Infinite product of fields The main source of inspiration for this question is this excerpt 

Recall: An ultrafilter on the set X gives you a maximal ideal in the ring of all real-valued functions, and these are the only prime ideals.

from this MathOverflow answer of Tom Goodwillie. (In  the same thread there is an outstanding answer of our friend Georges Elencwajg.)
Let $X$ be an infinite set, for each $x$ in $X$ let $K_x$ be a field, and let $A$ be the product of the $K_x$. 
Here is what I'd like to know: 
$(1)$ What are the maximal ideals of $A$, and what are the corresponding residue fields?
$(2)$ What are the non-maximal prime ideals of $A$, and what are the corresponding residue domains? 
$(3)$ What are the non-prime primary ideals of $A$, and what are the corresponding residue rings? 
$(4)$ How can one describe the Zariski topology on the prime spectrum of $A$? In particular, does this topology depend only on $X$, or does it really depend on the $K_x$?
For $f$ in $A$ and $x$ in $X$, write $f(x)$ for the $x$ component of $f$, and $f^{-1}(0)$ for the set of all $x$ in $X$ such that $f(x)=0$. 
Here are the few things I believe I know: 
$(5)$ To each $x$ in $X$ is attached an obvious maximal ideal $\mathfrak m_x$ with residue field $K_x$: the set of all $f$ vanishing at $x$. 
$(6)$ More generally, to each ultrafilter $U$ on $X$ is attached the set, which is in fact a prime ideal, of all $f$ such that $f^{-1}(0)$ is in $U$. Then (5) corresponds to the principal ultrafilters. But I have no idea about the residue domain associated with a non-principal ultrafilter. 
I believe that the answer to (2) is that all primes are maximal. 
I (perhaps naively) expect all primary ideals to be prime, and thus maximal. 
So, as you see, I need tons of help! Thank you very much in advance!
 A: For (1), there will be variety results depended on the fields.
（1）If $A=\prod_{i\in \mathbb{N}}\mathbb{F}_p$ then all the residue fields are equal to $\mathbb{F}_p$. (since $x^p=x$)
(2) If $A=\prod_{p}\mathbb{F}_p$, then for any prime $\mathfrak{m}$ containing $\oplus_p\mathbb{F}_p$ , then $\mathbb{Q}\subseteq A/\mathfrak{m}$ and the cardinality of $A/\mathfrak{m}$ is $\aleph$, thus $A/\mathfrak{m}$ is not an algebraic extension. It has transcendental elements over $\mathbb{Q}$.
Proof about the cardinality: define a map $\varphi$ from $\{0,1\}^{\mathbb{N}}$ to $A/\mathfrak{m}$ by sending $(a_i)_{i=1}^{\infty}$ to $(x_k)$,  $x_k=\sum_{j=1}^na_j2^{j-1}$ if $2^n-1<p_k\leq2^{n+1}-1$. Here index $k$ denotes $k$-th coordinate and $p_k$ denotes the $k$-th prime number. Then $\varphi$ is injective.

In this case, we want to ask if $A$ is a purely transcendental extension over $\mathbb{Q}$ ?

It is not true at least for some maximal ideal!
Consider the equation $x^2+1=0$. Then it can be solved in $\mathbb{F}_p$ if only if $p=1\mod 4$ or $p=2$. Since those primes are infinite, let $a=(a_{p_k})\in A$, $a_{p_k}=1$ if $p_k=1 \mod 4$, then we find a maximal ideal $\mathfrak{m}$ containing $\oplus\mathbb{F}_p$  such that $1-a\in \mathfrak{m}$. Hence $x^2+1=0$ can be solvable in $A/\mathfrak{m}$. 
But also we can find some maximal ideal $\mathfrak{m}$ such that the $x^2+1=0$ can not be solvable in $A/\mathfrak{m}$.

The residue fields are always not purely algebraic extension.
  Since  $(x^2+1)(x^2+2)(x^2-2)$ can be always solvable. 

(3) If $A=\prod_{i\in \mathbb{N}}\mathbb{Q}$, the discussion about the cardinality of the residue field is similar.
Those questions  relate to my recent thread the ring of Cauchy sequence.

Another  way to show that $\dim A=0$:
suppose we have a proper inclusion between prime ideals, say $\mathfrak{p}\subsetneq\mathfrak{m}$, then you can find an $x\in 
\mathfrak{m}-\mathfrak{p}$ such that the coordinate of $x$ is either $1$ or $0$, then we must have $1-x\in\mathfrak{p}\subset \mathfrak{m}$, contradiction! 
A: You are absolutely right when you write "I believe that the answer to (2) is that all primes are maximal",  Pierre-Yves : 
Proposition 1 The ring $A$ is zero-dimensional.
Proof Recall that a commutative ring $R$ is called von Neumann if  for all $ r\in R$  there exists an $s\in R$  such that  $r=sr^2$.
It is clear that a field is von Neumann and  that a product  of any family of von Neumann rings is von Neumann, so that your ring $A$ is von Neumann.
   But  a von Neumann ring has Krull dimension zero: indeed, if ${\mathfrak p } \subset A$ is prime, the quotient $D:=A/{\mathfrak p } $ is clearly von Neumann.
And to end the proof just  note that  a von Neumann domain $D$ is a field: if $d\neq 0\in D$, the  relation $d=rd^2$ implies $1=rd $ so that $d$ is invertible.  
Corollary For every prime (=maximal) ideal $\mathfrak p\subset A$ the residue field is  $\kappa (\mathfrak p)=A_{\mathfrak p}$
Proof Since the ring $A$ is reduced, so is its localization, the zero-dimensional (cf.Proposition) ring  $A_{\mathfrak p}$.
 But  then   ${\mathfrak p} A_{\mathfrak p}$ is the nilpotent radical of that reduced ring $A_{\mathfrak p}$ and so must be zero: ${\mathfrak p} A_{\mathfrak p}=0$ .
Hence  we get the (rather unusual !) formula  $\kappa (\mathfrak p)=A_{\mathfrak p}/ \mathfrak pA_{\mathfrak p}=A_{\mathfrak p}/0=A_{\mathfrak p} $ .  
And your other hope is true too:
Proposition 2 Every primary ideal of $A$ is maximal.
Indeed , since the ring $A$ is von Neumann 
 it is absolutely flat (Atiyah-Macdonald, Introduction to Commutative Algebra, Chapter  2, Exercise  27 ) and thus every primary ideal of $A$ is maximal. (Same reference, Chapter 4, Exercise  3)
