Let $\left(C(\left[0,1\right]),+,\cdot\right)$ be the commutative ring with unity, of all continuous functions $f:\left[0,1\right]\rightarrow \mathbb{R}$. For each $X\in\mathbb{P}(\left[0,1\right])-\left\{\varnothing\right\}$, define the set
$$I_{X}=\left\{f\in C(\left[0,1\right]): f(x)=0, \,\forall\,x\in X\right\}$$
a) Prove that the set $I_{X}$is an ideal of the ring $\left(C(\left[0,1\right]),+,\cdot\right)$ for each $X\in\mathbb{P}(\left[0,1\right])-\left\{\varnothing\right\}$.
b) If $X=\left\{x_0\right\}\subseteq \left[0,1\right]$ for some $x_0\in\left[0,1\right]$, then prove that the ideal $I_{X}$ is maximal.
Solution
a) Let $X\in\mathbb{P}(\left[0,1\right])-\left\{\varnothing\right\}$. Obviously, $I_{X}\neq \varnothing$ since the zero-function, which is the zero element of the ring $\left(C(\left[0,1\right],+,\cdot\right)$ , is an element of $I_{X}$. Let $f\,,g\in I_{X}$. Then, $f(x)=0, \;g(x)=0, \;\;\forall\,x\in X$ and thus:
$$\left(f+g\right)(x)=f(x)+g(x)=0, \;\;\; \forall x\in X$$
So $f+g\in I_{X}$ and
$$I_{X}\leq \left(C(\left[0,1\right],+\right) \tag{1}$$
Consider $g\in C(\left[0,1\right])$ and $f\in I_{X}$ . Then,
$$\left(g\cdot f\right)(x)=\left(f\cdot g\right)(x)=f(x)\cdot g(x)=0, \;\;\;\forall x\in X$$
which means that $f\cdot g\in I_{X}$. Therefore we conclude that:
$$\forall g\in C(\left[0,1\right]), \;\; \forall f\in I_{X}: f\cdot g=g\cdot f\in I_{X} \tag{2}$$
Combining $(1), \; (2)$ we have that $I_{X}$ is an ideal of the ring $\left(C(\left[0,1\right],+,\cdot\right)$.
b) Firstly, the ideal $I_{X}=I_{\left\{x_0\right\}}$ is a double (left and right) ideal and $I_{\left\{x_0\right\}}\neq C(\left[0,1\right]$ since the continuous function $a:\left[0,1\right]\rightarrow \mathbb{R}$ defined as $a(x)=x-x_0+1$ is not an element of $I_{\left\{x_0\right\}}$ because $a(x_0)=1\neq 0$. Let $J$ be a double ideal of the ring $\left(C(\left[0,1\right],+,\cdot\right)$ such that
$$I_{\left\{x_0\right\}}\subseteq J\subseteq C(\left[0,1\right])$$
Suppose that $I_{\left\{x_0\right\}}\subset J$. Then, there exists continuous function $g:\left[0,1\right]\rightarrow \mathbb{R}$ such that $g\in J$ and $g\notin I_{\left\{x_0\right\}}$, that is, $g(x_0)\neq 0$ . Since $I_{\left\{x_0\right\}}\subset J$, then $J$ is an ideal and $g^2\in J, \; f\in I_{\left\{x_0\right\}}$, where $f(x)=\left(x-x_0\right)^2\,,x\in\left[0,1\right]$, we get $h\in J$, where $h(x)=f(x)+g^2(x)>0, \; x\in\left[0,1\right]$ . So, the function $\frac{1}{h}$ is well defined and continuous with $1_{C(\left[0,1\right])}=\frac{1}{h}\cdot h\in J$ , so : $ J=C(\left[0,1\right])$.
Therefore, the ideal $I_{\left\{x_0\right\}}$ is a maximal ideal of the ring $\left(C(\left[0,1\right],+,\cdot\right)$.
The exercise can also be found in mathimatikoi.org
$$I_{X}=\left\{f\in C(\left[0,1\right]): f(x)=0, \,\forall\,x\in X\right\}$$
a) Prove that the set $I_{X}$is an ideal of the ring $\left(C(\left[0,1\right]),+,\cdot\right)$ for each $X\in\mathbb{P}(\left[0,1\right])-\left\{\varnothing\right\}$.
b) If $X=\left\{x_0\right\}\subseteq \left[0,1\right]$ for some $x_0\in\left[0,1\right]$, then prove that the ideal $I_{X}$ is maximal.
Solution
a) Let $X\in\mathbb{P}(\left[0,1\right])-\left\{\varnothing\right\}$. Obviously, $I_{X}\neq \varnothing$ since the zero-function, which is the zero element of the ring $\left(C(\left[0,1\right],+,\cdot\right)$ , is an element of $I_{X}$. Let $f\,,g\in I_{X}$. Then, $f(x)=0, \;g(x)=0, \;\;\forall\,x\in X$ and thus:
$$\left(f+g\right)(x)=f(x)+g(x)=0, \;\;\; \forall x\in X$$
So $f+g\in I_{X}$ and
$$I_{X}\leq \left(C(\left[0,1\right],+\right) \tag{1}$$
Consider $g\in C(\left[0,1\right])$ and $f\in I_{X}$ . Then,
$$\left(g\cdot f\right)(x)=\left(f\cdot g\right)(x)=f(x)\cdot g(x)=0, \;\;\;\forall x\in X$$
which means that $f\cdot g\in I_{X}$. Therefore we conclude that:
$$\forall g\in C(\left[0,1\right]), \;\; \forall f\in I_{X}: f\cdot g=g\cdot f\in I_{X} \tag{2}$$
Combining $(1), \; (2)$ we have that $I_{X}$ is an ideal of the ring $\left(C(\left[0,1\right],+,\cdot\right)$.
b) Firstly, the ideal $I_{X}=I_{\left\{x_0\right\}}$ is a double (left and right) ideal and $I_{\left\{x_0\right\}}\neq C(\left[0,1\right]$ since the continuous function $a:\left[0,1\right]\rightarrow \mathbb{R}$ defined as $a(x)=x-x_0+1$ is not an element of $I_{\left\{x_0\right\}}$ because $a(x_0)=1\neq 0$. Let $J$ be a double ideal of the ring $\left(C(\left[0,1\right],+,\cdot\right)$ such that
$$I_{\left\{x_0\right\}}\subseteq J\subseteq C(\left[0,1\right])$$
Suppose that $I_{\left\{x_0\right\}}\subset J$. Then, there exists continuous function $g:\left[0,1\right]\rightarrow \mathbb{R}$ such that $g\in J$ and $g\notin I_{\left\{x_0\right\}}$, that is, $g(x_0)\neq 0$ . Since $I_{\left\{x_0\right\}}\subset J$, then $J$ is an ideal and $g^2\in J, \; f\in I_{\left\{x_0\right\}}$, where $f(x)=\left(x-x_0\right)^2\,,x\in\left[0,1\right]$, we get $h\in J$, where $h(x)=f(x)+g^2(x)>0, \; x\in\left[0,1\right]$ . So, the function $\frac{1}{h}$ is well defined and continuous with $1_{C(\left[0,1\right])}=\frac{1}{h}\cdot h\in J$ , so : $ J=C(\left[0,1\right])$.
Therefore, the ideal $I_{\left\{x_0\right\}}$ is a maximal ideal of the ring $\left(C(\left[0,1\right],+,\cdot\right)$.
The exercise can also be found in mathimatikoi.org
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