By G Lefort

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**Extra info for Algebra & Analysis, Problems & Solutions**

**Sample text**

The ideals Ii are prime and they are uniquely determined by I. They are called the ideals associated with I. (46) Since f-1(rad J) = rad f-1(I) for any ring homomorphism f. (47) In such a case we say that the intersection J 1 n ... n Jk is irreducible. 35 §9. Noetherian rings The minimal elements of the set {II"'" h} are called the isolated ideals for I. They are precisely the minimal elements of the family of prime ideals containing I. Every prime ideal that contains I also contains an isolated ideal for I.

6). The same argument proves that the union of all ideals associated with an ideal I of the ring A is equal to the set of zero divisors in AI I. 3a. By a ring without nilpotents we mean a ring in which t k o (48) . = 0 =} t = PROPOSITION. If a noetherian ring A is without nilpotents) then its ring of fractions R is isomorphic to a finite Cartesian product of fields (49). PROOF . Take an irreducible primary decomposition of zero. 5) 0 = n~ Ii, where Ii are prime and (by deleting some of the Ii'S) we may assume that it is an irreducible primary decomposition.

Now let X E B. Then X = R( 0, where R E I<[X] (since L = I<[(]). The elements Xi = R((i), i = 1, ... , k, are also integral over A. ) Let us consider the following polynomial from L'[X]: k Q k = I:Xi II(X 1 v,pi (v) = I:R((i) II(X 1 (v). v,pi Its coefficients are values taken at ((1, ... , (n) by symmetric polynomials from I<[Xl"" ,Xk]' Hence (by the theorem on symmetric polynomials) they must belong to 1\. Moreover, they are integral over A (because Xi and (v are integral over A). , Q E A[X]. Consequently, xP'(O = Q(O E A[(].