For each of the following numbers $q$ in turn, consider a field $k$ of order $q$. In each case, determine the number of elements $\alpha$ in $k$ such that the smallest subfield of $k$ containing $\alpha$ is $k$ itself. $a)2^4, b)3^5, c)5^{10}, d) 7^{12}$.

**Solution**:**Results**: Let $\Bbb F_{p^n}$ be the finite field with $p^n$ elements. A finite field $\Bbb F_{p^m}$ is a subfield of $\Bbb F_{p^n}$ if and only if $m|n$. (**Note that the power has to divide**). Hence the number of subfields of $\Bbb F_{p^n}$ is equal to $d(n)$ the number of divisors of $n$.**option a**: Consider the finite field $\Bbb F_{2^4}$. This has $d(4)$ which is equal to $3$ subfields namely $\Bbb F_{2} \subset \Bbb F_{2^2} \subset \Bbb F_{2^4}$. Note that $\Bbb F_{2} \subset \Bbb F_{2^2}$ is true again because of the above result. Now, if we take an element from $\Bbb F_{2^2}$ then the small subfield containing that element cannot be larger than $\Bbb F_{2^2}$. But if we consider any element in $\Bbb F_{2^4}$ which lies outside $\Bbb F_{2^2}$ then the smallest field containing them will be the whole $\Bbb F_{2^4}$. Hence the required number is $2^4 - 2^2$.**option b**: We can use the same argument and the answer will be $3^5 -3^1$.**option c**: The subfields of $\Bbb F_{5^{10}}$ are $\Bbb F_{5^{10}}$, $\Bbb F_{5^5}$, $\Bbb F_{5^2}$ and $\Bbb F_{5^1}$. So one needs to conder the elements outside $\Bbb F_{5^5}$ and $\Bbb F_{5^2}$ so that the smallest field containing that element will be full $\Bbb F_{5^{10}}$. So one would expect this number will be $5^{10} -5^5-5^2$. But the fields $\Bbb F_{5^5}$ and $\Bbb F_{5^5}$ has the intersection and we have subtracted these elements twice. So we want to understand the intersection of two finite fields. In general, we have $$\Bbb F_{p^n} \cap \Bbb F_{p^m} = \Bbb F_{p^{gcd(m,n)}}.$$ So the subfields $\Bbb F_{5^5}$ and $\Bbb F_{5^2}$ have $5^1$ elements in common. Since we have twice subtracted them, we need to add it back once and the required answer will be $5^{10} -5^5-5^2 + 5$.**option****d**: We can use the same argument and the answer will be $7^{12} -7^6 - 7^4 + 7^2$. Note that, in this case, the intersection of the fields $\Bbb F_{7^6}$ and $\Bbb F_{7^4}$ will be $\Bbb F_{7^2}$ and this explains the addition of $7^2$ in the answer.**Share to your groups:**

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In the 4th line of option (a), the smallest subfield containing any element of order 4 field (outside order 2 field ) must be F4.

ReplyDeleteCorrect. Thanks for pointing out. I shall correct it :)

ReplyDeleteI have corrected it.

ReplyDelete