By Alexander Schmitt

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**Extra info for Algebra I: Commutative Algebra [Lecture notes]**

**Sample text**

We have to verify the inclusion “⊃”. Pick r ∈ I and s ∈ J with r + s = 1 and let a ∈ I ∩ J. Then, the equations a = a · 1 = a · (r + s) = a · r + a · s ∈I·J ∈I·J shows that a is an element of I · J. 7 Remarks. i) The operations of taking intersections, sums, and products of ideals are commutative and associative, and the following distributive law holds: For ideals I, J, K ⊂ R : I · (J + K) = I · J + I · K. The reader may verify this as an exercise. ii) Let I, J, K be ideals of the ring R and assume J ⊂ I or K ⊂ I.

5, ii), pi ∼ qτ(i) in R[x], This gives the assertion and concludes the proof. , t. 7. 1 Proposition. Let R be a ring and N := a ∈ R | a is nilpotent . i) The subset N of R is an ideal. ii) The quotient ring R/N has no nilpotent element other than 0. Proof. i) Clearly, 0 ∈ N. Suppose a, b ∈ N and choose exponents m ≥ 1 and n ≥ 1 with am = 0 and bn = 0. For 0 ≤ i < m, we have m + n − 1 − i ≥ n. This shows that every summand of the right hand sum in m+n−1 m+n−1 (a + b) = i=0 m+n−1 · ai · bm+n−1−i i is zero, so that (a + b)m+n−1 = 0 and a + b ∈ N.

Proof. Assume that R[x] is not noetherian. Let I ⊂ R[x] be an ideal which is not finitely generated and X the set of all finite subsets of I. It is obviously non-empty. For x, y ∈ X, we write x < y, if x ⊂ y, #y = #x + 1, and the unique element f ∈ y \ x satisfies f x ∀g ∈ I \ x : ∧ deg(g) ≥ deg( f ). Since I is not finitely generated, it is clear that we find for each x ∈ X an element y ∈ X with x < y. 2, there is a sequence (xk )k∈Æ with xk ∈ X and xk < xk+1 , k ∈ Æ. This defines the sequence ( fk )k≥1 with { fk } = xk \ xk−1 , k ≥ 1.