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7330hw4

# 7330hw4 - Math 7330 Functional Analysis Notes/Homework 4...

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Math 7330: Functional Analysis Fall 2002 Notes/Homework 4: Commutative Banach Algebras I A. Sengupta 1. Let R be a commutative ring with multiplicative identity e . A subset S R is an ideal of R if :(a) 0 S , (b) x + y S for every x, y S , and (c) rx S for every r R and x S . The ideal S is a proper ideal if S 6 = R . It is a maximal ideal if it is a proper ideal and if the only ideals containing S are S itself and the whole ring R . The ideal S is a prime ideal if for every x, y S if xy S then at least one of x and y must be in S . (i) Let I be an ideal of R . For any x R we write x + I be the set of all elements of the form x + i with i running over I . Let R/I be the set of all sets of the form x + I with x running over R : R/I def = { x + I : x R } Let p : R R/I : x 7→ x + I For any elements a, b R we have p ( a ) = p ( b ) if and only if a - b I Show that if x, x 0 , y, y 0 R are such that p ( x ) = p ( x 0 ) and p ( y ) = p ( y 0 ) then p ( x + x 0 ) = p ( y + y 0 ) and p ( xy ) = p ( yy 0 ). Thus there are well-defined operations of addition and multiplication on R/I given by p ( x ) + p ( y ) def = p ( x + y ) , p ( x ) p ( y ) def = p ( xy ) As is readily checked, these operations make R/I a ring and, of course, p : R R/I is a ring homomorphism. Commutativity of R implies that R/I is commu- tative. If e R is the identity of R then p ( e ) is the multiplicative identity in R/I . 1

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(ii) Suppose I is a maximal ideal of R . Show that then the commutative ring R/I is a field , i.e. every non-zero element has an inverse. Hint: Let x R be such that p ( x ) is a non-zero element of R/I , i.e. x R is not in the ideal I . The set Rx + I = { rx + y : r R, y
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7330hw4 - Math 7330 Functional Analysis Notes/Homework 4...

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