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Suppose that $F$ is a field of characteristic $p$. Let $K/F$ be a finite extension and $K=F(K^p)$, where $K^p:= \{x^p\mid x\in K\}$. Suppose $\{a_1,\ldots,a_n\} \subset K$ is linearly independent over $F$. Show that is $\{a_1^p,\ldots,a_n^p \}$ is also linearly independent over $F$.

Here is a sketch of my proof: Suppose $b_1a_1^p+\cdots +b_na_n^p=0$ where $b_i\in F$. I want to find $c_i$ such that $c_i^p=b_i$. So $(c_1a_1+\cdots +c_na_n)^p=0$ and hence $c_i=b_i=0$. But the problem is that I don't know how to prove the existence of $c_i$. Any ideas?

user26857
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user zero
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  • I think there's kind of circular definition here: $;K=F(K^p);$ ...? – DonAntonio May 05 '16 at 22:11
  • @Joanpemo Good to know: https://en.wikipedia.org/wiki/Composite_field_(mathematics) – user26857 May 06 '16 at 07:32
  • @user26857 Thank you, indeed good to know. Yet the notation still shows circular definition, in my opinion. – DonAntonio May 06 '16 at 11:15
  • I don't think so: there is a field extension $F\subset K$ and it's supposed that $K=F(K^p)$. I can't see anything circular here. – user26857 May 06 '16 at 11:29
  • @user26857 Thank you. The asker is defining $;K;$ by means of itself, with $;F(K^p);$ . I think that's circular: since $;F,,K^p\subset K;$ , I can't see how is it possible to define $;K=F(K^p)\subset K;$ , and worse: taking into account that the last inclusion may be sharp. Am I missing something here? – DonAntonio May 06 '16 at 12:56
  • No, he's not defining $K$ by means of itself. $K$ is already given, and then one has added the property of $K$ of being generated by two subfields: $F$ and $K^p$. – user26857 May 06 '16 at 13:12

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First notice that every element of $F(K^p)$ is a linear combination of elements from $K^p$ with coefficients in $F$.

We may assume that $\{a_1,\dots,a_n\}$ is an $F$-basis of $K$. The remark above tells us that $\{a_1^p,\dots,a_n^p\}$ spans $F$ over $K$, so it is a basis.

user26857
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  • Why linear combination? What about $\dfrac{f_1 k_1^p + f_2 k_2^p}{f_3 k_3^p + f_4 k_4^p}$ ? And what is "The remark above"? – darij grinberg Jul 01 '21 at 18:57
  • I have now elaborated on this answer at https://math.stackexchange.com/questions/4184233/if-the-elements-y-1-y-2-dots-y-r-in-e-are-linearly-independent-over-f-sho/4188099#4188099 . – darij grinberg Jul 01 '21 at 20:18
  • When one proves that $K(M)=K[M]$ for $M$ consisting of algebraic elements over $K$ is it shown that a rational algebraic fraction in some elements of $M$ is in fact a polynomial expression of elements of $M$? I never seen such a proof, but you can have a try. Long story short, $F(K^p)=F[K^p]$. Now since the answer has only two paragraphs which one can count as "above"? – user26857 Jul 02 '21 at 05:21
  • Oh, I thought you were referring to something in the original post or the comments underneath. I am not sure the OP knew that $K(M) = K[M]$. – darij grinberg Jul 02 '21 at 05:39
  • This is what a student learns in the first lectures in field theory. Moreover, since the answer has been accepted I believe the OP knew that. – user26857 Jul 02 '21 at 06:23
  • Well, the OP of https://math.stackexchange.com/questions/4184233/if-the-elements-y-1-y-2-dots-y-r-in-e-are-linearly-independent-over-f-sho/4188099#4188099 did not :) – darij grinberg Jul 02 '21 at 07:12
  • It is not my fault that some students are not aware of such basic things. They could have ask if felt the need of some help, but they did not. – user26857 Jul 02 '21 at 10:11