Let $f:R\to S$ be a ring homomorphism, and for a fixed prime ideal $\mathfrak p\subset R$, let $p:R\to k(\mathfrak p)$ be the canonical homomorphism, $k(\mathfrak p)$ being the fraction field of $R/\mathfrak p$. Take the following (pushout) square: $\require{AMScd}$ $$\begin{CD}
R@>>p> k(\mathfrak p)\\
@VVfV @VjVV\\
S@>i>>S\otimes_R k(\mathfrak p)
\end{CD}$$ and consider its image (in $\rm Top$) under the functor $\rm Spec$ (whose action I'll denote by $-^*$). So $p^*$ is injective, being its domain a singleton ($R\neq 0$), and also $i^*$ is injective: $i=l\circ q$, where $q:S\to S\otimes_R R/\mathfrak p$ is surjective, and $l: S\otimes _R R/\mathfrak p\to S\otimes_R k(\mathfrak p)$ is a localization (respect to the subset of elements of the form $1\otimes \bar r$, $r\notin \mathfrak p$) hence $q^*,l^*$ are injective. Until now, we can say that $i^*$ injects its domain into the fiber of $\mathfrak p\in \operatorname{Spec} R$; but I don't understand how to deduce the most important part, i.e. that the image of $i^*$ is the entire fiber (and so the square in $\rm Top$ is a pullback). I apologize for the third question of the same kind in two days, but this passage (that I found on Clark's Commutative Algebra, 4.3) is where my confusion actually started. Thanks for your patience
Fiber of a morphism of spectra
algebraic-geometrycommutative-algebra
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I do not agree with the first equation in your last equation environment. I guess you simply confused localization at an ideal with modding out by it for a moment. Viewing $(J/J^2)_{\mathfrak{p}}$ as a $B_{\mathfrak{P}}$-module, the equation should read: $$ (J/J^2)_{\mathfrak{p}}=J/J^2 \otimes_{B/J} (B/J)_{\mathfrak{p}}=J/J^2 \otimes_{B/J} (B/J)_{\mathfrak{P}}=J/J^2\otimes_{B/J} (B/J\otimes_B B_{\mathfrak{P}})=(J/J^2)_{\mathfrak{P}}=J_{\mathfrak{P}}/J_{\mathfrak{P}}^2. $$ Hence by Nakayama's Lemma, $(J/J^2)_{\mathfrak{p}}=0$ if and only if $J_{\mathfrak{P}}=0$.
By the way, (in case you were not aware of it) the finiteness condition on the ideal sheaf defining $X$ as a closed subscheme of some open in $X\times_S X$ is implied by the assumption on $f\colon X\to S$ being locally of finite presentation. Check out Tag 0818, Stacks Project.
I'm a bit confused by the argument that you made in your edit... I would dissuade you from trying to work with "nested" objects like $\mathbb{Z} \left [\sqrt{-3} \right ][x]$ unless absolutely necessary, because (at least for me) they're much trickier to reason about than classical polynomials. It sounds like you have the right general idea, but you might be a bit confused on the details. In particular it does matter what happens to $x^2 + 3$ mod $p$ for various $p$ when it comes to computing prime ideals. It only doesn't matter for this problem because we're interested in how the prime ideals in $A$ compare to those in $\overline{A}$.
To try to clear things up, I'll try to be fairly detailed in how you might attack this problem. Moreover, since you mention you aren't familiar with tensor products as a means of getting at the fibres, I'll include a more pedestrian approach. I can't help but try and explain why the tensor product "trick" works, though, so you'll have to forgive a quick digression.
With that out of the way, on with the show ^_^.
First, what are the (nonzero) primes in $A$? Well, if we write $A = \frac{\mathbb{Z}[x]}{x^2 + 3}$, then we see the prime ideals in $A$ are exactly the prime ideals in $\mathbb{Z}[x]$ that contain $x^2 + 3$. We know that these look like $(p,f)$ where $f$ is irreducible mod $p$, and so we're looking for ideals $(p,f)$ where $f \mid x^2 + 3$ (mod $p$). Of course, we'll write this polynomial $f$ as $f(\sqrt{-3})$ instead of $f(x)$ (since $x = \sqrt{-3}$ in our ring).
For example, let's compute the primes that lie over $(2) \in \mathbb{Z}$. That is, the primes where $p = 2$.
We want to know that
$$ \frac{A}{\left (2,f \left (\sqrt{-3} \right) \right )} \cong \frac{\mathbb{Z}[x]}{(x^2+3,2,f)} \cong \frac{\mathbb{F}_2[x]}{(x^2 + 1, f)} $$
is a field. Of course, the only way to do this is to choose $f = x+1$, which gives us the prime $\left ( 2, \sqrt{-3}+1 \right)$.
Doing this for a few other primes, we can draw a quick picture of $\text{Spec}(A)$:
(If it's not clear why this is the spectrum, you should work these out in a way analogous to the example for $p=2$ above!)
Now, let's take a look at $\overline{A} = \frac{\mathbb{Z}[y]}{y^2 - y + 1}$. Again, elements of the spectrum will look like $(p,f)$ where $f$ is irreducible mod $p$, but now we want $f \mid y^2 - y + 1$ (mod $p$).
Again, let's look at the case $p=2$. Then we want
$$ \frac{\mathbb{Z}[y]}{(y^2 - y + 1, 2, f)} \cong \frac{\mathbb{F}_2[y]}{(y^2 + y + 1, f)} $$
to be a field. Of course, we know that $y^2 + y + 1$ is $\mathbb{F}_2$-irreducible, so $f = y^2 + y + 1$ is our only choice, and we see our prime ideal is just $(2)$ (notice we don't get $(2,f)$, because we already quotiented by $f$ in the passage from $\mathbb{Z}[y]$ to $\overline{A} = \frac{\mathbb{Z}[y]}{f}$!).
Notice the spectrum for $\overline{A}$ looks superficially similar to the spectrum of $A$ above every prime except $(2)$. We haven't worked out anything to do with the map $\text{Spec}(\overline{A}) \to \text{Spec}(A)$, but now we might have an intuitive plan for what we should look for algebraically!
Inspired by the geometry above, we're led to consider the primes above $(2)$ separately from the other primes. We can get at just the odd primes by localizing at the "function" $2$ (that is, by formally inverting $2$ by adjoining $\frac{1}{2}$).
So, we localize and the inclusion $A \to \overline{A}$ becomes the inclusion $A \left [ \frac{1}{2} \right ] \to \overline{A} \left [ \frac{1}{2} \right ]$.
Of course, if we expand this out, we see we're looking at the map $\mathbb{Z} \left [ \sqrt{-3}, \frac{1}{2} \right ] \to \mathbb{Z} \left [ \frac{1 + \sqrt{-3}}{2}, \frac{1}{2} \right ]$ which sends $\sqrt{-3} \mapsto \sqrt{-3} = 2 \frac{1 + \sqrt{-3}}{2} - 1$. This is easily seen to be an isomorphism, with inverse given by the map sending $\frac{1 + \sqrt{-3}}{2} \mapsto \frac{1 + \sqrt{-3}}{2} = \frac{1}{2} \left ( 1 + \sqrt{-3} \right )$.
In particular, since this is an isomorphism, it identifies the primes away from $(2)$.
All that's left is to consider what happens over $(2)$. Now, as has been mentioned a few times now, we could handle this case via fibre products. The idea here is that we can find the fibre over $\mathfrak{p}$ by considering the pullback (you should convince yourself of this):
Of course, since the category of affine schemes is dual to the category of rings, we get a pushout of rings:
This tells us that $f^{-1}(\mathfrak{p})$ should be $\text{Spec} \left (\overline{A} \otimes_A A/\mathfrak{p} \right )$, which is exactly what the comments and the other answer suggest.
It seems like you're looking for an approach that avoids this machinery, though, so let's see another way to do this.
The fibre over $\left ( 2, 1 + \sqrt{-3} \right )$ in $\text{Spec}(A)$ will be exactly the primes in $\text{Spec}(\overline{A})$ which pull back to $\left ( 2, 1 + \sqrt{-3} \right )$ under the natural embedding $A \hookrightarrow \overline{A}$. Of course, there's only one prime that does the job: $(2)$.
But notice
$$ \mathbb{F}_2 \cong \frac{A}{\left ( 2, 1 + \sqrt{-3} \right )} \hookrightarrow \frac{\overline{A}}{(2)} \cong \mathbb{F}_4 $$
is not an isomorphism!
So taking spectra, we find the fibre is $\text{Spec}(\mathbb{F}_4) \to \text{Spec}(\mathbb{F}_2)$, which is a bijection, but not an isomorphism.
At the end of the day, we see every fibre is a singleton, and we get an isomorphism of over every prime except $(2)$.
I hope this helps ^_^
Best Answer
Let's translate this back into commutative algebra. Explicitly you need to show that
$$\{\mathfrak{q}\in\operatorname{Spec}S\mid f^{-1}(\mathfrak{q}) = \mathfrak{p}\}\subseteq i^\ast(\operatorname{Spec}S\otimes_R k(\mathfrak{p}));$$
i.e., that for every prime ideal $\mathfrak{q}\subseteq S$ which pulls back to $\mathfrak{p}\subseteq R,$ there exists a prime ideal $\mathfrak{Q}\subseteq S\otimes_R k(\mathfrak{p})$ with $i^{-1}(\mathfrak{Q}) = \mathfrak{q}.$
Here are some hints to help you finish this argument; the complete details are below under the spoiler tags.
Hint 1: Break the problem down into showing that $\mathfrak{q}$ is the inverse image of some prime $\mathfrak{q}'\subseteq S\otimes_R R/\mathfrak{p}$, and that this $\mathfrak{q}'$ is the inverse image of a prime $\mathfrak{Q}\subseteq S\otimes_R k(\mathfrak{p}).$
Hint 2:
Hint 3:
Full solution: