Vtyjkyuk

Problem

Let $R$ be a ring of continuous functions from $mathbbR$ to $mathbbR$. 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?

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.

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 $varphi colon R to mathbbR$ 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
$$ R/!A = R/ker(varphi) cong operatornameim(varphi) = mathbbR$$
is a field.
This shows that $A$ is maximal.

Hint: Use the first isomorphism theorem and a well chosen map $R to mathbbR$.

Notice that $C(mathbbR)$ is a vector space and that $A dotplus mathbbR1 = C(mathbbR)$.

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

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

$$1 = frac1lambda(f-g) in I$$

so $I = C(mathbbR)$.