Prove that $I$ is a maximal ideal of $\mathcal A$.

Question

Please, give-me a hint to prove this proposition:

Let $\mathcal A$ be the ring of all continuous real functions (with the usual operations of sum and multiplication) defined on the interval $[0,1]$, and $I= \{f \in \mathcal A;f(0)=0 \}$. Prove that $I$ is a maximal ideal of $\mathcal A$.

My approach to this problem was the following: consider $I \subset Y \subset \mathcal A$, with $Y \neq I$. Then I must prove that $Y=\mathcal A$. Since $I \subset Y$, there exists a continuous function $g$ such that $g \in Y$ and $g \notin I$, i.e., $g(0) \neq 0$, or $g(0)=a, a \neq 0$. Now, if I manage to show that $1 \in Y$, then I am done. So I ask: What should be the next step? Thanks in advance!

Answer 1

Easier would be to identify the quotient $\mathcal{A}/I$ and show that this is a field. (Hint: this will be isomorphic to $\mathbb{R}$). Thus $I$ is a maximal ideal.

Answer 2

Define $\phi : \mathcal{A} \to \mathbb{R}$ by $\phi (f)=f(0).$

Then clearly $\phi$ is a ring homomorphism. Also we see that $\forall r\in\mathbb{R}$ if we define $f:[0,1]\to\mathbb{R}$ by $f(x)=r\:\:, \forall x\in [0,1]$ then $f$ is continuous i.e $f \in \mathcal{A}$ and also $\phi(f)=f(0)=r$ So $\phi$ is onto.

Now $\ker\phi=\{f\in \mathcal{A}| \phi(f)=f(0)=0\}=I$

Then by First Isomorphism Theorem, $\mathcal{A}/I$ is isomorphic to $\mathbb R$

Since $\mathbb{R}$ is field, $\mathcal{A}/I$ is field. So $I$ is maximal.

Answer 3

You can do it using your approach. Suppose $I\subset J$ and that inclusion is proper. Then there is a continuous function $f\in J$ with $f(0)\not= 0$. Clearly $f(x)-f(0)\in I\subset J$. Thus $f(x)-\Big(f(x)-f(0)\Big)\in J$ and so $f(0)\in J$. which is to say $J$ contains $1.\:\:$