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Algebra I: Commutative Algebra [Lecture notes] by Alexander Schmitt

By Alexander Schmitt

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I) We have ∅ = V( 1 ) and nk = V( 0 ). 1, iv). We also have I · J ⊂ I ∩ J, so that V(I · J) ⊃ V(I ∩ J). Let us show that V(I · J) ⊂ V(I) ∪ V(J). To this end, let p ∈ V(I · J) and assume p V(I). Then, we find a polynomial f ∈ I with f (p) 0. For every g ∈ J, we obviously have f · g ∈ I · J, so that ∀g ∈ J : Since f (p) f (p) · g(p) = ( f · g)(p) = 0. 0 and k is a field, this means ∀g ∈ J : g(p) = 0 and shows that p ∈ V(J). iii) It is readily verified that Ik . V(Ik ) = V k∈K k∈K Ik . 1, v). 3). Therefore, we say that a subset Z ⊂ nk is Zariski closed, if it is algebraic.

31 I. 3 Remark. , the axiom of choice in its full strength. 5). 8 Operations on Ideals Here, we will discuss various ways to construct new ideals from given ones. These constructions and their properties will be used throughout the following text. Intersections, Sums, and Products Let R be a ring, S an index set and (Is )s∈S a family of ideals in R indexed by the set S . Then, it is a straightforward task to verify that the intersection Is s∈S is also an ideal. 1 Lemma. Let R be a ring and X ⊂ R.

Since V(I(S )) is closed, we also find S ⊂ V(I(S )). Let S ⊂ Z ⊂ nk be a closed subset. , xn] with Z = V(I). We see Z = V(I) ⊃ V I(Z) ⊃ V I(S ) . The first inclusion is a consequence of I ⊂ I(Z) and the second one of I(Z) ⊂ I(S ) which, in turn, follows from Z ⊃ S . , xn ] −→ Algebraic sets in I −→ V(I). n k 43 I. 5, ii), shows Ψ ◦ Φ = id . In particular, Φ is injective and Ψ is surjective. 4 asserts that, if k is algebraically closed, then also Φ ◦ Ψ = id . Then, we have translated the theory of algebraic sets into the theory of ideals in rings.

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