L$ generates $L$ over $K$, unless $x\in M_1$ or $x\in M_2$. The intermediate fields $M_1$ and $M_2$ intersect at $K$, so this happens for $$ N=|M_1\cup M_2|=|M_1|+|M_2|-|M_1\cap M_2|=q^5+q^2-q $$ elements $x$.

Therefore the correct answer is that $L=K(x)$ for $$q^{10}-N=q^{10}-q^5-q^2+q=3^{60}-3^{30}-3^{12}+3^6$$ choices of $x$.

The general formula involves the Möbius function via the usual exclusion/inclusion business.

share|cite|improve this answer

Your Answer


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.