Let $K$ be a compact subset of $\Bbb C$. Let $\mathcal P(K):=$ closure in $||\cdot||_{\infty}$ of all complex polynomials on $K$. Note that $\mathcal P(K) = C(K)$ do not generally hold since Stone-Weierstrass do not apply to complex polynomial.
What is the character space $\Phi_{\mathcal P(K)}$ of $\mathcal P(K)$?
It is commonly known that $\Phi_{C(K)}=\{ \delta_z|z\in K\}$ where $\delta_z$ is the pointwise evaluation at $z$. Since $\mathcal P(K)\subset C(K)$ we have $\Phi_{C(K)}\subset\Phi_{\mathcal P(K)}$. It appears that the converse do not hold in general but I can't think of an example here.
PS. I would also really appreciate if anyone could point me to a good reference about this subject. My lecture note is really short and hand-wavy-ish.
Any value $\Phi(z)$ determines a unique multiplicative linear map on polynomials, which has a unique continuous (multiplicative linear) extension onto $\mathcal P(K)$ iff $|\Phi(p)|≤\|p\|$ for all polynomials $p$.
As noted in the question, $\delta_x$ for $x\in K$ determines a character in $\mathcal P(K)$, but there are others.
An example:
Let $K$ be the annulus $\overline{B_2(0)-B_1(0)}$. Since for polynomials/holomorphic functions maximal values can never lie on interior points one has that the maximal value of $p$ on $\overline{B_2(0)}\supset K$ must lie on the circle of radius $2$. In particular $\|p\|_K≥|p(0)|=|\delta_0(p)|$ for all $p$ and $\delta_0$ extends to a character on $\mathcal P(K)$, even though $0\notin K$.
This means that $\Phi_{\mathcal P(K)}\neq \Phi_{C(K)}$. More generally one can see the following:
The proof is the same, let $U$ be the open set containing $x$ with boundary $\gamma$. Since $\bar U$ is compact and polynomials must take the maximal value on the boundary one has by virtue of $\gamma \subset K$ that: $$\|p\|_K≥\|p\|_\gamma=\|p\|_{U}≥|p(x)|=|\delta_x(p)|$$ and $\delta_x$ extends to a character.
Not all $\delta_x$ extend to characters: If $B_r(y)$ is a ball that contains $K$ and $x$ is outside of this ball, then the polynomial $z-x$ is invertible in $\mathcal P(K)$, having inverse $\sum_n\left(\frac {z-y}{x-y}\right)^n$ (sum converges because $|x-y|=r≥|z-y|$ for all $z\in B_r(y)$). From this it follows that there can be no character with $\Phi(z)=x$.
This is as much as I know unfortunately, but I would hypothesise the following:
(The lemma from before shows the converse, namely that if $x\notin U$ then $\delta_x\in \Phi_{\mathcal P(K)}$).