Let $R$ be the ring of continuous real-valued functions on the interval $(0, 1)$ and $I=\{f∈R:f (1/3)=0\}$. Prove that is a maximal ideal in $R$.
I am honestly lost here, i have studied abour ideals in rings in general, but i still am not very much Familiar with the maximal ideals in abstract algebra, except that to prove if I is a maximal ideal meaning its the largest proper subset of a ring, i need to show there is other ideal larger than I that contains I then ita no other than the ring itself , that's the definition i understood, but i cant seem to apply it on this question,maybe i am doing it wrong. A detailed proof would really help me out to understand more. Thank you
Let $N$ be an ideal of $R$ properly containing $I$ ($N\neq I$). Then there exists $f\in N$ but not in $I.$ This means $f(1/3)\neq 0$ and say$f(1/3)=c\neq 0.$ Let $g\in R$ by $g(x)=c, x\in(0,1).$ Consider $h\in R$ such that $h=f-g.$ But then $h(1/3)=0$ and hence $h\in I\subset N.$ Then $f-h=g\in N$. Now define $i(x)=1/c, x\in(0,1). $ But then $i(x)\in R$ and hence $ig\in N.$ Observe that $i(x)g(x)=1$ for all $x\in(0,1).$ Thus unity element of $R$ is in $N$ and hence $N=R.$
The proving of I is an ideal of R, is left to you. Complete it.