- #1

Mr Davis 97

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## Homework Statement

Let ##R## be the ring of all continuous real-valued functions ##f : [0,1] \to \mathbb{R}## with pointwise addition and pointwise multiplication of functions as its two operations. Let ##c \in [0,1]## and denote ##M_c = \{f\in R : f(c) = 0\}##.

a) Show that any ##f\in R## such that ##f(c) \not = 0## for all ##c\in [0,1]## is a unit in ##R##.

b) Show that ##M_c## is a maximal ideal of ##R##.

## Homework Equations

## The Attempt at a Solution

I need some pointers for this.

For a), it's clear that I need to produce some inverse function ##g## such that ##f \circ g = id## and ##g\circ f = id##. But I'm not seeing why the condition that ##f(c) \not = 0## for all ##c\in [0,1]## guarantees that ##f## has an inverse.

For b), I think I might have figured this out.

For each fixed ##c\in [0,1]## the map ##E_c: R \to \mathbb{R}## such that ##E_c(f) = f(c)## (called evaluation at ##c##) is a ring homomorphism because the operations in ##R## are pointwise addition and multiplication of functions. The kernel of ##E_c## is given by ##M_c = \{f\in R : f(c) = 0\}##. Also, ##E_c## is surjective: given any ##a\in \mathbb{R}## the constant function ##f(x)=a## maps to ##a## under evaluation at ##c##. Thus ##R/M_c \cong \mathbb{R}##. Since ##\mathbb{R}## is a field, ##M_c## is a maximal ideal.

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