# $F(a_1,\ldots,a_{n-1})(a_n)=F(a_1,\ldots,a_n)$

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.

• $F(S)$ is the field generated by $F \cup S$, not $F \cap S$. Commented Nov 29, 2023 at 21:39
• Okay, maybe this is a typo in my teacher's notes, thank you. Commented Nov 29, 2023 at 21:42
• 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? Commented Nov 29, 2023 at 21:46
• 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. Commented Nov 29, 2023 at 22:07
• Actually, it should work in the same way as $F(S) \subseteq F(T)$. Commented Nov 29, 2023 at 22:19

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 $$F(a_1,\ldots,a_n)\subseteq F(a_1,\ldots,a_{n-1})(a_n).$$
• @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 $F(a_1,\ldots,a_{n-1})(a_n)$.