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In paragraph V.1 of Algebra proposition 1.7 Lang claims that the class of algebraic extensions is distinguished. I know that if $F/k$ and $E/F$ are algebraic extensions than so is the $E/k$ - that is easy to prove. However, the second required property is that if $E/k$ is algebraic and $F/k$ is arbitrary, then (assuming the compositum is defined) $EF/F$ is algebraic. Lang more or less skips the proof of this, only saying that "an element remains algebraic under lifting, and hence does the extension."

However, there are no finiteness conditions here and no element given in advance. Although $EF$ is generated over $F$ by a set of elements which are algebraic over $F$ (namely, all the elements of $E$), there are no facts known (up to this point in the book at least) about the infinitely generated extensions.

What am I missing?

Thank you.

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marked as duplicate by user26857 abstract-algebra Jun 23 at 5:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If it helps, every element $\alpha$ of $EF/F$ is contained in a finitely-generated extension of $F$ (e.g. $F(\alpha)$) and can be expressed in terms of finitely many elements from $E$. – Hurkyl Jan 14 '13 at 13:18
That $F(\alpha)$ is finitely generated is clear. Why can it be expressed in terms of finitely many elements from $E$? And how? – Artem Jan 14 '13 at 13:33
up vote 4 down vote accepted

In Lang's definition of distinguished, both conditions (2)-(3) require that $\,k\subset E,F\,$ and both $\,E,F\,$ are contained in some object $\,R\in\mathcal C\,$ , so $\,F/k\,$ cannot be "arbitrary":

$\,\Bbb Q/\Bbb Q\,$ is algebraic and $\,\Bbb Q(\pi)/\Bbb Q\,$ is any extension, but $\,\Bbb Q\cdot\Bbb Q(\pi)/\Bbb Q=\Bbb Q(\pi)/\Bbb Q\,$ is not algebraic.

Added: The elements of $\,EF\,$ can be seen as rational functions in the elements of $\,E\,$ with coefficients of $\, F\, $ (and the other way around, of course).

Let $\,x\in EF\,$. Then there exists a finite number of elements in $\,E\,$ , say $\,e_1,...,e_r\,$ which actually appear in the above rational expression of $\,x\,$, and thus $\,x\in F(e_1,...,e_n)\subset EF\,$.

Since clearly $\,F(e_1,...,e_n)/F\,$ is algebraic we're done.

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My book (revised 3rd edition, don't know which printing, but I got it new a couple of months back from amazon, if that's any help) "(2) If $k\subset E$ is in $\mathcal C$, if $F$ is any extension of $k$, and E,F are both contained in some field, then $F\subset EF$ is in $\mathcal C$". – Artem Jan 14 '13 at 13:35
I sent it unfinished by mistake. There is now a quote. The property (2) does not require that both $E,F$ are members of an element of $\mathcal C$, only that they are both subfields of some other field. – Artem Jan 14 '13 at 13:39
In the example you gave above Langs property (2) states that $\mathbb Q\cdot\mathbb Q(\pi)/\mathbb Q(\pi)$ is algebraic. It does not require that $\mathbb Q\cdot\mathbb Q(\pi)/\mathbb Q$ be algebraic. – Artem Jan 14 '13 at 13:42
That's true, @Artem. I realized that after your completed first message appeared. – DonAntonio Jan 14 '13 at 14:03
Thank you, @DonAntonio! – Artem Jan 14 '13 at 15:39

If $E/F$ is an algebraic extension, $K/F$ is an extension. Prove that $EK/K$ is algebraic.

Assume $H=\{a\in EK|a \text{ is algebraic over } K\}$. It is easy to see that $H$ is a field and $K<H<EK$. $E/F$ is algebraic, so all the elements in $E$ are algebraic over $F$, also over $K$; so $E\subseteq H$; then we have $EK\subseteq H$, which means $EK=H$. $EK$ is an algebraic extension over $K$. ■

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The proof given by the previous answer is quite clear: when add (adjoint, if you prefer) algebraic elements to a field, then get an algebraic extension. Moreover, this fact is easily proven than "$H$ is a field". – user26857 Jun 23 at 5:31

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