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Let $E/F$ be a finite field extention.

Then we assume that $[E:F]=n\in \Bbb{N}$ and the set $A:=\{a_1,...,a_n \}\subseteq E$ is basis of the $F$-vector space $E$. So, $E=\langle A\rangle$. Can we say that $$F(A)=\langle A \rangle\ ?$$

It's obvious that $\langle A \rangle\ \subseteq F(A)$ because if we take the element $f(a_1,...,a_n):=k_1a_1+...+k_na_n \in \langle A \rangle,\ k_i\in F\implies f(a_1,...,a_n)\in F(A)$. But, what happens for the other inclusion?

How can we proove that every element in the form $\frac{f(a_1,...,a_n)}{g(a_1,...,a_n)}\in F(A) $ belongs to $\langle A \rangle $?

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    $\begingroup$ What's your notation $\left<A\right>$ mean? $\endgroup$ – Lord Shark the Unknown Aug 10 '17 at 19:15
  • $\begingroup$ @LordSharktheUnknown Thanks for your comment. We mean $$\langle A \rangle= \{ k_1a_1+...+k_na_n \in E:k_i \in F,\ i=1,...,n \}\subseteq E$$ $\endgroup$ – Chris Aug 10 '17 at 19:19
  • $\begingroup$ Clearly $F(A)\subseteq \left<A\right>$ as $F(A)\subseteq E\subseteq \left<A\right>$. Also your quotient is in $\left<A\right>$ since $\left<A\right>$ is a field. $\endgroup$ – Lord Shark the Unknown Aug 10 '17 at 19:22
  • $\begingroup$ Can we say that: $\langle A \rangle = F[A]:=\{f(a_1,...,a_n) :f(x_1,...,x_n) \in F[x_1,...,x_n] \}$, but every finite field extension is algebraic, so $F[A]=F(A)$? $\endgroup$ – Chris Aug 10 '17 at 19:28
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The assumption that $[E:F]$ is finite is totally irrelevant here. By definition, $F(A)$ is the smallest subfield of $E$ containing $F$ and $A$. In particular, this means $F(A)\subseteq E=\langle A\rangle.$

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  • $\begingroup$ Thank you for your comment. It's clear. But, is there a way to proove it with the way I asked? (to write every element of $F(A)$ in the form of an arbitrary element of $\langle A \rangle $ ? $\endgroup$ – Chris Aug 10 '17 at 20:27
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    $\begingroup$ No, not really. At some point you must use the fact that $E=\langle A\rangle$. $\endgroup$ – Eric Wofsey Aug 10 '17 at 20:29
  • $\begingroup$ Why this happens? Also, is $F[A]=\langle A\rangle$ ? $\endgroup$ – Chris Aug 10 '17 at 20:31
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    $\begingroup$ For any subset $A$ of $E$, we have $\langle A\rangle\subseteq F[A]\subseteq F(A)\subseteq E$ (this is immediate from all the definitions). But in this case we have chosen $A$ such that $\langle A\rangle=E$, so trivially all these sets are equal. $\endgroup$ – Eric Wofsey Aug 11 '17 at 2:46

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