By Owen Biesel

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**Example text**

This G-closure canonically induces an H-closure of A over R, namely 0 @A⌦n Proof. Indeed, ' (A⌦n )H O R, ' (A⌦n )H (A⌦n )H 1 A. : (A⌦n )H ! R is normative because ' is, and the above is the resulting H-closure. In particular, every G-closure induces the canonical Sn -closure. In contrast, a 1closure rarely exists—there must be a normative R-algebra homomorphism A⌦n ! R, whereas typically there is no homomorphism at all—but if an R-algebra A does have a 1-closure, then A also has a G-closure for every subgroup G ✓ Sn .

R. (A⌦n )Sn This alternative parametrization will be very useful in Chapter 6, when we explic- itly compute this tensor product to check for homomorphisms to R. But for now, we will prove a generalization pointed out by L. 1. Let R be a ring, and let A be a degree-n extension of R. Let G be a subgroup of Sn . For each R-algebra R0 , there is a canonical correspondence between isomorphism classes of G-closures of R0 ⌦ A over R0 and R-algebra homomorphisms O (A⌦n )G R ! R0 . (A⌦n )Sn In other words, the functor F from R-algebras to sets, where F sends R0 to the set O of isomorphism classes of G-closures of R0 ⌦A over R0 , is represented by (A⌦n )G R.

E⇡n )]) : ⇡ 2 Hom([n], S)} . 35 With s([(e⇡1 , . . , e⇡n )]) evaluated according to the first part of the lemma, and employing the isomorphism (RS )⌦n ⇠ = RHom([n],S) , we can write this generating set as 80 < @ : X ⇡2Bij([n],S) 1 e⇡ A 9 ( ! ) = X 1 [ e⇡ : O 2 Hom([n], S)/Sn , O = 6 Bij([n], S) . ; ⇡2O We show that this ideal is equal to J := (e⇡ : ⇡ 2 / Bij([n], S)). To show that e⇡ 2 I whenever ⇡ is not a bijection, note that X 1 ⇡ 0 2Bij([n],S) 0 @ e⇡ e⇡0 2 I, so X ⇡ 0 2Bij([n],S) 1 e⇡0 A e⇡ 2 I, but the second term vanishes because e⇡0 e⇡ = 0 whenever ⇡ 0 6= ⇡, which here is always the case since ⇡ 0 is a bijection while ⇡ is not.