Let $A = C^\infty(S^1)$ be the ring of smooth functions on the circle (if you prefer, you can see it as the ring of smooth $2\pi$-periodic functions $\mathbb R \to \mathbb R$).
First, $A$ isn't Noetherian : the ideal $I_{\mathscr V(0)}$ of functions vanishing on a neighbourhood of $0$ isn't finitely generated.
But the maximal ideals of $A$ are exactly the $$\mathfrak m_p = \left\{ f \in A\, \Big |\, f(p) = 0 \right \},$$ for $p \in S^1$, which are generated by the two functions $(x,y) \mapsto x-x_p$ and $(x,y) \mapsto y - y_p$. (If you think of $A$ as a set of trigonometric functions, $x$ is $\cos$ and $y$ is $\sin$).
Proof of the various claims:
- $I_{\mathscr V(0)}$ isn't f.g. : Suppose ad absurdum that $I_{\mathscr V(0)} = (f_1, \ldots, f_r)$ where each $f_i$ vanishes on a neighbourhood $V_i$ of $0$. Then any function of $(f_1, \ldots, f_r)$ vanishes on $V = \bigcap V_i$, which is a fixed neighbourhood of 0. But it is easy to construct functions of $A$ vanishing on a neighbourhood of $0$ as small as desired (in particular, strictly smaller than $V$), a contradiction.
- $\mathrm{Max}(A) = \left\{ \mathfrak m_p \, \Big | \, p \in S^1 \right\}$ : Let $I$ be an ideal of $A$. We are going to prove that either $I$ is contained in some $\mathfrak m_p$ or $I = A$. The negation of “$I$ is contained in some $\mathfrak m_p$” is “forall $p \in S^1$, there is a function $f$ s.t. $f(p) \neq 0$”. Since the set on which a function doesn't vanish is open and $S^1$ is compact, that implies the existence of finitely many functions $f_1, \ldots, f_r \in I$ such that $\forall p \in S^1, \exists i : f_i(p) \neq 0$. Then, $f = f_1^2 + \cdots + f_r^2 \in I$ is everywhere nonzero, so it is invertible in $A$ and $I = A$.
- $\mathfrak m_p = (x-x_p, y-y_p)$ : The inclusion $\supseteq$ is clear. Let $f \in \mathfrak m_p$. By definition of a smooth function on a submanifold, $f$ is the restriction of a smooth function $F \in C^1(V)$ for some neighbourhood $V$ of $p$ in $\mathbb R^2$. Of course, $F$ still vanishes on $p$. The claim then follows from Hadamard's lemma.
PS : All this seems to indicate that $A$ has some strange (in particular non f.g.) prime ideals. I must confess I cannot really understand who they are.
Let $k$ be a field and $\bar{k}$ its algebraic closure. An algebraic zero of a subset $\Phi$ of $k[x_1,\dots,x_n]$ is an element $(a_1,\dots,a_n) \in \bar{k}^n$ such that $f(a_1,\dots,a_n)=0, \, \forall f \in \Phi$. Then Hilbert's Nullstellensatz says that if $g \in k[x_1,\dots,x_n]$ vanishes at every algebraic zero of $\Phi$, then $g$ is inside the radical of the ideal generated by $\Phi$ (Matsumura, Theorem 5.4).
The next key thing to observe is that given an ideal $I$ of $k[x_1,\dots,x_n]$, there is a $1-1$ correspondence between algebraic zeros of $I$ and maximal ideals of $k[x_1,\dots,x_n]$ that contain $I$. To see that, note that if $m$ is a maximal ideal that contains $I$ and we define $a_i$ to be the class of $x_i$ mod $m$, then $(a_1,\dots,a_n)$ is an algebraic zero of $I$ by the Zariski Lemma. Conversely, if $(a_1,\dots,a_n)$ is an algebraic zero, then $k[a_1,\dots,a_n] = k(a_1,\dots,a_n)$ and the kernel of the $k$-algebra homomorphism $k[x_1,\dots,x_n] \rightarrow k[a_1,\dots,a_n]$ that sends $x_i$ to $a_i$ is a maximal ideal (Matsumura, Theorem 5.1).
Finally, we clearly have that $I \subset \cap_{m \supset I} m$. Conversely, let $f \in \cap_{m \supset I} m$. Then $f$ vanishes at every algebraic zero of $I$ and by Hilbert's Nullstellensatz $f \in \sqrt{I}$.
Best Answer
In the following argument $\mathbb{F}_p$ denotes either a finite field or the field $\mathbb{F}_0 = \mathbb{Q}$.
Yes. Let $\phi : R \to S$ be a morphism of finitely-generated $\mathbb{Z}$-algebras and let $m$ be a maximal ideal of $S$. Then $S/m$ is a finitely generated (as a ring) field.
Lemma: Finitely generated fields are finite fields.
Proof. Any finitely generated field $F$ is finitely generated over $\mathbb{F}_p$ for some $p$. By Noether normalization $F$ is a finite integral extension of $\mathbb{F}_p[x_1, ... x_n]$ for some $n$, and since it is a field we must have $n = 0$. Hence $F$ is either a finite field or a number field, but the latter is impossible as rings of integers in number fields have infinitely many primes.
It follows that the image of $R$ in $S/m$ is a subring of a finite field, hence also a finite field. Hence $m$ is sent to a maximal ideal of $R$.