$I:=\{f(x)\in R\mid f(1)=0\}$ is a maximal ideal? 
Let $R$ be a ring of the real functions with "+" and "x" and let $I_1:=\{f(x)\in R\mid f(1)=0\}$, $I_1$ is a maximal ideal?

I think yes, but don't know how to prove it. 
 A: An ideal $I \subset R$ is maximal if and only if the quotient $R/I$ is a field. Consider the map
$$\phi : R \to \mathbb{R}$$
defined by $\phi(f) = f(1).$
A: Consider $\varphi : \mathcal{R}\to\Bbb{R}$ such that $\varphi(f)=f(1)$. One verifies that $\varphi$ is a ring morphism that is obviously surjective and that $\ker{\varphi}=I_1$. So $\mathcal{R}/I_1$ and $\Bbb{R}$ are isomorphic as rings. Therefore $\mathcal{R}/I_1$ is a field and $I_1$ is a maximal ideal.
A: The previous proofs are fine, and are probably more generalizable, but I'll give another one, because it's always nice to have different proofs at hand.
Let $g\notin I_1$. Then $g(1)\neq 0$. Let $f$ be any function on $\mathbb{R}$ that only cancels at $1$ (i.e. $f(x)\neq 0$ for $x\neq 1$). 
Then defining $h(x) := \frac{f(x)}{f^2(x)+g^2(x)}$ and $k(x):=\frac{g(x)}{f^2(x) + g^2(x)}$ (well-defined since $f$ and $g$ are never both $0$), one has $hf + gk =1$. Therefore $1\in I_1 +(g)$, so that $I_1 +(g) = R$.
This is for any $g\notin I_1$, so $I_1$ is maximal.
