By V. A. Krechmar

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P. The composite AG ! A ! AG is then just multiplication by |G|. Changing the N base along AG ! R, the composite R-linear map R ! A AG R ! R is still just 57 multiplication by |G|. Since |G| is a non-zerodivisor in R, this R-linear map R ! R N is injective, and therefore the map R ! A AG R must be injective as well. 4. Let A = R[x]/(m(x)) be a monogenic degree-n extension of R, and let G ✓ Sn with |G| a non-zerodivisor in R. Then isomorphism classes of G-closures of A over R correspond to R-algebra homomorphisms R[x1 , .

R[x1 , . . , xn ] ! A⌦n ! B = R[x1 , . . , xn ]G ! R[x1 , . . , xn ] ! B = R[x1 , . . , xn ]G ⇣ R ,! B. The image of (A⌦n )G in B is thus contained in R, and we obtain a factorization of R[x1 , . . , xn ]G ! R as R[x1 , . . , xn ]G ! (A⌦n )G ! R. Then the R-algebra homomorphism (A⌦n )G ! 1, the corresponding G-closure of A is B. 2, we need criteria for proving that R ! R[x1 , . . ,xn ]G is injective. 3. Let R be a ring, let n be a natural number, and let G be a finite group acting on an R-algebra A.

6, the Sn -closure of RS is the quotient of (RS )⌦n ⇠ = RHom([n],S) by the ideal I generated by {e([(e⇡1 , . . , e⇡n )]) s([(e⇡1 , . . , 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)).

### A Problem Book in Algebra by V. A. Krechmar

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