Let $R = \mathbb{Z}[2x]$, how does a nonzero prime ideal look like?
An element of $R$ is of the form $$f(x) = a_0 + a_1(2x) + a_2(2x)^2 +... + a_n(2x)^n, a_i \in \mathbb{Z}.$$ Comparing with the case of $S = \mathbb{Z}[x]$, a nonzero prime ideal of $S$ is of the form $(p)$ for a prime $p$ or $(f(x))$ for irreducible $f$ or $(p,f(x))$ for $f$ irreducible modulo $p$, as seen in Mumford's drawing.
I believe that $(p)$ remain as prime ideals in $R$ for a prime $p$ but a polynomial like $2x^2+2x+2$ is questionably prime because we cannot write it as $2\cdot(x^2+x+1)$ since the latter term doesn't belong to $R$, and by this reasoning even $(2x)$ has to be prime. Or did I get some obvious things wrong? I'm so confused.
EDIT. In the comment by user26857 below, $2x^2+2x+1$ obviously cannot be inside $R$, since there is no way to obtain the term $2x^2$. By the isomorphism $S \rightarrow R$, given by $f(x) \mapsto f(2x)$, it seems like there isn't much work to be done to classify all prime ideals of $R$.