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Let $X=(0,1)$ be the open unit interval and $C(X,\mathbb R)$ be the ring of continuous functions from $X$ to $\mathbb R.$For any $x\in (0,1),$let $I(x)=\{f\in C(X,\mathbb R):f(x)=0\}.$Then which of the following is true?

$(1)I(x)$ is a prime ideal.

$(2)I(x)$ is a maximal ideal.

$(3)$Every maximal ideal of $C(X,\mathbb R)$ is equal to $I(x)$ for some $x\in X.$

$(4)$ $C(X,\mathbb R)$ is an integral domain.

$I(x)$ is a Prime ideal

let $f(x),g(x)\in C(X,\mathbb R)$.For any $x\in (0,1),$ let $f(x)g(x)\in I(x)\implies f(x)g(x)=0\implies $ either $f(x)=0$ or $g(x)=0$ $\implies$ either $f(x)\in I(x)$ or $g(x)\in I.$

$I(x)$ is a prime ideal

$I(x)$ is Not an Integral domain

$f(x) = \begin{cases} 0, & \text{if 0<x $\le$ 1/2} \\ x-1/2, & \text{if 1/2$\le$ x<1} \end{cases}$

$g(x) = \begin{cases} x-1/2, & \text{if 0<x $\le$ 1/2} \\ 0, & \text{if 1/2$\le$ x<1} \end{cases}$

then $f(x)\neq 0$ and $g(x)\neq 0$,but $f(x)g(x)=0$

So, $C(X,\mathbb R)$ is Not an integral domain.

Please comment on option (2) and option (3),also check whether my arguments are correct or not?

Thank you!

Hilbert
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  • No hint given for (3) ? – reuns Feb 11 '23 at 15:02
  • @reuns:there is no hint given... – Hilbert Feb 11 '23 at 15:04
  • 1-3 are answered many times on the site already. Poke around a bit. if you don't find anything ping me and I'll find them when I have a chance. 4 is easy to produce a counterexample for. – rschwieb Feb 11 '23 at 16:50
  • @rschwieb: I found this https://math.stackexchange.com/questions/375400/maximal-ideals-in-the-ring-of-real-functions-on-0-1 ,but here X =[0,1] but in my case X=(0,1)...Will this make any difference? – Hilbert Feb 11 '23 at 17:38
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    I think your proof for (1) is incorrect. You are saying $f(x).g(x) = 0 \implies f(x) = 0 \lor g(x) = 0$, but as in your second proof, you have shown that this ring can have zero divisors so both $f,g$ can be non-zero – Axo Feb 11 '23 at 22:21
  • @Axo:Yes,i did'nt noticed it...can you point out the mistake? – Hilbert Feb 12 '23 at 05:57
  • @Hilbert Yes, it makes a difference. Here is a solution relevant to you. – rschwieb Feb 12 '23 at 14:08
  • @Hilbert Denote function $f$ in $C(X, \mathbb{R})$ as $f(X)$ to make notation unambiguous. Now if $f(X) \cdot g(X) \in I(x)$, it implies the fixed point $x \in X$ is a root of $f(X) \cdot g(X)$ or $f(x) \cdot g(x) = 0$. Now, you can claim that $f(x), g(x) \in \mathbb{R}$ which has no zero divisors so either $f(x) = 0 \lor g(x) = 0$, thus showing $I(x)$ is prime. In your writeup, this distinction isn't made between $f(x) \in \mathbb{R}$ and $f(X)$. – Axo Feb 12 '23 at 14:14

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