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Let $R$ be the ring of continuous functions on $[0,1]$ under pointwise addition and multiplication, $c \in [0,1]$, and $M_c$ the ideal defined by the set of $f\in R$ that vanish at $c$.

It is true that $M_c$ is a maximal ideal (I've seen several proofs), but I am having trouble explicitly identifying the inverse of the nonzero elements of $R/M_c$.

Suppressing the dependence on $c$, the element $(M + f)$ has inverse $(M + 1/f)$ only if $f \notin M$ and if $f$ is non-zero on all of $[0,1]$. We only know it's nonzero at $c$, and I don't know how to show if $f$ is not in $M$, then it's nonzero on all of $[0,1]$. Thanks!

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up vote 4 down vote accepted

Note that $f+M_c=g+M_c$ if and only if $f(c)=g(c)$, so $f+M_c=f(c)+M_c$, where $f(c)$ is used in the last equation to denote the constant function whose value is $f(c)$. Then $\frac{1}{f(c)}+M_c$ is the inverse when $f(c)\neq 0$, which is equivalent to $f\not\in M_c$.

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You can also just see that it is the kernel of a homomkrphism $\phi:R\rightarrow \mathbb R$ defined as $\phi(f) = f(c)$. It's easy to show that this is a homomorphism, that it is onto $\mathbb R$, and that the kernel is your ideal. Since the reals are a field, that ideal is maximal.

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From the question: "It is true that $M_c$ is a maximal ideal (I've seen several proofs)...." I wonder if this is one of them. But this proof could also be turned into an answer to the question asked here from a slightly different perspective, since we know how to find inverses in $\mathbb R$. The inverse of $\phi(f)$ is $\frac{1}{\phi(f)}$ in $\mathbb R$, and to find $g$ such that $\phi(g)=\frac{1}{\phi(f)}$, we only need $g(c)=\frac{1}{f(c)}$. For such $g$, $g+M_c=(f+M_c)^{-1}$. – Jonas Meyer Feb 10 '12 at 15:06
Ah, yes, that's the problem with reading a question on the way home, then answering it an hour later without re-reading it. :) But this proof shows the nature of your proof - that is, the fact that the map $\phi$ is onto is the key, and that is shown with constant functions, which is also how you show that inverses exist... – Thomas Andrews Feb 10 '12 at 15:41

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