Prove that every homomorphism of $H(D)$ onto the field of complex numbers is evaluation at a point of $D$. Let $H(D)$ be the full ring of analytic function on some open set $D$ in the complex plane. Prove that every homomorphism of $H(D)$ onto the field of complex numbers is evaluation at a point of $D$. If $D$ is non-empty prove the kernels of these homomorphism do not exhaust the maximal ideals of the ring $H(D)$.
Please help me calculating. Thanks for help.
 A: If you mean ring homomorphisms, then the first statement is not true as it stands. For instance, let $D$ be the open unit disk and define $\phi:H(D)\to C$ by $$\phi(f)=\overline{f(0)}.$$ Then if $f,g$ are analytic on $D$, we have $$\phi(f\cdot g)=\overline{(f\cdot g)(0)}=\overline{f(0)\cdot g(0)}=\overline{f(0)}\cdot\overline{g(0)}=\phi(f)\cdot\phi(g),$$ and likewise $\phi(f+g)=\phi(f)+\phi(g),$ so $\phi$ is a homomorphism. However, if $f$ is the constant function $f(z)=i$ then $\phi(f)=-i\neq i=f(z)$ for all $z\in D$, so $\phi$ is not an evaluation homomorphism.
For each $w\in\Bbb C$, define $c_w:D\to\Bbb C$ by $c_w(z)=w$ for all $z\in D$. If $\phi:H(D)\to\Bbb C$ is a surjective homomorphism and $f\in H(D)$, then you should be able to show that one of the following must hold:

(1) $\phi(c_w)=w$ for all $w\in\Bbb C$, or
(2) $\phi(c_w)=\overline w$ for all $w\in\Bbb C$.

[Hint: First show $\phi(c_0)=0$ and $\phi(c_1)=1,$ whence $\phi(c_{-1})=-1$, and inductively $\phi(c_n)=n$ for all $n\in\Bbb Z$. From there, show that $\phi(c_q)=q$ for all $q\in\Bbb Q$, then use limit arguments to show $\phi(c_x)=x$ for all $x\in\Bbb R$. Finally, show that $\phi(c_i)=i$ or $\phi(c_i)=-i$ whence $\phi(c_w)=w$ or $\phi(c_w)=\overline w.$ Surjectivity will be necessary in one of those steps, so if you don't see where you used it, then you have something to fix.]
Now, (2) will make the first statement false, for the same reason as the example above fails, so we need the added condition that we are only considering homomorphisms that act as evaluations on the subring of constant functions $D\to\Bbb C$.
I'm not familiar with the convention for what exactly $H(\emptyset)$ is, but it seems to me that constant functions can be said to be analytic on $\emptyset$, vacuously. If that is so, then we must further assume that $D$ is non-empty, or the first statement will be false. If $H(\emptyset)$ is finite, then the first statement is vacuously true. In any case, I consider $D\neq\emptyset$ in the following.
Finally, note that $D$ needs to be connected. (I assume you just omitted that condition.)

We must show that if $\phi:H(D)\to\Bbb C$ is a (surjective) homomorphism such that $\phi(c_w)=w$ for all $w\in\Bbb C$, then there is some $\alpha\in D$ such that for all $f\in H(D)$ we have $\phi(f)=f(\alpha).$
Set $\alpha=\phi(z)$. Then $$0=\phi(z)-\alpha=\phi(z)-\phi(c_\alpha)=\phi(z-c_\alpha)=\phi(z-\alpha).$$ Now, if $\alpha\notin D$, then $\frac1{z-\alpha}\in H(D),$ but then $$1=\phi(c_1)=\phi\left((z-\alpha)\cdot\frac1{z-\alpha}\right)=\phi(z-\alpha)\cdot\phi\left(\frac1{z-\alpha}\right)=0\cdot\phi\left(\frac1{z-\alpha}\right)=0,$$ so we must have $\alpha\in D$.
Now, by definition of $\alpha$, it's clear that for any polynomial $p$ we have $\phi(p)=p(\alpha).$ Take any non-constant $f\in H(D)$, and let $g(z)=f(z)-f(\alpha).$ Then $g\in H(D)$ is non-constant and $g(\alpha)=0.$ Now, since $g$ is non-constant, then we can't have $g^{(n)}(\alpha)=0$ for all $n\ge0$, for then $g$ would be identically zero, so there is some least such $n$ (necessarily greater than $0$) such that $g^{(n)}(\alpha)\neq 0$. Put $\beta=\frac{g^{(n)}(\alpha)}{n!}.$ Define $h:D\to\Bbb C$ by $h(\alpha)=\beta$ and $$h(z)=\frac{g(z)}{(z-\alpha)^n}$$ for $z\ne\alpha$. It can be shown that $h$ is analytic on $D,$ and putting $p(z)=(z-\alpha)^n,$ it's clear that $f(z)=f(\alpha)+g(z)=f(\alpha)+p(z)h(z)$ for all $z\in D,$ or equivalently, $$f=c_{f(\alpha)}+p\cdot h.$$ Thus, $$\begin{align}\phi(f) &= \phi\left(c_{f(\alpha)}+p\cdot h\right)\\ &= \phi(c_{f(\alpha)}+\phi(p)\phi(h)\\ &= f(\alpha)+p(\alpha)\cdot\phi(h)\\ &= f(\alpha)+0\cdot h\\ &= f(\alpha),\end{align}$$ as desired.

For your second question, you should try a similar approach to this answer.
