Proof on showing an ideal of the set of all real functions is maximal Let $R$ denote the ring of all real valued functions $\{f \colon\mathbb{R}\to\mathbb{R} \}$
Let $S$ denotes the ideal of all functions such that $f(t) = 0$ at $t$
Goal: Show $S$ is maximal ideal using definition
i.e. Let $T$ be a set such that $S\subset T$, then we want to show $T = R$ by showing the constant $f(x) = 1$ function exists in $T$
** My proof so far:
(1) By how $R$ is defined, I assume the functions in $R$ don't have to be continuous (although I think that will make things much easier)
(2) Let $T$ be a set such that  $S\subset T$.
Then I claim there exists a function $g$ in the set $T - S$ (or $T\setminus S$ in other notation), such that $g(t)$ is not $0$ at $t$
(3) Let $f(x) \in S$.  Thus $f(t) = 0$ at $t$
WLOG, I assume in (2), $g(t) > 0$.
So I have $f(t) - g(t) < 0$
(4) Now this is the part that I'm not so sure about.
I plan to say the following:

for $f(x) - g(x)$ not being $0$, i.e. $f(x) - g(x) > 0$  or $f(x) - g(x) < 0$
==> the multiplicative inverse for $f(x) - g(x)$, namely $1 / [f(x) - g(x)] $ must exist. (*)

And this inverse is in $T - S$.  As a result, $1$ must be in $T$, and therefore $T = R$
But I don't know if the idea in (*) is valid.  Well, if $R$ only contain continuous, real-valued functions, I think I can use the "delta-epsilon" idea to argue that if a function $f(x)$ is not $0$, the inverse function of $f(x)$ must exist.  But since in this case, the functions in $R$ may not be continuous, can I still use that argument ??
Would someone please help me on this?  Thank you in advance
^_^
 A: Here's a much simpler (imo) proof which works also for the ring of continuous, or even derivable, functions:
1) Define $\,\phi:R\to\Bbb R\;\;,\;\;\;\;\phi(f):=f(t)\,$ . Check this is is an onto ring homomorphism.
Now apply the first isomorphism theorem and remember that in a commutative unitary ring an ideal is maximal iff the corresponding quotient ring is a field...
A: Your step at (*) is indeed not fully satisfying.
The same argument (when corrected, see below) works if we consider the ring $R$ of functions $f\colon X\to F$ where $X$ is an arbitrary set, $t\in X$ an arbitrary element, and $F$ is an arbitrary field. It also works for any subring of $R$ (e.g. continuous functions, periodic, Lebesgue integrable, ...) , provided for each $a\in F$ there exists $f\in R$ with $f(t)=a$ (§).
Of course the much simpler proof of maximality is to observe that $\phi\colon R\to F$, $f\mapsto f(t)$ has $S$ has kernel and $F$ (i.e. a field) as image.
What you are supposed to find here as "direct" proof from the definition does not really differ:
You consider an ideal $T$ with $S\subset T\subseteq R$.
The set $I:=\{g(t)\mid g\in T\}$ is an ideal of the ring $F$:
It is closed under addition (if $i_1, i_2\in I$, there are $g_1, g_2\in T$ with $i_1=g_1(t)$ and $i_2=g_2(t)$; then $i_1+i_2=(g_1+g_2)(t)$ with $g_1+g_2\in T$) and under scalar multiplication with elements of $F$ (if $a\in F$, there exists by assumption (§) above (this is a very crucial step) an $f\in R$ with $f(t)=a$; if $i\in I$ and $g\in T$ with $g(t)=i$, then also $fg\in T$, hence $f(t)g(t)=ai\in I$). 
By definition of $S$ and because $T$ is strictly bigger, we see that $I\ne\{0\}$.
Because $F$ is field and only has the ideals $0$ and $F$, we conclude $I=F$.
Now let $f\in R$ be arbitrary.
Then by definition of $I$, there exists some $g\in T$ with $g(t)=f(t)$.
Then $h:=f-g$ has the property that $h(t)=f(t)-g(t)=0$, hence $h\in S$ and also $h\in T$. Then also $f=g+h\in T$.
As $f$ was an arbitrary element of $R$, we see that $T=R$.
A: This still works for continuous functions. Note that $1/x$ is continuous on $\mathbb R\setminus \{0\}$, so if $f:\mathbb R\to\mathbb R$ is continuous and everywhere nonzero then $1/x\circ f = \frac{1}{f(x)}$ is continuous.
