By Papaioannou A.
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Extra resources for Solutions to Atiyah and MacDonald's Introduction to Commutative Algebra
Example text
Mr be the maximal ideals that contain a. Let x0 be a non-zero element of a and let m1 , m2 , . . mr+s be the maximal ideals that contain x0 . Since mr+1 , r + 2, . . , mr+s do not contain a, there exist xj ∈ a, 1 ≤ j ≤ s, such that xj ∈ / mj . Since each Amj is Noetherian, the extension Amj a of a in Amj is finitely generated; let xi , x2 , . . , xt be the elements of a whose images generate Ami a, 1 ≤ i ≤ r. Let a0 = (x0 , . . , xt ); we observe that a0 and a have the same extension in Am for al maximal ideals m (since they do in the finite number of ideals m1 , .
Therefore, if L = F ∩ G is any locally closed subset of X, then the proper subspace U ∩ X0 of U , locally satisfies condition (3), therefore in particular the intersection of the set (F ∩ U ), which is closed in U , with U ∩ X0 is non-empty. But then we have ∅ = (F ∩ U ) ∩ (U ∩ X0 ) = L, and this yields the desired result. A subset X0 of X that fulfills these conditions is called very dense. For the algebraic equivalences, we have: (i) ⇒ (ii) Let L be any locally closed set in Spec(A). This can be precisely characterized as the set of prime ideals of A that contain some fixed ideal α of A but do not contain some fixed a ∈ A.
Xn−1 ], hence there exists a polynomial f with coefficients in k such that f (x1 , x2 , . . , xn ) = 0. We may of course assume that the homogeneous part F with the largest degree d of f is monic in the last argument (xn ), because, since k is infinite, there are λ1 , λ2 , . . , λn−1 ∈ k such that F (λ1 , λ2 , . . , 1) = 0, and this will be the coefficient of xn in F so we may divide by it. Putting xi = xi − λi xn , we obtain from F a monic polynomial in k[x1 , x2 , . . , xn−1 ] that vanishes at xn , hence A = k[x1 , x2 , .