In $\mathbb{C}[x]$ is it true that $F_{a,b}=\{p\in\mathbb{C}[x] : p(a)=p(b)\}$ for $a\neq b$ is a maximal subring? The problem is in the title. It is clear that $F_{a,b}$ is a ring, but it is not so clear to me that it is maximal in $\mathbb{C}[x]$. I tried to consider it as a vector space and show that it has codim=1 but I didn't go far with that. Does anyone have any ideas?
 A: This answer is based on the hints and insights of Qiaochu Yuan. 
We can consider the rings and subrings as vector spaces. If $T\leq \mathbb{C}[x]$ then $T$ is a subspace of the vector space $\mathbb{C}[x]$. So, in order to prove that $T$ is a maximal subring of $\mathbb{C}[x]$, we can just prove that $\operatorname{codim}T = 1$. Indeed, suppose by contradiction that $\operatorname{codim}T = 1$ and $T$ is not a maximal subring of $\mathbb{C}[x]$. In this case we have that there exists $V$ a subring of $\mathbb{C}[x]$ such that $T\subsetneq V\subsetneq \mathbb{C}[x]$. But in the context of vector spaces, that means that $\operatorname{codim}T> \operatorname{codim}V\geq 1$ which implies that $\operatorname{codim}T>1$. Contradiction! So it is true that $$\operatorname{codim}T = 1 \Rightarrow T \text{ is a maximal subring of } \mathbb{C}[x].$$
In our case we choose the linear functional $f:\mathbb{C}[x]\rightarrow \mathbb{C}$ defined by $f(p)=p(a)-p(b)$. Clearly $F_{a,b} \equiv \ker f$. Furthermore $\operatorname{Im}f=\mathbb{C}$. So
$$\operatorname{codim}\ker f=\dim\mathbb{C}[x]/\ker f=\dim\operatorname{Im}f = \dim\mathbb{C}=1.$$
From what we proved above, since $\operatorname{codim}F_{a,b}=1$ we have that $F_{a,b}$ is a maximal subring of $\mathbb{C}[x].$
