For a commutative ring $R$ with identity it is well-known that if $\mathfrak m$ is a maximal ideal of $R$ then $\mathfrak{m}[x]$ is also a maximal ideal of $R[x]$. Thus, the set $\mathcal{A}=\{\mathfrak m[x]\mid \mathfrak m \in \mathrm{Max}(R)\}$ is a subset of $\mathrm{Max}(R[x])$. When we consider $\mathrm{Max}(R)$ and $\mathcal{A}$ as two subspaces of $\mathrm{Max}(R)$ and $\mathrm{Max}(R[x])$, respectively, with respect to the Zariski topology on these spaces, I want to show that $\mathrm{Max}(R)$ is homeomorphism to $\mathcal{A}$. So I define $\phi:\mathrm{Max}(R)\rightarrow \mathcal{A}$ by $\varphi(\mathfrak m)=\mathfrak m[x]$. Clearly $\phi$ is bijective. How can we show that $\phi$ is continuous and open?
(Note that $\mathrm{Max}$ means the set of all maximal ideals of a ring and $\mathfrak m[x]$ is the set of all polynomial in $R [x] $ whose coefficients are in $\frak {m} $.)
1) $\frak {m}$ $[x]$ is never a maximal ideal of $R[x]$.
Indeed, denoting by $k$ the field $R/\frak m$, the kernel of the canonical quotient morphism $R[x]\to k[x]$ is $\mathfrak m[x]$ so that $R[x]/\mathfrak m$$ [x]=k[x]$, which is never a field, proving that $\frak m$$ [x]$ is never maximal in $R[x]$.
2) Anyway $ \operatorname {Specmax} R$ is not homeomorphic to $ \operatorname {Specmax} R[x]$ in general.
For example if $R=\mathbb C$, the space $ \operatorname {Specmax}\mathbb C=\{0\}$ has only one point, whereas the space $\operatorname {Specmax } \mathbb C[x]=\{\langle x-z\vert z\in \mathbb C\}$ has the power of the continuum.
3) However there is a map...
The surjective "evaluation at zero" map $ev_0:R[x]\to R:P(x)\mapsto P(0)$ induces a continuous map $$ \operatorname {Specmax} R\to \operatorname {Specmax } R[x]:\mathfrak m \to ev_0^{-1}(\mathfrak m)=\mathfrak m +XR[X]$$ Since $ev_0$ is surjective this map is a closed immersion (and thus injective) but it is not a bijection in general, as follows from the counterexample in 2).