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Let $L$ be a field, and $\{ E_i \}_{i \in I}$ be a family of subfields of $L$, that is, $E_i \subset L$ for each $i \in I$. Then, we define the compositum of the $E_i$'s to be the smallest field contained in $L$ that contains each of the $E_i$'s. More precisely, we define the compositum of the $E_i$'s to be $$ \bigcap_{\substack{E_i \subset K \subset L \\ \text{for all } i \in I}} K, $$ and it is easy to show that this is a field which is contained in any subfield of $L$ that contains $E_i$ for all $i \in I$.

My question is whether there is any standard notation to denote the compositum of a family of fields as above. I already know the following:

  1. If $I = \{ 1, \dots, n \}$, then the compositum of $E_1,\dots,E_n$ is usually denoted by $$E_1 \cdots E_n.$$
  2. If each $E_i$ is an extension of a field $k$, then one can denote the compositum by $$k(S),\quad \text{where } S = \bigcup_{i \in I} E_i.$$

But, if I don't have such a $k$ and if $I$ is an arbitrary indexing set, then is there a standard notation for the compositum? I have been reading Serge Lang's Algebra (3rd edition) and Patrick Morandi's Field and Galois Theory and no notation for the compositum (in the general scenario) is given in them.

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  • $\begingroup$ You can always choose $k$ as the prime field of $L$ and then use your second notation. But that is not super succinct. $\endgroup$ – Eike Schulte Aug 24 '18 at 19:48
  • $\begingroup$ @EikeSchulte ah yes, that didn't occur to me... A notation without making reference to the prime subfield of $L$ would be great, but your suggestion might be the next best thing. $\endgroup$ – Brahadeesh Aug 24 '18 at 19:54
  • $\begingroup$ @EikeSchulte if you make your comment into an answer then I'll accept it :) It doesn't look like this question will receive any more attention. $\endgroup$ – Brahadeesh Sep 5 '18 at 10:17
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You can always choose a $k$ for your second notation: both the prime field of $L$ and the intersection of all the $E_i$ would work.

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