$R=C[0,1]$ be a commutative ring with identity. Is my proof correct?

Question

So Today , I face a problem , Here it is. Let $R$ be a commutative ring with unity and $R=C[0,1]$. For each $c\in[0,1]$ Let, $M_{c}=\{f\in C[0,1] | f(c)=0\}$ Show that ,

NOTE BEFORE SOLUTION :All $M_{c}$ are non empty just Define the zero function

1. $M_{c}$ is a maximal ideal of $R$
2. Any Maximal ideal of $R$ is of the form $M_{c}$
3. $M_{b}\neq M_{c}$ if $b\neq c$

My try.

1. Clearly, The quotient ring $R/M_{c}$ is a commutative ring with unity. Infact Suppose there is some $g+M_{c}\in R/M_{c}$ is nonzero that is $g(c)\neq 0$ for some $c\in[0,1]$ because $g\in R$ Choose $f$ such that , $f(c)=\frac{1}{g(c)}$ such $f$ exist as take $f(x)=\frac{x+1}{(c+1)g(c)}$ for Define $h(x)=g(x)f(x)-1$ Clearly, $h\in M_{c}$ so we have $$g(x)f(x)+M_c=1+M_c$$ or, $$(f+M_c)(g+M_c)=1+M_c.$$ Thus $R/M_{c}$ is a field implying $M_c$ is a maximal Ideal.

Part 2: I am thinking.

3. Suppose for $b\neq c$ Choose, $t\neq 0$ and $t\in \mathbb{R}$ and $f(c)=0$(Because for the note) such functions exist in $C[0,1]$ ,$f(x)=\frac{t(x-c)}{b-c}$ Thus $f\in M_{c}$ But $f\notin M_{b}$

My Question: Is my proof correct? If it is, is there a better proof exist? Then please mention.

Thanks in advance.

Well:I wanna thank Both who comments and I finally cooked up the proper proof, due to them only ,for 1 and 3 , I will try 2 later.

Answer 1

A simpler way to prove 1 is to consider the "evaluation at $c$" function, $\epsilon_c\colon R\to\mathbb{R}$ given by $\epsilon_c(f) = f(c)$. This is easily verified to be a ring homomorphism, which is onto a field. Therefore, the kernel must be a maximal ideal, and the kernel is exactly $M_c$.

I am not convinced about your proof for item 2. How do you know such a $k$ exists? There is no justification. And there has to be one, because closely related rings do not satisfy condition 2: consider the ring $S$ given by $$S = \{f\colon\mathbb{R}\to\mathbb{R}\mid f\text{ is continuous}\},$$ and let $J$ be the functions in $R$ of compact support, where a function $g$ has "compact support" if and only if there exists $N\gt 0$ such that $g(x)=0$ if $|x|\gt N$. It is easy to verify that $J$ is a proper ideal of $S$, and thus is contained in a maximal ideal $M$. However, for every $c\in\mathbb{R}$ there exists $f\in J$ such that $f(c)\neq 0$, and therefore the same holds for $M$, so $M\neq M_c$ for every $c\in\mathbb{R}$. Where exactly does the argument you present fail for this $M$?

Also, you show that there exists a function in $J$ that takes the value $0$ at $k$. (And we already knew such a function existed anyway: the zero function, which definitely lies in $J$). But why are all functions that take the value $0$ at $k$ in $J$? You do not justify this claim either, and that is a problem.

For 3 you likewise need to explain why such an $f$ exists; it is not hard, but it needs to be done. For example, assuming without loss of generality that $b\lt c$, we can take the function $$f(x) = \left\{\begin{array}{lll} 0 & \text{if }{0\leq x\leq b}\\ \frac{x-b}{c-b} & \text{if }b\lt x\lt c\\ 1&\text{if }c\leq x\leq 1. \end{array}\right.$$ Then we can easily verify that $f$ is continuous, $f\in M_b$, but $f\notin M_c$. But an explicit function or an explicit justification for why such a function exists really needs to be given.

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Here is a standard proof of 2. Let $I$ be an ideal of $R$ that is not contained in $M_c$ for any $c\in [0,1]$. We prove that $I=R$.

For each $c$ let $f_c\in I$ be such that $f(c)\neq 0$. It exists, since $I\not\subseteq M_c$. Since $f_c$ is continuous, there exists $\delta_c\gt 0$ such that if $x\in [0,1]$, $|x-c|\lt \delta_c$, then $f_c(x)\neq 0$.

Now the intervals $(c-\delta_c,c+\delta_c)$ form an open cover of $[0,1]$. Since the latter is compact, there is a finite subcover, corresponding to the points $c_1,\ldots,c_n$. Now let $$f=f_{c_1}^2+\cdots+f_{c_n}^2.$$ This is a sum of elements of $I$, hence lies in $I$. And since for every $x\in[0,1]$ there exists $i$ such that $f_{c_i}(x)\neq 0$ (since we have a cover), it follows that $f(x)\neq 0$ for all $x\in [0,1]$. Then $f$ is invertible in $R$, so $I$ contains a unit and thus $I=R$, as desired.

Answer 2

Part 2 is incorrect, sorry. You must use the fact that $[0,1]$ is closed and bounded, because the statement is false for $C(A)$ where $A$ is not a closed and bounded subset of $\mathbb{R}$.

^{The maximal spectrum of $C(A)$ is homeomorphic to the Stone-Čech compactification of $A$, but you're not supposed to know it. In particular, $C(\mathbb{R})$ has a lot of maximal ideals that aren't of the form $M_c$.}

Let $M$ be a maximal ideal of $R$. Then every $f\in M$ must vanish somewhere; define $$ Z(f)=\{x\in[0,1]:f(x)=0\} $$ (the zero-set of $f$). Suppose $f,g\in M$ and that $Z(f)\cap Z(g)=\emptyset$. Then $Z(f)$ and $Z(g)$ are closed and disjoint. Consider $F=f^2+g^2$. Then $F(x)>0$ for every $x\in[0,1]$, because there's no point where $f$ and $g$ vanish together, so $F$ is invertible in $R$. However $F\in M$, which is a contradiction. This can be easily extended to any finite number of functions $f_1,f_2,\dots,f_n\in M$, so $$ Z(f_1)\cap Z(f_2)\cap\dots\cap Z(f_n)\ne\emptyset $$ Hence the family of zero-sets of the functions $f\in M$ is a family of closed sets with the finite intersection property, so it has nonempty intersection $C$ (this is what fails for $C(A)$ when $A$ is not compact).

Now prove that $C$ cannot contain two distinct points and finish up.