In the ring of continuous functions $\mathbb{R} \to \mathbb{R}$, the set of all $f$ with $f(0) = 0$ is a maximal ideal.

Question

Problem

Let $R$ be a ring of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Show that $A = \{f \in R \mid f(0)=0\}$ is a maximal ideal of $R$.

If $f,g \in A$, then $f-g(0) = f(0)-g(0) = 0$ . So $f-g$ is in $A$. Also $f(0) h(0) = 0$ for all $h \in R$. So $A$ is an ideal.

Attempt to show $A$ is maximal

If $R/A$ is a field then $A$ is maximal. Let $f \notin A$. Then $f(0) \neq 0$. Let $g(x)=0$. Then $f-g \in A$. So $f+A = g+A$. $1/g$ is the inverse of $g$. So $f$ has unity (not sure about this).

Is this correct?

Answer 1

Your proof that $A$ is an ideal works, although strictly speaking, you also need to check that $0 \in A$ (which shouldn’t be a problem).

Your proof that $A$ is maximal does not work in its current form. You’re right that it sufficies to show that $R/\!A$ is a field, and that for this you need to show that for $f \in R$ with $f \notin A$ the element $f + A \in R/\!A$ has an inverse. The problem is that your choice of $g$ and how you work with it makes no sense:

• You choose $g(x) = 0$, i.e. $g = 0$ is the (constant) zero function. But then $$ (f - g)(0) = f(0) - g(0) = f(0) \neq 0 $$ and therefore $f - g \notin A$ (and thus also $f + A \neq g + A$).

• The function $1/g$ does not exists.

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Your approach can be fixed by making the right choice for $g$:

For $c := f(0)$ you choose $g$ as the constant function $g(x) = c$. Then $$(f - g)(0) = f(0) - g(0) = c - c = 0,$$ so that $f - g \in A$ and thus $f + A = g + A$. The function $g$ has in $R$ a multiplicative inverse $1/g$ given by $(1/g)(x) = 1/c$ because $c \neq 0$. It follows that $1/g + A$ is a multipicative inverse to $g + A = f + A$ in $R/\!A$.

A different approach to the problem would be to use a suitable isomorphism theorem:

The set $A$ is the kernel of the ring homomorphism $\varphi \colon R \to \mathbb{R}$ given by $\varphi(f) = f(0)$. This already shows that $A$ is an ideal. The ring homomorphism $\varphi$ is surjective, so it follows from the first isomorphism theorem that $$ R/\!A = R/{\ker(\varphi)} \cong \operatorname{im}(\varphi) = \mathbb{R}$$ is a field. This shows that $A$ is maximal.

Answer 2

Hint: Use the first isomorphism theorem and a well chosen map $R \to \mathbb{R}$.

Answer 3

Notice that $C(\mathbb{R})$ is a vector space and that $A \dotplus \mathbb{R}1 = C(\mathbb{R})$.

Let $I$ be an ideal in $C(\mathbb{R})$ such that $A \subsetneq I$. Notice that $I$ is also a subspace of $C(\mathbb{R})$, because $\lambda f = (\lambda 1) f$.

Pick $f \in I \setminus A$. There exist unique $g \in A$ and $\lambda \in \mathbb{R}$ such that $f = g + \lambda 1$. Since $f \notin A$ we have $\lambda \ne 0$. Therefore

$$1 = \frac1{\lambda}(f-g) \in I$$

so $I = C(\mathbb{R})$.