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It's well-known that $\mathbb Q(\sqrt{2},\sqrt{3}\,\dots \sqrt{n},\dots)$ is an algebraic extension over $\mathbb{Q}$, but how to prove it?. Here is my attempt: actually, I guess this statement is true: if $\alpha_1,\alpha_2,\dots,\alpha_n,\dots$ are countable infinitely algebraic elements over $F$, then $F(\alpha_1,\alpha_2,\dots,\alpha_n,\dots)$ is algebraic extension over $F$.

It's true when we have just finite elements. For countable infinitely case, let $\beta\in F(\alpha_1,\alpha_2,\dots,\alpha_n,\dots)$ be arbitrary element, it must be written by finite many elements in $F\cup\{\alpha_1,\alpha_2,\dots,\alpha_n, \dots\}$(i.e. it looks like $(\alpha_1+\alpha_2)^{-1}\alpha_3-\alpha_4)$, thus $\beta\in F(\alpha_{i_1},\alpha_{i_2},\dots,\alpha_{i_m})$, thus $\beta$ is algebraic element. Then we have$F(\alpha_1,\alpha_2,\dots,\alpha_n,\dots)$ is algebraic extension over $F$.

$\mathbb Q(\sqrt{2},\sqrt{3}\,\dots \sqrt{n},\dots)$ is an algebraic extension over $\mathbb{Q}$, since for every $n$, $\sqrt{n}$ is algebraic element.

Is the statement true? I have something strange feeling about my proof, "any element can be written by finite many elements in $F\cup\{\alpha_1,\alpha_2,\dots,\alpha_n, \dots\}$" seems weird, I claimed this by using intuition and lack of explanation. Is there another way to interpret the seemingly obvious fact?

Thanks for your attention, any answers will be appreciated!

Greg Martin
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MGIO
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    That is just the definition of the countable extension. – Qiaochu Yuan Jan 18 '25 at 19:01
  • @Qiaochu Yuan Please forgive my ignorance, I don't know about countable extension, and I can't find definition of it from Abstract Algebra by Dummit, can you provide some online resourece about it? – MGIO Jan 18 '25 at 19:13
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    Just consider any set $A$ of algebraic numbers, so that $A$ is countable and then $\mathbb{Q} (A) $ is algebraic extension of rationals. This is because every element of this extension is algebraic. – Paramanand Singh Jan 18 '25 at 19:14
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    In general, if $S$ is a set (doesn't even have to be countable) then $F(S)=\bigcup F(S')$ where the union goes over the finite subsets $S'\subseteq S$. Try to prove this. And then, if all elements of $S$ are algebraic then $F(S)/F$ is an algebraic extension, because any $x\in F(S)$ belongs to some $x\in F(S')$ for a finite $S'$, and so algebraic over $F$. – Mark Jan 18 '25 at 19:23
  • @Mark: Your comment would make a nice answer. – Lee Mosher Jan 18 '25 at 19:29

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In general, if $S$ is a set (doesn't even have to be countable) then $F(S)=\bigcup F(S')$ where the union goes over the finite subsets $S'\subseteq S$. Try to prove this. And then, if all elements of $S$ are algebraic then $F(S)/F$ is an algebraic extension, because any $x\in F(S)$ belongs to some $x\in F(S')$ for a finite $S'$, and so algebraic over $F$.

Mark
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    Please don't answer duplicate questions. This is a basic question about algebraic extensions. Using site search will easily locate several duplicate targets. Please also read the meta announcement regarding quality standards. (And while Lee's comment is basically correct that comments should not be used to give answers, answers should only be given for new questions.) – Martin Brandenburg Jan 18 '25 at 19:42
  • Thanks for your useful answer and your kindness! But I can only prove it by using the same argument above: every elements in $F(S)$ can be written finite elements in $F\cup S$. Is there another way to prove it? – MGIO Jan 18 '25 at 20:16
  • @MGIO The union is clearly contained in $F(S)$ by definition. For the converse, you might show that the union is by itself a field which contains $F$ and $S$, and hence contains $F(S)$. This is simple. For example, if $x\in F(S')$ and $y\in F(S'')$ for some finite $S', S''\subseteq S$ then $x+y\in F(S'\cup S'')$, and so the union is closed under addition. And so on... – Mark Jan 18 '25 at 21:39
  • I understand it! 1. $\bigcup F(S')$ is a field, since for $x\in F(S_1), y\in F(S_2)$, $x-y\in F(S_1\cup S_2)$, the same as multiplication, and we also have $x^{-1}\in F(S_1)$. (we need not to check addition right?)2.$\bigcup F(S')$ contains $F$ obviously, and contains $S$, since every single elements is finite set. So $\bigcup F(S')$ is a field contains $F$ and $S$, and by definition of $F(S)$, we have $F(S)\subset \bigcup F(S')$. Thanks for your kindness and patience again! – MGIO Jan 18 '25 at 22:05