Assignment 5 – Part 1 – Math 413/612 (All-1) the Total Ring Of

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Assignment 5 { Part 1 { Math 413/612 (All-1) The total ring of fractions. Let R be a commutative ring and let Z ⊆ R be the set of zero-divisors in R. Let S := R n Z. Show that S is a multiplicative subset of R. (Remark: the ring S−1R is called the total ring of fractions of R because it is the smallest ring containing (an isomorphic copy of) R in which every element of R that can be inverted is invertible. Indeed, one can show that if R is a subring of R0 and u is an element of R that is invertible in R0, then u cannot be a zero-divisor; furthermore, we saw in class that the natural homomorphism R ! S−1R is injective if and only if S contains no zerodivisors. When R is an integral domain, the total ring of fractions is the field of fractions.) (All-2) Localization of a module. Let M be an R-module and let S ⊆ R be a multiplicative subset. In class, we defined a relation ∼ on M × S via (m; s) ∼ (n; t) if there is a σ 2 S such that σ(tm − sn) = 0: Show that this is indeed an equivalence relation on M × S. (All-3) (WI) Local rings. Let R be a commutative ring. It is called a local ring if it has a unique maximal ideal. So, for instance, every field is a local ring since 0 is a maximal ideal in that case. Also, Z is not a local ring since 2Z and 3Z are both maximal ideals. (a) Let R be a commutative ring and let P E R be a prime ideal. Show that the localization RP of R at P is a local ring with maximal ideal PRP (i.e. the extension of P to RP ). 2 (b) Let F be a field and let I E F [x] be the principal ideal I = (x ). Let := x + I 2 F [x]=I. Show that F [x]=I is a local ring with maximal ideal the principal ideal generated by . (Hint: show that every element in F [x]=I can be written as a + b with a; b 2 F and that a + b is a unit in F [x]=I if and only if a 6= 0.) (Remark: the ring F [x]=I is usually denoted F [] and called the ring of dual numbers.) 413 only: None (yet). 612 only: None (yet). 1.

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