In EGA IV, Lemma, the first sentence is

Soit $R$ un anneau composé direct d'un nombre fini de corps.

I'm guessing this means

Let $R$ be a ring that is the direct product of a finite number of fields,

but I'm not positive. For one thing, "produit direct" seems to be the more usual term for "direct product." I've tried looking up "composé direct" on Google and on Wikipedia.fr, and I have not been able to find a definitive reference. I'm also confused by the fact that "direct" appears to be an adjective in the French, but I can't see what noun it might be modifying if my translation is correct. Google translate gives

Let R be a ring composed of a direct finite number of bodies.

Obviously, "bodies" should be "fields," but it still does not seem to be helpful. Thus, my question:

Question: What does "composé direct" mean in this context? In particular, why is "direct" being used as an adjective?

Secondary Question: Is there a good reason that "composé direct" is being used here rather than another, more searchable term such as "produit direct"?

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    $\begingroup$ I'm French. I believe your translation is correct. I don't know why the term "produit direct" is not used (maybe just to avoid monotony) but the adjective direct obviously modifies the noun "compose" exactly as in "direct product" in english. I guess what is meant should be clear if you look at the proof of the lemma. $\endgroup$ Nov 27 '11 at 20:11
  • $\begingroup$ Geoffroy: Thanks. I did not know that "composé" could be a noun; I was assuming it was a participle, as in the Google translation. $\endgroup$ Nov 27 '11 at 20:15
  • $\begingroup$ Dear Charles, I can't find the Lemma you are talking about. Could you please tell in what volume of EGA IV it is and on what page? $\endgroup$ Nov 28 '11 at 17:12
  • $\begingroup$ Georges: I had the wrong lemma; it's, not It is in EGA IV, volume 2, pages 169-170. $\endgroup$ Nov 29 '11 at 17:50

In Bourbaki 'composé direct' is the analog for algebras of 'somme direct' (direct sum) for modules (not direct product of modules). As pointed out by OP this makes sense as Bourbaki's algbras do not need to have a multiplicative unit [see also below]. In particular, while for the precise situation of OP 'direct product' would work as translation (regarding the meaning) this is (only) due to the fact that the number of fields is finite; if it were infinite the 'composé direct' would not be the direct product; but what I would call the direct sum of the fields. [As an aside: it seems to me, though I am in no way an authority on this, that Bourbaki uses rather just 'produit' and not 'produit direct.']

In more detail in Théorie des ensembles (E IV.32, §3) after the existance of a certain 'application universel' is established (under certain conditions) it is said that this applies to groups, modules and algebras. And

... est appelé produit libre des $A_i$ dans le cas des groupes, sommes directe dans le cas des modules, composé direct dans le cas des algèbres.

Meaning: ... is called 'produit libre' of the $A_i$ in the case of groups, 'somme direct' (direct sum) in the case of modules, 'composé direct' in the case of algebras.

In an earlier version of this answer, I possibly caused confusion, by saying this is also used for rings and fields. What I meant is that, as in the given passage or, e.g., in Bourbaki's Algèbre Commutative one can find the 'composé direct' terminology applied when the constituing structures are fields or rings. However, the quoted definition/terminology is just for algebras so that for this construction the ring or the field is merely considered as an algebra and the resulting structure (thus) does not have to be a field or ring.

Moreover, it seems to me that the property of this above mentioned 'application universel' is that of the coproduct in the relevant category. However, since I am not too firm in all category stuff and in particular rather unfamiliar with certain terminology of Bourbaki used there I do not vouch for this. (The reference is here, corrections welcome.) Yet, it is not the product; for various reasons including the soft ones that it is dicussed earlier and this would not make sense since for modules this precise construction is 'somme direct' and this certainly means only finitely many 'non-zero coordinates.'

  • $\begingroup$ Thanks for the explanation! It makes the terminology a bit more transparent. It also helps me know, before I put <i>too</i> much effort into understanding the proof, that I'm looking at the right statement. $\endgroup$ Nov 27 '11 at 21:32
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    $\begingroup$ Concerning 'direct sum' versus 'direct product': It seems to be quite common to talk about a ring being a 'finite direct sum of fields' (for instance, Mumford does this somewhere in the Red Book), but I'm pretty sure the term ought to be direct product rather than direct sum. My reasoning is quite simple: We want this "composé" to be a ring (algebra, field). For me, and most others working in algebraic geometry or commutative algebra, this means it needs to include $1$. An infinite direct product of rings contains $1$; an infinite direct sum does not. $\endgroup$ Nov 27 '11 at 21:32
  • $\begingroup$ As I understand it, Bourbaki does not require his algebras to have a unit, so perhaps you are correct that Bourbaki's definition corresponds to direct sum rather than direct product. But for the sort of rings that Grothendieck is discussing, I'm pretty sure that 'direct product' is a better choice of English terminology, since the direct product of rings is naturally a ring with unit but the direct sum is not. $\endgroup$ Nov 27 '11 at 21:37
  • $\begingroup$ You are welcome, my edit was before your comments; I will try to check the definition more carefully, but there is no room for inteprretation: either what I said is right and 'composé' is the coproduct in the category or it is the product and I am wrong. But in any case I would say 'composé direct' is what is defined there. $\endgroup$
    – user9072
    Nov 27 '11 at 21:40
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    $\begingroup$ The coproduct of $A,B$ in the category of rings without unit is $A\otimes B\oplus A\oplus B$. $\endgroup$
    – user2035
    Nov 28 '11 at 6:19

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