1. ## fields

the set of real numbers is an infinite field.
so R[x] is infinite
does this mean that R[x]/<p>
for some irreducible polynomial p will always be an infinite field or are there exceptions?

2. the answer is yes. in fact $\mathbb{R}[x]/ \langle p \rangle$ is either $\mathbb{R}$ or $\mathbb{C}.$ the reason is that the degree of an irreducible polynomial over $\mathbb{R}$ is at most 2.

also note that for any field $k$ and any non-constant polynomial $f \in k[x],$ the ring $k[x]/\langle f \rangle$ contains a copy of $k$. the reason is that the map $\varphi : k \longrightarrow k[x]/\langle f \rangle$ defined by

$\varphi(\alpha)=\alpha + \langle f \rangle$ is an injective ring homomorphism. so if $k$ is infinite, $k[x]/\langle f \rangle$ will be infinite too.

4. i alreday answered your question! for any field $k$ (so you may choose $k=\mathbb{Q}$) and any non-constant polynomial $f \in k[x]$, not necessarily irreducible, the ring $V=k[x]/\langle f \rangle$ contains
a copy of $k.$ so you may consider $V$ as a vector space over $k.$ it is clear that $1 \leq \dim_k V \leq \deg f.$ of course if $f$ is irreducible over $k,$ then $\dim_k V = \deg f.$
as a result, if $k$ is infinite, then $|V|=\text{card}(k).$ thus if $k$ is a countable (resp., uncountable) field, then $V$ is a countable (resp., uncountable) ring.
if $k$ is finite, then $|k| \leq |V| \leq |k|^{\deg f}.$ if $k$ is finite and $f$ is irreducible, then $|V|=|k|^{\deg f}.$