Finding Annihilator of an Ideal in $\mathscr{C} \left[ 0, 1 \right]$.

Question

I am trying to construct examples of rings, ideals and annihilators, and have taken the example of $\left( \mathscr{C} \left[ 0, 1 \right], +, \cdot \right)$, where $\mathscr{C} \left[ 0, 1 \right]$ denote the set of all continuous functions from the interval $\left[ 0, 1 \right]$ to $\mathbb{R}$, $+$ denotes point - wise addition and $\cdot$ denotes point - wise multiplication.

One can easily show that this forms a commutative ring.

The next thing I tried to obtain is an ideal $I_{x_0} = \left\lbrace f \in \mathscr{C} \left[ 0, 1 \right] | f \left( x_0 \right) = 0 \right\rbrace$. Again, it is easy to prove that this is an ideal of $\mathscr{C} \left[ 0, 1 \right]$.

However, when I tried to find out annihilator of $I_{x_0}$, I got stuck (rather confused). As such, my guess is that the annihilator should be $\left\lbrace 0 \right\rbrace$, where $0: \left[ 0, 1 \right] \rightarrow \mathbb{R}$ is the function $\forall x \in \mathbb{R}, 0 \left( x \right) = 0$ and is the addtivie identity of the ring $\mathscr{C} \left[ 0, 1 \right]$.

Although this guess is completely intuitive, I am not able to prove it! I would like help in proving the guess or disproving it!

Answer 1

Suppose $f \in I_{x_0}$. Especially take $$ f(x) = |x - x_0|, $$ then since $h \cdot f = 0$, $h(x) = 0$ except $x_0$ [since $f(x)\neq 0$ whenever $x \neq x_0$]. Since $h \in \mathcal C[0,1]$, $h(x_0) = \lim_{x\to x_0} h(x) = 0$, so the annihilator is just $ \{0\} $.

Answer 2

Take $f(x) = (x-x_0)$ on $[0,1]$. Clearly, $f \in I_{x_0}$. Now consider $h \in Ann(I_{x_0})$. For $x \neq x_0$, $h(x)$ must be $0$ for $h(x)f(x)$ to be $0$.

So $h$ is a continuous functions such that $h(x) = 0$ for any $x \neq x_0$. By the continuity of $h$, we then have $h(x) = 0$ for any $x \in [0,1]$