By Drozd Yu.

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**Example text**

Replacing B by B/q and A by A/(q ∩ A) , we may suppose that q = { 0 } and both B and A are integral. Then we have to show that p ∩ A = { 0 } . Take any non-zero element a ∈ p and consider an equation am + b1 am−1 + · · · + bm with bi ∈ A of the smallest possible degree. Then bm = 0 and bm ∈ p ∩ A . As m is maximal, p ∩ A = m for every prime ideal p ⊂ B containing m . Take a maximal ideal n ⊇ p . , all prime ideals from B containing m are maximal. Thus, maximal ideals √ containing m are just minimal among the prime √ ideals containing mB , or, the same, the prime components of mB (cf.

The main origin of local rings in algebraic geometry are the stalks of the sheaves of regular functions. Remind the corresponding definition. 2. Let F be a sheaf on a topological space X , p ∈ X . The stalk Fp of the sheaf F at the point p is, by definition, the direct limit limU p F(U ) . In other words, Fp is defined as the set of −→ the equivalence classes of U p F(U ) under the following equivalence relation: a ∼ b , where a ∈ F(U ), b ∈ F(V ), if and only if there is U V W ⊆ U ∩ V such that FW (a)FW (b) .

4 is an obvious consequence of 3 . We shall also use the following result. 4. Let Q be a finitely generated extension of an algebraically closed field K , n = tr. deg(Q/K) and R = K( x1 , . . , xn ) . Then either Q R or Q R(α) , where α is algebraic and separable over R . Proof. Let Q = K( α1 , . . , αm ) . We use the induction on m . For m = 1 , the claim is obvious. Suppose that it holds for L = K( α1 , . . , αm−1 ) . Put l = tr. deg(L/K) and S = K( x1 , . . , xl ) . Then we can suppose that either L = S or L = S(β) with β algebraic and separable over S .