= Solution
{c}
Here the coefficient <fields> are $\mathbb R$ and $\mathbb C$. A <polynomial ring> in one variable over a <field> is a <Euclidean domain>, hence a <principal ideal domain>; its nonzero <prime ideals> are generated by <irreducible polynomials>. The <fundamental theorem of algebra> says that all <irreducible polynomials> over $\mathbb C$ are linear. Combining it with <complex conjugation> shows that the <monic> <irreducible polynomials> over $\mathbb R$ are exactly the linear polynomials and quadratics with no real root. Thus
$$
\boxed{\operatorname{Spec}\mathbb R[x]
=\{(0)\}\cup\{(x-a):a\in\mathbb R\}
\cup\{((x-a)^2+b^2):a\in\mathbb R,\ b>0\}.}
$$
Every nonzero <prime ideal> in this list is a <maximal ideal>; $(0)$ is a <prime ideal> because $\mathbb R[x]$ is an <integral domain>. Likewise,
$$
\operatorname{Spec}\mathbb C[x]=\{(0)\}\cup\{(x-z):z\in\mathbb C\}.
$$
The map is <contraction of an ideal> along the inclusion. It sends $(0)$ to $(0)$. For $z=a+ib$, the contraction of $(x-z)$ consists of the real polynomials vanishing at $z$. If $b=0$, the <factor theorem> gives the <ideal> $(x-a)$. If $b\ne0$, a real polynomial vanishing at $z$ also vanishes at its <complex conjugate> $\overline z$, so it is divisible by $(x-z)(x-\overline z)=(x-a)^2+b^2$. Division by this real quadratic gives the converse and hence
$$
\boxed{f((x-z))=
\begin{cases}
(x-a),&z=a\in\mathbb R,\\
((x-a)^2+b^2),&z=a+ib,\ b\ne0.
\end{cases}}
$$
Thus each real linear <prime ideal> has one preimage, each real quadratic <prime ideal> has the two conjugate preimages $(x-a-ib)$ and $(x-a+ib)$, and $(0)$ has only $(0)$ above it. In particular, $f$ is surjective.
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