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Let $K \to L$ be a finite field extension with $[L:K] = n$ and $K$ a finite field of cardinality $q$. A proper subextension is a field extension $K \to M$ such that $M \subsetneq L$. I want to show that $L$ can not be written as the union of its proper subextensions.

I've already shown that I can write any proper subextension $M$ as $M = \{x \in L\;|\; x^{q^m} - x = 0\}$ for some $m \in \mathbb{N}$. Using this, I can show that I can't write $L$ as union of such subextensions because that would lead to a contradiction to the fact that the multiplicative group of a finite field is cyclic (it would imply that $L$ can not contain an element of order $q^n -1$). However, we have not shown that fact in class yet so I'm wondering if I'm overlooking something much simpler.

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  • $\begingroup$ Do you know how the subextensions are related to the divisors of $n$? $\endgroup$ Nov 16, 2016 at 23:20
  • $\begingroup$ @DanielFischer whoops, I corrected that. I think so: If $M$ is a subextension containing $q^m$ elements, then $m$ must divide $n$. Conversely, if $m$ divides $n$, the splitting field of $x^{q^m} -x$ is a subextension of $L$ containing $q^m$ elements. $\endgroup$
    – user159517
    Nov 16, 2016 at 23:25
  • $\begingroup$ And how many subextensions corresponding to each divisor are there? $\endgroup$ Nov 16, 2016 at 23:34
  • $\begingroup$ @DanielFischer well one, up to isomorphism. I still dont quite see what you're getting at. $\endgroup$
    – user159517
    Nov 16, 2016 at 23:39
  • $\begingroup$ Not only up to isomorphism. Then we have a cardinality argument $\downarrow$. $\endgroup$ Nov 16, 2016 at 23:42

1 Answer 1

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The subextensions - or intermediate fields - $K \subset M \subset L$ are in bijection with the divisors of $n$, to each $d \mid n$ there corresponds a unique intermediate field $M_d$ with $[M_d : K] = d$ - this is the set of zeros of $X^{q^d} - X$ in $L$. Thus we have

\begin{align} \operatorname{card} \Biggl(\bigcup_{\substack{d\mid n \\ d < n}} M_d\Biggr) &\leqslant \sum_{\substack{d\mid n \\ d < n}} \operatorname{card} M_d \\ &= \sum_{\substack{d\mid n \\ d < n}} q^d\\ &\leqslant \sum_{k = 1}^{n-1} q^k \\ &< q^n, \end{align}

since $q \geqslant 2$.

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  • $\begingroup$ I strongly suspected that there was some easy argument I just didn't see. Thanks for pointing it out to me. $\endgroup$
    – user159517
    Nov 16, 2016 at 23:45
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    $\begingroup$ As usual, it's only easy when you know where to look. $\endgroup$ Nov 16, 2016 at 23:47

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