By Genov G.K.
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Extra info for A basis of identities of the algebra of third-order matrices over a finite field
N; we can write all the fractions κi,j with a common denominator s (outside p); consequently there exists a family (λi,j ) of elements of K such that the following equalities hold for a suitable t also lying outside p : r (st)xj = λi,j xi for j = r + 1, r + 2, . . , n; i=1 if the prime ideal q does not belong to V(Kst), then st is invertible in Kq and therefore x1 , x2 , . . , xr still give a family of generators of Mq (not necessarily a minimal family); this proves that rk(q, M ) ≤ r for all q in the open subset complementary to V(Kst).
3)). Let N be a free module with basis (e1 , e2 , . . , en ), and f the surjective morphism N → M that maps each ei to xi ; let us consider R = Ker(f ). Since the kernel of N → M → M/mM is exactly mN , we know that R ⊂ mN . Let us prove that the injectiveness of the morphism m ⊗ M → M implies R = mR, since this immediately leads to the awaited conclusions: R = 0, therefore f is an isomorphism, M is free, and the minimal generating family (x1 , x2 , . . , xn ) is a basis. Indeed since R ⊂ mN , for n every y ∈ R there exist µ1 , µ2 , .
Proof. 7) we derive (b)⇔(a)⇒(c), and it is obvious that (d)⇔(e), because for a subset of elements s of K the following three assertions are equivalent: – it is contained in no maximal ideal m; – it generates K as an ideal; – it contains a ﬁnite subset generating K as an ideal. Therefore it suﬃces to prove (c)⇒(d) and (e)⇒(a). Let us prove (c)⇒(d). Let m be a maximal ideal, r the rank of P at m, and (x1 , x2 , . . , xn ) a ﬁnite family of generators of P such that (x1 /1, x2 /1, . . , xr /1) is a minimal family of generators of Pm .