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Let $E/F$ be a field extension, and $a_1,\ldots,a_n \in E$. Is it true that $F(a_1,\ldots,a_{n-1})(a_n)=F(a_1,\ldots,a_n)$? Recall that for any $S\subset E$, $F(S)$ is the subfield of $E$ generated by $S$ over $F$, i.e, $F(S)=(F\cup S)$.

If $S=\{a_1,\ldots,a_{n-1}\}$ and $T=\{a_1,\ldots,a_n\}$, then clearly $S\subset T$, so that $F(S)\subset F(T)$ and therefore $(F(S)\cup\{a_n\})\subset F(T)$ by minimality, that is to say, $F(S)(a_n)\subset F(T)$. For reverse inclusion I have been trying to use a similar argument but getting nowhere. Thanks in advance.

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  • $\begingroup$ $F(S)$ is the field generated by $F \cup S$, not $F \cap S$. $\endgroup$
    – Tzimmo
    Commented Nov 29, 2023 at 21:39
  • $\begingroup$ Okay, maybe this is a typo in my teacher's notes, thank you. $\endgroup$
    – isaac098
    Commented Nov 29, 2023 at 21:42
  • $\begingroup$ Can you clarify how are you getting nowhere in the reverse inclusion? All you need is all $a_i$ to be in $F(S)(a_n)$. Where are you stuck? $\endgroup$
    – Tzimmo
    Commented Nov 29, 2023 at 21:46
  • $\begingroup$ So you're saying that the reverse inclusion follows from $T\subset F(S)(a_n)$? I'm stuck in the sense that I can't think of how to prove such inclusion. Let me see if I get anywhere with your suggestion. $\endgroup$
    – isaac098
    Commented Nov 29, 2023 at 22:07
  • $\begingroup$ Actually, it should work in the same way as $F(S) \subseteq F(T)$. $\endgroup$
    – Tzimmo
    Commented Nov 29, 2023 at 22:19

1 Answer 1

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Note that $F(a_1,\ldots,a_{n-1})(a_n)$ is a field containing $F(a_1,\ldots,a_{n-1})$ and $a_n$. In particular it is a field containing $F$ and $\{a_1,\ldots,a_n\}$. Since $F(a_1,\ldots,a_n)$ is the smallest field containing $F$ and $\{a_1,\ldots,a_n\}$ then necessarily $$F(a_1,\ldots,a_n)\subseteq F(a_1,\ldots,a_{n-1})(a_n).$$

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  • $\begingroup$ What about the opposite inclusion? $\endgroup$
    – Emptymind
    Commented Feb 3 at 18:12
  • $\begingroup$ @Emptymind Since $F(a_1,\ldots,a_n)$ is a field containing $F(a_1,\ldots,a_{n-1})$ and $a_n$, it is also a field containing the minimum field containing $F(a_1,\ldots,a_{n-1})$ and $a_n$, i.e., containing $F(a_1,\ldots,a_{n-1})(a_n)$. $\endgroup$
    – Deif
    Commented Feb 4 at 19:55

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