Question
Prove the ideal $\mathrm I=\{f \in \mathrm R| f(2)=0 \}$ of $\mathrm R=\{f(x) | f: \Bbb R \to \Bbb R $ is continue} is maximal.
DO NOT use the $1$st isomorphism theorem.
I tried.
(1) $\mathrm I \neq \mathrm R$
(2) Suppose that if "$\mathrm I \subseteq \mathrm J \subseteq \mathrm R$ where $J$ is ideal of $\mathrm R$".
For all $f(x) \in \mathrm I$, $f(x) \in \mathrm J=\{ j(x)h(x)$ | $ h(x) \in \mathrm R\}$
Since $f(2)=0$, $f(x)=j(x)h(x)$ for some $h(x) \in \mathrm R$. (ie. $j(2)h(2)=0$)
If $j(2)=0$, then $j(x) \in \mathrm I$. so $\mathrm I = \mathrm J $.
And we show that if $\mathrm I \neq \mathrm J$, then $\mathrm J = \mathrm R $.
If $f(2)=j(2)h(2)=0$ and $j(2) \neq 0$ then, there exist $g(x) \in \mathrm J$ s.t $g(2) \neq 0$.
moreover there exist $g(x) \in J$ s.t $g(2)=1$. And $g(x)$ is continues.
So we also take $g(x)=1$.
$g(x)=1 \in \mathrm J$. ( ie. $\mathrm J$ contains unit.)
threrfore $\mathrm J=\mathrm R$. ($\mathrm I \neq \mathrm R$)
we conclude $\mathrm I$ is maximal ideal of $\mathrm R$.
is it all right? please fix it or hint. thanks
$(x-2)^2\in I$. If there is $g(x) \in J$ such that $g(2)\neq 0$ then
$$(x-2)^2+(g(x))^2 \in J$$ and this function has no zeros so is invertible.